MATH MADE EASY - MathMaster

, the premier engineering university in Pakistan. At NUST, I pursued a Bachelor's degree in Avionics Engineering, focusing on key subjects such as Electronics, Signal Processing, Radar, and Antennas. This immersive academic experience not only provided me with a strong theoretical foundation but also exposed me to practical applications of mathematics in the realm of avionics.\n\nDuring my time at university, I also engaged in various educational courses and classes that enriched my understanding of mathematics and its diverse applications. This multifaceted approach to learning enabled me to not only excel academically but also to develop a comprehensive grasp of mathematical concepts.\n\nFurthermore, my love for mathematics transcended the confines of the classroom. I embraced the challenge of complex problem-solving and developed a proficiency in English, which ultimately led me to a unique opportunity. For the past two years, I have been contributing as a proficient solver and reviewer for Photomath.inc, an experience that has further honed my mathematical skills and analytical thinking.\nmake it a bit more concise\n",4.5,103,{"data":36},[37,38,39,40,41,42],{"id":6,"title":7,"slug":8},{"id":10,"title":11,"slug":8},{"id":13,"title":14,"slug":8},{"id":16,"title":17,"slug":8},{"id":19,"title":20,"slug":8},{"id":22,"title":23,"slug":8},{"data":44,"links":117,"meta":119},[45,54,61,68,75,82,89,96,103,110],{"id":46,"category":47,"text_question":48,"photo_question":49,"text_answer":50,"step_text_answer":8,"step_photo_answer":8,"views":51,"likes":52,"slug":53},538059,"general","Calculate the area of a square with P= 20mm","","1. Calculate the side length using the perimeter: \u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> s = \\frac{20}{4} = 5 \\, \\text{mm} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Use the side length to find the area: \u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> A = 5^2 = 25 \\, \\text{mm}^2 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. The area of the square is: \u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 25 \\, \\text{mm}^2 \u003C/math-field>\u003C/math-field>",1051,210,"calculate-the-area-of-a-square-with-p-20mm",{"id":55,"category":47,"text_question":56,"photo_question":49,"text_answer":57,"step_text_answer":8,"step_photo_answer":8,"views":58,"likes":59,"slug":60},538004,"Prime factor of 1150","1. Start by dividing 1150 by the smallest prime number, which is 2:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>1150 \\div 2 = 575\u003C/math-field>\u003C/math-field> \u003Cbr />\n So, the first prime factor is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>2\u003C/math-field>\u003C/math-field>.\u003Cbr />\n\u003Cbr />\n2. Next, we check 575. It is not divisible by 2, so we test the next smallest prime number, which is 3. It is not divisible by 3 either because the sum of its digits is 17, which is not divisible by 3.\u003Cbr />\n\u003Cbr />\n3. The next smallest prime number is 5. Since 575 ends in 5, it is divisible by 5:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>575 \\div 5 = 115\u003C/math-field>\u003C/math-field>\u003Cbr />\n Thus, a prime factor is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>5\u003C/math-field>\u003C/math-field>.\u003Cbr />\n\u003Cbr />\n4. Next, we verify 115. It ends with 5, so divide by 5 again:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>115 \\div 5 = 23\u003C/math-field>\u003C/math-field>\u003Cbr />\n So, another factor is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>5\u003C/math-field>\u003C/math-field>.\u003Cbr />\n\u003Cbr />\n5. Finally, 23 is already a prime number.\u003Cbr />\n\u003Cbr />\nTherefore, the prime factors of 1150 are: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>2 \\times 5^2 \\times 23\u003C/math-field>\u003C/math-field>.",369,74,"prime-factor-of-1150",{"id":62,"category":47,"text_question":63,"photo_question":49,"text_answer":64,"step_text_answer":8,"step_photo_answer":8,"views":65,"likes":66,"slug":67},538001,"Express 2.50 as a percentage","Solution:\\\u003Cbr />\n1. Understand the problem: \\\u003Cbr />\n- Convert the number 2.50 into a percentage.\\\u003Cbr />\n2. To express a number as a percentage, multiply it by 100%.\\\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>2.50 \\times 100\\% = 250\\%\u003C/math-field>\u003C/math-field>",1339,268,"express-2-50-as-a-percentage",{"id":69,"category":47,"text_question":70,"photo_question":49,"text_answer":71,"step_text_answer":8,"step_photo_answer":8,"views":72,"likes":73,"slug":74},537971,"8/12 – 4/8 reduced to the lowest terms","Solution:\u003Cbr />\n1. Simplify each fraction individually:\u003Cbr />\n - \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{8}{12}\u003C/math-field>\u003C/math-field> can be simplified by dividing the numerator and the denominator by their greatest common divisor, which is 4:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{8 \\div 4}{12 \\div 4} = \\frac{2}{3}\u003C/math-field>\u003C/math-field>\u003Cbr />\n - \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{4}{8}\u003C/math-field>\u003C/math-field> can be simplified by dividing the numerator and the denominator by their greatest common divisor, which is 4:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{4 \\div 4}{8 \\div 4} = \\frac{1}{2}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Subtract the simplified fractions:\u003Cbr />\n - Find a common denominator for \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{2}{3}\u003C/math-field>\u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{1}{2}\u003C/math-field>\u003C/math-field>, which is 6.\u003Cbr />\n - Convert \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{2}{3}\u003C/math-field>\u003C/math-field> to have a denominator of 6:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{2}{3} = \\frac{2 \\times 2}{3 \\times 2} = \\frac{4}{6}\u003C/math-field>\u003C/math-field>\u003Cbr />\n - Convert \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{1}{2}\u003C/math-field>\u003C/math-field> to have a denominator of 6:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{1}{2} = \\frac{1 \\times 3}{2 \\times 3} = \\frac{3}{6}\u003C/math-field>\u003C/math-field>\u003Cbr />\n - Perform the subtraction:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{4}{6} - \\frac{3}{6} = \\frac{4 - 3}{6} = \\frac{1}{6}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. The result is already in its lowest terms.",468,94,"8-12-4-8-reduced-to-the-lowest-terms",{"id":76,"category":47,"text_question":77,"photo_question":49,"text_answer":78,"step_text_answer":8,"step_photo_answer":8,"views":79,"likes":80,"slug":81},537926,"given the following sequence find a50 2,8,14,20,26","\u003Cdiv>\n \n \u003Cmath-field style=\"font-size: 16px;padding: 8px;border-radius: 8px;border: 1px solid rgba

