Which properties must you use to add or subtract complex numbers?

For which of the following conditions will the sum of integers m and n always be an odd integer?

lisov135 [29]

Answer:

option K cuz

Step-by-step explanation:

if we take m=7 and n=2

7 + 2= 9

and 9 is an odd number

5 0

3 years ago

Read 2 more answers

Word Problem 3 Priscilla works 43 hours a week for eight weeks. If she makes $14.73 per hour, how much money did she make in the

VLD [36.1K]

Answer:

$5067,12

Step-by-step explanation:

14,73*43*8

4 0

2 years ago

Nate is making a picture frame by cutting a small rectangular prism out of a larger rectangular prism. When the project is compl

dybincka [34]

Surface area= [15*3*2]+[20*3*2]+[10*3*2]+[14*3*2]+[15*3*2]+[14*2.5*2]

surface area=90+120+60+84+90+70=514 in²

the answer is 514 in²

6 0

3 years ago

Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver’s license. Twenty-five years later (in 2008) that perc

Dima020 [189]

Answer:

a) $ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019$

b) $ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245$

c) On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of $\hat p$ the margin of error change for part a and b.

Step-by-step explanation:

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

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

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)

If solve n from equation (a) we got:

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

Part a

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 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=\pm 1.96$

If we replace the values into equation (a) for 1983 we got:

$ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019$

Part b

Since is the same confidence level the z value it's the same.

If we replace the values into equation (a) for 2008 we got:

$ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245$

Is the margin of error the same in parts (a) and (b)? Why or why not?

On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of $\hat p$ the margin of error change for part a and b.

3 0

3 years ago

URGENTTTT HELPPPPPPP please

Ahat [919]

Given that ABC is a right triangle.

The length of AC = 10 and AB = 17

We need to determine the m∠B

<u>The measure of ∠B:</u>

The side opposite to ∠B is AC and the hypotenuse of the triangle is AB

The m∠B can be determined using the trigonometric identity,

$sin B=\frac{opp}{hyp}$