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3 years ago

The perimeter of a rectangle is 36 feet. The length is 2 less than 3 times the width. What is the length? What is the width?

1 answer:

3 0

Automatically, you can rule out 13 and above, as they are way too large. 12 can be ruled out too, because 12•3-2=34, and 12+34 is above 36. What if the length is 10? 10+2=12, 12÷3=4. 10+4 is too low. If the width is 8, 8•3=24, 24+8=32. So close!!! Let's bump it up to 9. 9•3=27, 27-2=25. 25+9=34. Just one more! If the width is 10, 10•3=30, 30-2=28, 28+10=38. 9.5 should work. 9.5•3=28.5. 28.5-2=26.5. 26.5+9.5=36. There's your answer! The width is 9.5 and the length is 26.5.

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Solve for the missing side in the following right triangle: hypotenuse = 17, leg = 8, leg = ?.

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3 years ago

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An educator claims that the average salary of substitute teachers in school districts is less than $60 per day. A random sample

valentina_108 [34]

Answer:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

Step-by-step explanation:

Information given

60, 56, 60, 55, 70, 55, 60, and 55.

We can calculate the mean and deviation with these formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Replacing we got:

$\bar X=58.875$ represent the mean

$s=5.083$ represent the sample standard deviation for the sample

$n=8$ sample size

$\mu_o =60$ represent the value that we want to test

$\alpha=0.1$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is less than 60, the system of hypothesis would be:

Null hypothesis: $\mu \geq 60$

Alternative hypothesis: $\mu < 60$

The statistic would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info we got:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

3 0

3 years ago