$0, 0, 0, .3$ ;box-shadow: 0 0 0 rgba

$0, 0, 0, .2$ \n\" read-only>\u003Cmath-field read-only>a_{50}=2+6\\times 49\u003C/math-field> \u003C/math-field>\n \u003Cbr>\n \u003C/div>\n \n \u003Cdiv>\n \n \u003Cmath-field style=\"font-size: 16px;padding: 8px;border-radius: 8px;border: 1px solid rgba

$0, 0, 0, .3$ ;box-shadow: 0 0 0 rgba

$0, 0, 0, .2$ \n\" read-only>\u003Cmath-field read-only>a_{50}=2+294\u003C/math-field> \u003C/math-field>\n \u003Cbr>\n \u003C/div>\n \n \u003Cdiv>\n \n \u003Cmath-field style=\"font-size: 16px;padding: 8px;border-radius: 8px;border: 1px solid rgba

$0, 0, 0, .3$ ;box-shadow: 0 0 0 rgba

$0, 0, 0, .2$ \n\" read-only>\u003Cmath-field read-only>a_{50}=296\u003C/math-field> \u003C/math-field>\n \u003Cbr>\n \u003C/div>",637,127,"given-the-following-sequence-find-a50-2-8-14-20-26",{"id":83,"category":47,"text_question":84,"photo_question":49,"text_answer":85,"step_text_answer":8,"step_photo_answer":8,"views":86,"likes":87,"slug":88},537876,"You have a 1.5 L solution of hydrochloric acid

