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Setler79 [48]
2 years ago
8

The drama club is selling short-sleeved shirts for $5 each, and long-sleeved shirts for $10 each. They hope to sell all of the s

hirts they ordered, to earn a total of $1,750. After the first week of the fundraiser, they sold StartFraction one-third EndFraction of the short-sleeved shirts and StartFraction one-half EndFraction of the long-sleeved shirts, for a total of 100 shirts.
This system of equations models the situation.
5x + 10y = 1,750
x + A system of equations. 5 x plus 10 y equals 1,750. StartFraction one-third EndFraction x plus StartFraction one-half EndFraction y equals 100.y = 100
Let x represent the number of short-sleeved shirts ordered and let y represent the number of long-sleeved shirts ordered.
How many short-sleeved shirts were ordered?

How many long-sleeved shirts were ordered?
Mathematics
1 answer:
liraira [26]2 years ago
6 0

Answer:

<u>The drama club ordered 150 short-sleeved shirts and 100 long-sleeved shirts</u>

Step-by-step explanation:

1. Let's review all the information provided for answering the questions properly:

5x + 10y = 1,750

1/3x + 1/2y = 100

x = number of short-sleeved shirts ordered

y = number of long-sleeved shirts ordered

Resolving the 1st equation:

5x + 10y = 1,750

5x = 1,750 - 10y

x = 350 - 2y (Dividing by 5 at both sides)

Resolving  the 2nd equation:

1/3x + 1/2y = 100

1/3 (350 - 2y) + 1/2y = 100

350/3 -2y/3 + 1/2y = 100

700 - 4y + 3y = 600 (Lowest common denominator = 6)

-y = 600 - 700 (Subtracting 700 at both sides)

<u>y = 100</u> (Dividing by - 1)

Finding the value of x in the 1st equation:

5x + 10 (100) = 1,750

5x + 1,000 = 1,750

5x = 1,750 - 1,000 (Subtracting 1,000 at both sides)

5x = 750

<u>x = 150</u> (Dividing by 5)

2. Proving that x = 150 and y = 100 are correct

1/3x + 1/2y = 100

1/3 * 150 + 1/2 * 100 = 100

150/3 + 100/2 = 100

50 + 50 = 100

100 = 100

<u>We proved that x = 150 and y = 100 are correct</u>

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Answer:

We need to conduct a hypothesis in order to test the claim that the true proportion of employess are late to work is 0.16, so we need to apply a one sample proportion test:  

Null hypothesis:p=0.16  

Alternative hypothesis:p \neq 0.16  

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p_v =2*P(z>0.546)=0.585  

Step-by-step explanation:

Data given and notation

n=25 represent the random sample taken

\hat p=0.2 estimated proportion  

p_o=0.16 is the value that we want to test

\alpha represent the significance level

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 the true proportion of employess are late to work is 0.16, so we need to apply a one sample proportion test:  

Null hypothesis:p=0.16  

Alternative hypothesis:p \neq 0.16  

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 requires we can replace in formula (1) like this:  

z=\frac{0.20 -0.16}{\sqrt{\frac{0.16(1-0.16)}{25}}}=0.546  

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 next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z>0.546)=0.585  

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Answer:

(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953

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And the confidence interval for the difference is between:

27.953 \leq \mu_1 -\mu_2 \leq 32.047

Step-by-step explanation:

We have the following info given:

\bar X_1 = 240 sample mean for medium Pizzas from Prim's

\bar X_2 = 210 sample mean for medium Pizzas from Pizza Place

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The confidence interval for the difference of means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

And for the 95% confidence we need a significance level of \alpha=1-0.95=0.05 and \alpha/2 =0.025, the degrees of freedom are given by:

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And the critical value would be

t_{\alpha/2}= 1.984

And replacing we got:

(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953

(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047

And the confidence interval for the difference is between:

27.953 \leq \mu_1 -\mu_2 \leq 32.047

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