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3 years ago

Measure of the supplement of angle

1 answer:

8 0

Two supplemental angles add up to 180°
So we have (A + B = 180°)
The measure of one of them is always 180° - the other.
A = 180° - B
B = 180° - A

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6. Benny earns $12.00 per hour as a lifeguard. He earns a bonus of $75.00

Vesna [10]

Answer:

75+12h ≥ 399

He must work at least 27 hours

Step-by-step explanation:

How much does benny earn

He makes 75 for the course plus 12 dollars an hour

75+ 12h

This must be at least 399

75+12h ≥ 399

Subtract 75 from each side

75-75+12h ≥ 399-75

12h ≥ 324

Divide each side by 12

12h/12 ≥ 324/12

h ≥ 27

He must work at least 27 hour

3 0

3 years ago

Read 2 more answers

Simplify (9 + 8 – 6)(4 – 5).

Temka [501]

Answer:

The answer would be - 11.

Step-by-step explanation:

9 + 8 = 17

17 - 6 = 11.

4 - 5 = 4 + - 5 = - 1

11(- 1) = 11 × - 1 = - 11

Therefor, your answer would be - 11.

Hope this helps!

5 0

4 years ago

Read 2 more answers

Ben is 444 times as old as Ishaan and is also 666 years older than Ishaan.

Roman55 [17]

Let B = age on Ben
and age of Ishaan =I
B =4I
B = 6+I
Gives 6+I = 4I
6 = 3I
I =2 years and B = 4x2 = 8years
Age of Ishaan = 2 years

6 0

3 years ago

The three altitudes of ΔABC intersect at point F. Point F is the

zlopas [31]

D.) Orthocenter.....

5 0

3 years ago

Read 2 more answers

Sample Size for Proportion As a manufacturer of golf equipment, the Spalding Corporation wants to estimate the proportion of gol

Dima020 [189]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56$

And rounded up we have that n=2663

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.025$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume an estimated proportion of $\hat p =0.5$ since we don't have prior info provided. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56$

And rounded up we have that n=2663

6 0

4 years ago