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2 years ago

You buy 3 shirts and 1 pair of pants for $32. Your friend buys 2 shirts and 3 pairs of pants for $40.

1 answer:

5 0

You buy 3 shirts and 1 pair of pants for $32. Your friend buys 2 shirts and 3 pairs of pants for $40. How much did each item cost?

First let’s define a shirt as X and pants as Y!

In the first scenario, the equation will be 3X + Y =$32

In the second scenario, the equation will be 2X + 3Y = $40

Now because we have two different variables (X & Y) let’s solve for one of them (Ex: Y) to simplify the equations!

3X + Y= $32—> subtract 3X on both sides of the equation

-3X. - 3X

Net result: Y= 32- 3X

Now we can insert this into the second equation 2X + 3Y = $40 in order to solve for X variable

2X + 3(32-3X) = $40

2X + 96 - 9X = $40 —> simplify the equation

-7X + 96 = $40 —> substrate 96 from both sides

-7X = -56 —> divide both sides by -7

X= 8

Now reinsert 8= X into 3X + Y = $32

3(8) + Y = $32

24+ Y = $32 —> substract 24 from both sides

y = 8

ANSWER: X= 8, Y= 8

If I am wrong I am sorry >_<

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Step-by-step explanation:

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2 years ago

A store buys a digital watch for $80 and sells it for $95 what is the percent of increase

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2 years ago

We are conducting a hypothesis test to determine if fewer than 80% of ST 311 Hybrid students complete all modules before class.

Juli2301 [7.4K]

Answer:

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

Null hypothesis: $p\geq 0.8$

Alternative hypothesis: $p < 0.8$

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

$p_v =P(Z$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

n=110 represent the random sample taken

X=97 represent the students who completed the modules before class

$\hat p=\frac{97}{110}=0.882$ estimated proportion of students who completed the modules before class

$p_o=0.8$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) 2) Concepts and formulas to use We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.8 or 80%: Null hypothesis:[tex]p\geq 0.8$

Alternative hypothesis: $p < 0.8$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

4) 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 assumed $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(Z$

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

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3 years ago