$HCI$ with a molarity of 0.80 M.F\nmany moles of HCl are present in the solution?","1. Use the formula for calculating the number of moles: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> n = C \\times V \u003C/math-field>\u003C/math-field>\u003Cbr />\n2. Substitute the given values into the formula: \u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> n = 0.80 \\, \\text{M} \\times 1.5 \\, \\text{L} \u003C/math-field>\u003C/math-field>\u003Cbr />\n3. Calculate the number of moles: \u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> n = 1.2 \\, \\text{moles} \u003C/math-field>\u003C/math-field>\u003Cbr />\n4. Therefore, the number of moles of HCl present in the solution is:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 1.2 \\, \\text{moles} \u003C/math-field>\u003C/math-field>",905,181,"you-have-a-1-5-l-solution-of-hydrochloric-acid-hci-with-a-molarity-of-0-80-m-f-many-moles-of-hcl-are-present-in-the-solution",{"id":90,"category":47,"text_question":91,"photo_question":49,"text_answer":92,"step_text_answer":8,"step_photo_answer":8,"views":93,"likes":94,"slug":95},537836,"How many pounds of butterfat are in 1 gal of whole milk that weighs 8.6 lb and contains 4% butterfat?","1. Determine the weight of butterfat in the milk by multiplying the total weight of the milk by the percentage of butterfat: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>8.6 \\, \\text{lb} \\times 0.04 = 0.344 \\, \\text{lb}\u003C/math-field>\u003C/math-field> \u003Cbr />\n\u003Cbr />\n2. Thus, the pounds of butterfat in the milk is: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>0.344 \\, \\text{lb}\u003C/math-field>\u003C/math-field>",487,97,"how-many-pounds-of-butterfat-are-in-1-gal-of-whole-milk-that-weighs-8-6-lb-and-contains-4-butterfat",{"id":97,"category":47,"text_question":98,"photo_question":49,"text_answer":99,"step_text_answer":8,"step_photo_answer":8,"views":100,"likes":101,"slug":102},537746,"add 10 5/6 + 7 3/8","Multiply \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$10$ \u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$6$ \u003C/math-field> to get \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$60$ \u003C/math-field>.\n \n \u003Cdiv>\n \n \u003Cmath-field style=\"font-size: 16px;padding: 8px;border-radius: 8px;border: 1px solid rgba

$0, 0, 0, .3$ ;box-shadow: 0 0 0 rgba

$0, 0, 0, .2$ \n\" read-only>\u003Cmath-field read-only>\\frac{60+5}{6}+\\frac{7\\times 8+3}{8}\u003C/math-field> \u003C/math-field>\n \u003Cbr>\n \u003C/div>\n \n Add \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$60$ \u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$5$ \u003C/math-field> to get \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$65$ \u003C/math-field>.\n \n \u003Cdiv>\n \n \u003Cmath-field style=\"font-size: 16px;padding: 8px;border-radius: 8px;border: 1px solid rgba

$0, 0, 0, .3$ ;box-shadow: 0 0 0 rgba

$0, 0, 0, .2$ \n\" read-only>\u003Cmath-field read-only>\\frac{65}{6}+\\frac{7\\times 8+3}{8}\u003C/math-field> \u003C/math-field>\n \u003Cbr>\n \u003C/div>\n \n Multiply \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$7$ \u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$8$ \u003C/math-field> to get \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$56$ \u003C/math-field>.\n \n \u003Cdiv>\n \n \u003Cmath-field style=\"font-size: 16px;padding: 8px;border-radius: 8px;border: 1px solid rgba

$0, 0, 0, .3$ ;box-shadow: 0 0 0 rgba

$0, 0, 0, .2$ \n\" read-only>\u003Cmath-field read-only>\\frac{65}{6}+\\frac{56+3}{8}\u003C/math-field> \u003C/math-field>\n \u003Cbr>\n \u003C/div>\n \n Add \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$56$ \u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$3$ \u003C/math-field> to get \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$59$ \u003C/math-field>.\n \n \u003Cdiv>\n \n \u003Cmath-field style=\"font-size: 16px;padding: 8px;border-radius: 8px;border: 1px solid rgba

$0, 0, 0, .3$ ;box-shadow: 0 0 0 rgba

$0, 0, 0, .2$ \n\" read-only>\u003Cmath-field read-only>\\frac{65}{6}+\\frac{59}{8}\u003C/math-field> \u003C/math-field>\n \u003Cbr>\n \u003C/div>\n \n Least common multiple of \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$6$ \u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$8$ \u003C/math-field> is \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$24$ \u003C/math-field>. Convert \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$\\frac{65}{6}$ \u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$\\frac{59}{8}$ \u003C/math-field> to fractions with denominator \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$24$ \u003C/math-field>.\n \n \u003Cdiv>\n \n \u003Cmath-field style=\"font-size: 16px;padding: 8px;border-radius: 8px;border: 1px solid rgba

$0, 0, 0, .3$ ;box-shadow: 0 0 0 rgba

$0, 0, 0, .2$ \n\" read-only>\u003Cmath-field read-only>\\frac{260}{24}+\\frac{177}{24}\u003C/math-field> \u003C/math-field>\n \u003Cbr>\n \u003C/div>\n \n Since \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$\\frac{260}{24}$ \u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$\\frac{177}{24}$ \u003C/math-field> have the same denominator, add them by adding their numerators.\n \n \u003Cdiv>\n \n \u003Cmath-field style=\"font-size: 16px;padding: 8px;border-radius: 8px;border: 1px solid rgba

