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3 years ago

How would I find the unit rate of 1/5 miles over 1/65 hours

Mathematics

1 answer:

cestrela7 [59]3 years ago

8 0

Answer:

1mph

Step-by-step explanation:

Showing Work

This is a fraction equal to

1 miles ÷ 1 hours

We want a unit rate where

1 is in the denominator,

so we divide top and bottom by 1

1 miles ÷ 1

1 hours ÷ 1

1 miles

1 hour

1 miles

hour

= 1 miles per hour

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Someone help me please and thank you!! ASAP :(

pantera1 [17]

<h3>Answer: Choice D) </h3>

Work Shown:

$(f+g)(x) = f(x)+g(x)\\\\(f+g)(x) = \frac{x-23}{x^2+9x-36}+\frac{1}{x+12}\\\\(f+g)(x) = \frac{x-23}{(x+12)(x-3)}+\frac{1(x-3)}{(x+12)(x-3)}\\\\(f+g)(x) = \frac{x-23+1(x-3)}{(x+12)(x-3)}\\\\(f+g)(x) = \frac{x-23+x-3}{x^2+9x-36}\\\\(f+g)(x) = \frac{2x-26}{x^2+9x-36}\\\\$

We must require that $x \ne -12$ and $x \ne 3$ to avoid having 0 in the denominator. This is why choice D is the answer.

3 0

3 years ago

4)In order to set rates, an insurance company is trying to estimate the number of sick daysthat full time workers at an auto rep

bagirrra123 [75]

Answer:

A sample of 18 is required.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.92}{2} = 0.04$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.04 = 0.96$ , so Z = 1.88.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

A previous study indicated that the standard deviation was 2.2 days.

This means that $\sigma = 2.2$

How large a sample must be selected if the company wants to be 92% confident that the true mean differs from the sample mean by no more than 1 day?

This is n for which M = 1. So

$M = z\frac{\sigma}{\sqrt{n}}$

$1 = 1.88\frac{2.2}{\sqrt{n}}$