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Nikitich [7]
3 years ago
11

A muffin recipe, which yields 12 muffins, calls for 2/4 cup of milk for every 1 3/4 cups of flour. The same recipe calls for 1/4

cup of coconut for every 3/4 cup of chopped apple.
To yield a batch of 30 muffins, how much flour will be needed in the mix?
a) 6 6/7
b) 4 3/8
c)52 1/2
d) 17 1/7
Mathematics
1 answer:
mezya [45]3 years ago
7 0

Answer:

b) 4 3/8

Step-by-step explanation:

we know that

12 muffins calls for 1 3/4 cups of flour

using proportion

Find out how much flour will be needed in the mix for a bath of 30 muffins

\frac{12}{1\frac{3}{4}}=\frac{30}{x}\\\\x=30(1\frac{3}{4})/12

Convert mixed number to an improper fraction

1\frac{3}{4}=1+\frac{3}{4}=\frac{7}{4}

substitute

x=30(\frac{7}{4})/12\\\\x=\frac{210}{48}\ cups\ of\ flour

convert to mixed number

\frac{210}{48}=\frac{192}{48}+\frac{18}{48}=4\frac{3}{8}\ cups\ of\ flour

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A cellphone provider has the business objective of wanting to estimate the proportion of subscribers who would upgrade to a new
stiv31 [10]

Answer:

z=\frac{0.27 -0.2}{\sqrt{\frac{0.2(1-0.2)}{500}}}=3.913  

p_v =P(z>3.913)=0.000046  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of subscribers that would upgrade to a new cellphone at a reduced cost is significantly higher than 0.2 or 20%

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=135 represent the subscribers that would upgrade to a new cellphone at a reduced cost

\hat p=\frac{135}{500}=0.27 estimated proportion of subscribers that would upgrade to a new cellphone at a reduced cost

p_o=0.2 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that true proportion is higher than 0.2 or not.:  

Null hypothesis:p \leq 0.2  

Alternative hypothesis:p > 0.2  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info required we can replace in formula (1) like this:  

z=\frac{0.27 -0.2}{\sqrt{\frac{0.2(1-0.2)}{500}}}=3.913  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(z>3.913)=0.000046  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of subscribers that would upgrade to a new cellphone at a reduced cost is significantly higher than 0.2 or 20%

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