$0, 0, 0, .3$ ;box-shadow: 0 0 0 rgba

$0, 0, 0, .2$ \n\" read-only>\u003Cmath-field read-only>\\frac{260+177}{24}\u003C/math-field> \u003C/math-field>\n \u003Cbr>\n \u003C/div>\n \n Add \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$260$ \u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$177$ \u003C/math-field> to get \u003Cmath-field read-only default-mode=\"inline-math\" style=\"padding-left:5px; padding-right:5px; display:inline-block; border-radius:4px; border: 1px solid rgba

$0, 0, 0, .3$ \">

$437$ \u003C/math-field>.\n \n \u003Cdiv>\n \n \u003Cmath-field style=\"font-size: 16px;padding: 8px;border-radius: 8px;border: 1px solid rgba

$0, 0, 0, .3$ ;box-shadow: 0 0 0 rgba

$0, 0, 0, .2$ \n\" read-only>\u003Cmath-field read-only>\\frac{437}{24}\u003C/math-field> \u003C/math-field>\n \u003Cbr>\n \u003C/div>",365,73,"add-10-5-6-7-3-8",{"id":104,"category":47,"text_question":105,"photo_question":49,"text_answer":106,"step_text_answer":8,"step_photo_answer":8,"views":107,"likes":108,"slug":109},537694,"Considering that the monthly fixed cost of sandwich production by a company\n fast food is R

$100,000.00, that the average variable cost is R$ 25.00 and that the unit cost is\n sold for R

$48.00, answer the questions:\n 1) What quantity is sold to cover the company's monthly costs?\n 2) Consider that the company's fixed cost increases to R$ 112,000.00 per month. What is the\n new balance point?\n 3) Based on question 2, consider that the average variable cost has increased to R

$\n 35.00. What is the new break-even point?\n 4) From question 3, consider that prices have increased to R$ 53.00,\n unit sold. What is the new break-even point?","1) O ponto de equilíbrio é encontrado quando a receita é igual aos custos. Definido por:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n \\times

$p - cv$ = CF\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>Substituindo os valores dados:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n \\times

$48 - 25$ = 100000\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n \\times 23 = 100000\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n = \\frac{100000}{23} = 4347,826 ≈ 4348 \\text{ unidades}\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>2) Com o novo custo fixo:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n \\times

$48 - 25$ = 112000\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n \\times 23 = 112000\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n=\\frac{112000}{23}=4869,565≈4870\\text{ unidades}\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>3) Com o novo custo fixo e custo variável:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n \\times

$48 - 35$ = 112000\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n \\times 13 = 112000\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n=\\frac{112000}{13}=8615,384≈8615\\text{ unidades}\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>4) Com o novo preço:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n \\times

$53 - 35$ = 112000\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n \\times 18 = 112000\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n = \\frac{112000}{18} = 6222,222 ≈ 6222 \\text{ unidades}\u003C/math-field>\u003C/math-field>",974,195,"considering-that-the-monthly-fixed-cost-of-sandwich-production-by-a-company-fast-food-is-r-100-000-00-that-the-average-variable-cost-is-r-25-00-and-that-the-unit-cost-is-sold-for-r-48-00-answer",{"id":111,"category":47,"text_question":112,"photo_question":49,"text_answer":113,"step_text_answer":8,"step_photo_answer":8,"views":114,"likes":115,"slug":116},537625,"1. Case Study: Prevalence of Mild Anemia in School Children in a Rural Community in Peru\n\n Introduction: Mild anemia is a public health problem in Peru, especially in rural areas where access to adequate nutrition may be limited. This study seeks to investigate the prevalence of mild anemia in schoolchildren in a rural community in Peru and its possible association with access to a balanced diet.\n Problem: What is the prevalence of mild anemia in schoolchildren in a rural community in Peru and how is it related to access to a balanced diet? / Determine whether there is a significant association between dichotomous variables.\n Anemia

$Present/absent$ .\n Balanced diet

$Yes / No$ .\n Data were collected from 95 schoolchildren in a rural community in the country, recording their anemia status

