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nasty-shy [4]
3 years ago
6

ANSWER PLEASE PLEASE Don’t MIND MY PENCIL WORK

Mathematics
1 answer:
morpeh [17]3 years ago
3 0
You need to make two equations-
Use b for butternut squash and z for zucchini.
2b+4z = $6.96
3b=6z
Solve for one of the variables in the second equation ( I will solve for b)
3b/3=6z/3
b=2z 
Plug it in-
2(2z)+4z =$6.96
4z+4z=$6.96
8z =$6.96
8z/8 = 6.96/8
z= .87
Plug in again - (b=2z)
b= (2)(.87)
b= $1. 74
So the zucchini is 87 cents and the butternut squash is  $1.74 (each)
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Suppose Brianna decided that the error margin of her 95% confidence interval was too large and wanted an error margin of .05 whi
svetoff [14.1K]

Answer:

n=\frac{0.26 (1-0.26)}{(\frac{0.05}{1.96})^2} =295.65 \approx =296 (b)  

Rounded up would be n =296

Step-by-step explanation:

Assuming this previous info:  She randomly interested in the p selects 100 engineering majors, and 26 of them said they earned an A in Calc 1.

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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Solution to the problem

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}} (a)  

If solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)  

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:  

z_{\alpha/2}=\pm 1.96  

Thes estimated proportion is \hat p = \frac{26}{100}=0.26

And replacing we got:

n=\frac{0.26 (1-0.26)}{(\frac{0.05}{1.96})^2} =295.65 \approx =296 (b)  

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