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1 year ago

an ostrich can run at a rate of 50 miles in 60 minutes at this rate how long would it take an ostrich to run 15 miles

Mathematics

1 answer:

vova2212 [387]1 year ago

5 0

This is direct proportion - it takes less time to run 15 miles than 50 miles. Show the comparison:

<span>x=18</span>minutes

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Which equation could be used to find the length of the hypotenuse?

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2 years ago

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Solve the inequality. Using a verbal statement, in simplest terms, describe the solution of the inequality. Be sure to include t

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-2x + 3 > 3(2x - 1)

the inequality sign > means 'greater than'.

First simplify the right side of the equation by distributing the 3 over the parentheses:-

-2x + 3 > 6x - 3

Now add -6x to both sides of the inequality which gives us
-8x + 3 > -3
Adding -3 to both sides we have
-8x > -6

The next step is to divide both sides by -8 so x is isolated on the left side and this gives us the solution. However there is a rule with inequalities that if you divide the variable term by a negative the inequality is flipped. So in this case , greater that (>) becomes less than (<):-

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2 years ago

The scale for the drawing of a rectangular playing field is 2 inches= 5 feet. Write an equation you can use to find the dimensio

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Answer:

Step-by-step explanation:

Given:

Scale

2 inches : 5 feet

Actual dimensions

Let x be the dimension of the scale drawing (in inches) and y, the corresponding dimension of the actual field (in feet).

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1 year ago

In measuring reaction time, a psychologist estimates that a standard deviation is .05 seconds. How large a sample of measurement

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Answer:

Step-by-step explanation:

We are asked to find the size of sample to be 95% confident that the error in psychologist estimate of mean reaction time will not exceed 0.01 seconds.

We will use following formula to solve our given problem.

$n\geq (\frac{z_{\alpha/2}\cdot\sigma}{E})^2$ , where,

$\sigma=\text{Standard deviation}=0.05$ ,

$\alpha=\text{Significance level}=1-0.95=0.05$ ,

$z_{\alpha/2}=\text{Critical value}=z_{0.025}=1.96$ .

$E=\text{Margin of error}$

$n=\text{Sample size}$

Substitute given values:

$n\geq (\frac{z_{0.025}\cdot\sigma}{E})^2$

$n\geq (\frac{1.96\cdot0.05}{0.01})^2$

$n\geq (\frac{0.098}{0.01})^2$

$n\geq (9.8)^2$

$n\geq 96.04$

Therefore, the sample size must be 97 in order to be 95% confident that the error in his estimate of mean reaction time will not exceed 0.01 seconds.

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2 years ago

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Answer:

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