What is Y=-6x+2 ????????

SCORPION-xisa [38]

Answer:

15 x 1 + 5 x 3 = 30

Step-by-step explanation:

If you turn the x into a one then it multiplies 15 by one and then it remains at 15, then you multiply 5 and 3 so that's 15 and then 15 + 15 is 30.

3 0

3 years ago

Read 2 more answers

- 6 = z + 3<br>solve each equation for the unknown variable

Nataly [62]

Solution: Z= -9

Reasoning: -6 - 3 = -9

4 0

3 years ago

Completely Factor the expression -18 P - 9

jekas [21]

can i have brainliest pls tyyy :)))

Answer:

-9(2p+1)

Step-by-step explanation:

-18p - 9

find the GCF (greatest common factor) for the expression.

-9

then, divide each term by the GCF

-18p / -9 -9 / -9

= 2p = 1

next, put the two terms you get when dividing into the final form of the answer:

GCF on the outside and quotients on the inside

-9(2p+1)

hope this helped! ;)

5 0

3 years ago

Help no links please

Ludmilka [50]

Answer:

180

Step-by-step explanation:

The difference between 68 and 23 is 45. The company charging 23$ has to be 45$ more for the amount of hours. When you multiply 180 by .4 you get 72, and when you multiply 180 by .65 you get 117. The difference is 45.

Sorry that I didn't show my work, I don't really know how to do this other than plugging in random numbers and 180 fits so...

Hope this helps!

3 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