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Korolek [52]
3 years ago
14

A softball coach has ordered softballs for two different leagues. The Junior League uses

Mathematics
1 answer:
dedylja [7]3 years ago
5 0

The number of 11-inch softball is 70 and the number of 12-inch softball is 50.

<u>Step-by-step explanation</u>:

<u>Given that,</u>

  • The cost of 11-inch softball = $2.50
  • The cost of 12-inch softball = $3.50

<u>Let us assume,</u>

  • The number of 11-inch softball be 'x'.
  • The number of 12-inch softball be 'y'.

<u>Forming the equation to solve x and y values :</u>

  • The total number of softball ordered = 120
  • The total cost for 120 softballs = $350

x + y = 120  -------(1)

2.5x + 3.5y = 350   --------(2)

<u>Multiply eq(1) by 2.5 and subtract eq(2) from eq(1)</u>,

 2.5x +2.5y = 300

-<u>(2.5x +3.5y = 350)</u>

  <u>        -1y     =  -50</u>

Therefore the value of y = 50.

The number of 12-inch softball is 50.

<u>Substitute y=50 in eq(1),</u>

x+50 = 120

x = 120-50

x = 70

The number of 11-inch softball is 70.

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A local marketing company wants to estimate the proportion of consumers in the Oconee County area who would react favorably to a
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Answer:

C. n=423

Step-by-step explanation:

1) Notation and important concepts

Margin of error for a proportion is defined as "percentage points your results will differ from the real population value"

Confidence=90%=0.9

\alpha=1-0.9=0.1 represent the significance level defined as "a measure of the strength of the evidence that must be present in your sample before you will reject the null hypothesis and conclude that the effect is statistically significant".

\hat p=0.5 represent the sample proportion of consumers in the Oconee County area who would react favorably to a marketing campaig. For this case since we don't have enough info we use the value of 0.5 since is equiprobable the event analyzed.

z_{\alpha/2} represent a quantile of the normal standard distribution that accumulates {\alpha/2} on each tail of the distribution.

2) Formulas and solution for the problem

For this case the margin of error for a proportion is given by this formula

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

For this case the confidence level is 90% or 0.9 so then the significance would be

\alpha=1-0.9=0.1 and \alpha/2=0.05

With \alpha/2=0.05, we can find the value for z_{\alpha/2} using the normal standard distribution table or excel.

The calculated value is z_{\alpha/2}=1.644854

Now from equation (1) we need to solve for n in order to answer the question.

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

Squaring both sides:

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

And solving for n we got:

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

Now we can replpace the values

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.644854})^2}=422.739

And rounded up to the nearest integer we got:

n=423

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