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3 years ago

An angle with measure of 71° is bisect at what angle?

Mathematics

1 answer:

Korolek [52]3 years ago

6 0

Answer:

Step-by-step explanation: just divide 71 into 2 which gives you 35.5 making a your answer.

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It is worked out that 4 ladles full of soup each will feed 160 people. The customers have complained in the past that the portio

polet [3.4K]

160 people x 4 ladle/1 person = 640 ladles full
640 ladles x 1 person/5 ladles = 128 people fed

7 0

3 years ago

Is this statement true or false -7 -2 > 6 - 14

exis [7]

Answer:

false

Step-by-step explanation:

(-7-2) > (6-14)

-7-2 = -9

6-14 = -8

-9 is not > -8, because -9 is farther to the left on a number line. So, this statement would be false.

6 0

3 years ago

Let f(x)=x2 and g(x)=x-3. Evaluate (f•g)(0)

Galina-37 [17]

Answer:The answe would be 5 cause the 0 doesn't count

Step-by-step explanation:

3 0

2 years ago

Find the midpoint between these two points (-7,-6) and (2,5) PLEASE ANSWER!

QveST [7]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-7}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{2-7}{2}~~,~~\cfrac{5-6}{2} \right)\implies \left( \cfrac{-5}{2}~~,~~\cfrac{-1}{2} \right)\implies (-2.5,-0.5)$

4 0

4 years ago

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

son4ous [18]

Answer:

D. 31

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:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

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 population standard deviation is 2.8 days.

This means that $\sigma = 2.8$

How large a sample must be selected if the company wants to be 95% confident that the true mean differs from the sample mean by no more than 1 day?

This is n for which M = 1. So

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

$1 = 1.96\frac{2.8}{\sqrt{n}}$