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3 years ago

How many 2/3- cup servings are in 6 cups of yogurt

Mathematics

2 answers:

Citrus2011 [14]3 years ago

8 0

6 / (2/3)
= 6 * 3/2
= 18/2
= 9

answer

there are 9 (2/3 cup) servings in 6 cups of yogurt

Dafna1 [17]3 years ago

8 0

Answer: The required number of servings in 6 cups of yogurt is 9.

Step-by-step explanation: We are given to find the number of $\dfrac{2}{3}$ cup servings in 6 cups of yogurt.

We will be using the UNITARY method to solve the given problem.

We have

Number of servings in $\dfrac{2}{3}$ cups of yogurt = 1.

So, the number of servings in 1 cup of yogurt will be

$\dfrac{1}{\frac{2}{3}}=\dfrac{3}{2}.$

Therefore, the number of servings in 6 cups of yogurt is given by

$\dfrac{3}{2}\times6=9.$

Thus, the required number of servings in 6 cups of yogurt is 9.

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3 years ago

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VikaD [51]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$P(X = 8) = 0.0037$

$P(X < 5) = 0.805$

$P(X > 6) = 0.0206$

I would be surprised because the value is very small , less the 0.05

Step-by-step explanation:

From the question we are told that

The probability a randomly selected individual will not cover his or her mouth when sneezing is $p = 0.267$

Generally data collected from this study follows binomial distribution because the number of trials is finite , there are only two outcomes, (covering , and not covering mouth when sneezing ) , the trial are independent

Hence for a randomly selected variable X we have that

$X \ \ \~ \ \ { B ( p , n )}$

The probability distribution function for binomial distribution is

$P(X = x ) = ^nC_x * p^x * (1 -p) ^{n-x}$

Considering question a

Generally the the probability that among 12 randomly observed individuals exactly 8 do not cover their mouth when sneezing is mathematically represented as

$P(X = 8) = ^{12} C_8 * (0.267)^8 * (1- 0.267)^{12-8}$

Here C denotes combination

$P(X = 8) = 495 * 0.000025828 * 0.28867947$

$P(X = 8) = 0.0037$

Considering question b

Generally the probability that among 12 randomly observed individuals fewer than 5 do not cover their mouth when sneezing is mathematically represented as

$P(X < 5 ) =[P(X = 0 ) + \cdots + P(X = 4)]$

=> $P(X < 5 ) =[ ^{12} C_0 * (0.267)^0 * (1- 0.267)^{12-0} + \cdots + ^{12} C_4 * (0.267)^4 * (1- 0.267)^{12-4} ]$

=> $P(X < 5 ) = 0.02406 + 0.10516 + 0.21067 + 0.25580 + 0.20964$

=> $P(X < 5) = 0.805$

Considering question c

Generally the probability that fewer than half(6) covered their mouth when sneezing(i.e the probability the greater than half do not cover their mouth when sneezing) is mathematically represented as

$P(X > 6) = 1 - p(X \le 6)$

=> $P(X > 6) = 1 - [P(X = 0) + \cdots + P(X =6)]$

=> $P(X > 6)=1 - [^{12} C_0 * (0.267)^0 * (1- 0.267)^{12-0}+ \cdots + ^{12} C_4 * (0.267)^6 * (1- 0.267)^{12-6} ]$

=> $P(X > 6)= 1 - [0.02406 + \cdots + 0.0519 ]$

=> $P(X > 6) = 0.0206$

I would be surprised because the value is very small , less the 0.05

8 0

3 years ago