$present or absent$ and their access to a balanced diet

$yes or no$ , as assessed by parents or guardians. The data are presented in the following table:\n Table 1: Type of diet and anemia in children in a rural community, Peru 2022.\n\n Balanced diet Anemia\n Present Absent\n No 32 12\n Yes 9 42\n\n With a significance level of 0.05, can we conclude that anemia and balanced diet are independent?\n Apply the Chi Square test to determine if there is a significant association between these two variables in the sample of school children from the rural community.\n Interpret the results obtained and draw conclusions about the relationship with food in the community studied.\n Reflection Question: After performing the Chi-Square analysis and obtaining the results, reflect on what other variables could influence reducing the prevalence of anemia that were not included in this study?","1. **Datos Observados:**\u003Cbr />\n\u003Cbr />\n - Sin alimentación balanceada: \u003Cbr />\n - Anemia presente: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>O_{11} = 32\u003C/math-field>\u003C/math-field>\u003Cbr />\n - Anemia ausente: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>O_{12} = 12\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n - Con alimentación balanceada: \u003Cbr />\n - Anemia presente: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>O_{21} = 9\u003C/math-field>\u003C/math-field>\u003Cbr />\n - Anemia ausente: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>O_{22} = 42\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. **Totales:**\u003Cbr />\n\u003Cbr />\n - Filas:\u003Cbr />\n - Sin alimentación balanceada: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>32 + 12 = 44\u003C/math-field>\u003C/math-field>\u003Cbr />\n - Con alimentación balanceada: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>9 + 42 = 51\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n - Columnas:\u003Cbr />\n - Anemia presente: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>32 + 9 = 41\u003C/math-field>\u003C/math-field>\u003Cbr />\n - Anemia ausente: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>12 + 42 = 54\u003C/math-field>\u003C/math-field>\u003Cbr />\n \u003Cbr />\n - Total general: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>44 + 51 = 95\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. **Valores Esperados

$\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>E_{ij}\u003C/math-field>\u003C/math-field>$ :**\u003Cbr />\n\u003Cbr />\n - \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>E_{11} = \\frac{44 \\times 41}{95} \\approx 19\u003C/math-field>\u003C/math-field>\u003Cbr />\n - \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>E_{12} = \\frac{44 \\times 54}{95} \\approx 25\u003C/math-field>\u003C/math-field>\u003Cbr />\n - \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>E_{21} = \\frac{51 \\times 41}{95} \\approx 22\u003C/math-field>\u003C/math-field>\u003Cbr />\n - \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>E_{22} = \\frac{51 \\times 54}{95} \\approx 29\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. **Cálculo de \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\chi^2\u003C/math-field>\u003C/math-field>:**\u003Cbr />\n\u003Cbr />\n - \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\chi^2 = \\sum \\frac{

$O_{ij} - E_{ij}$ ^2}{E_{ij}}\u003C/math-field>\u003C/math-field>\u003Cbr />\n - \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\chi^2 = \\frac{

$32 - 19$ ^2}{19} + \\frac{

$12 - 25$ ^2}{25} + \\frac{

$9 - 22$ ^2}{22} + \\frac{

$42 - 29$ ^2}{29}\u003C/math-field>\u003C/math-field>\u003Cbr />\n - \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\chi^2 \\approx 27.01\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. **Valor P y Grados de Libertad:**\u003Cbr />\n - Grados de libertad \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>=

$n_{\\text{filas}} - 1$ \\times

$n_{\\text{columnas}} - 1$ = 1\u003C/math-field>\u003C/math-field>\u003Cbr />\n - Valor P \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>= 0.0000002024\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n6. **Conclusión:**\u003Cbr />\n - Dado que el valor P es mucho menor que el nivel de significancia de \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>0.05\u003C/math-field>\u003C/math-field>, se rechaza la hipótesis nula.\u003Cbr />\n - La conclusión es que existe una asociación significativa entre la presencia de anemia y el acceso a una alimentación balanceada.",1201,240,"1-case-study-prevalence-of-mild-anemia-in-school-children-in-a-rural-community-in-peru-introduction-mild-anemia-is-a-public-health-problem-in-peru-especially-in-rural-areas-where-access-to-adequ",{"first":6,"last":118,"prev":8,"next":10},11,{"current_page":6,"from":6,"last_page":118,"links":120,"path":151,"per_page":145,"to":145,"total":34},[121,124,127,129,131,133,135,138,141,144,147,149],{"url":6,"label":122,"active":123},"1",true,{"url":10,"label":125,"active":126},"2",false,{"url":13,"label":128,"active":126},"3",{"url":16,"label":130,"active":126},"4",{"url":19,"label":132,"active":126},"5",{"url":22,"label":134,"active":126},"6",{"url":136,"label":137,"active":126},7,"7",{"url":139,"label":140,"active":126},8,"8",{"url":142,"label":143,"active":126},9,"9",{"url":145,"label":146,"active":126},10,"10",{"url":118,"label":148,"active":126},"11",{"url":10,"label":150,"active":126},"Next 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