Ask question

Ask question

All categories

3 years ago

14

Help me please ???????

1 answer:

FinnZ [79.3K]3 years ago

5 0

Answer:

it is the last one give me 5 stars thanks

Step-by-step explanation:

You might be interested in

Determine the # of rollovers and total cost # of rollovers

miv72 [106K]

Answer:

7747747474774774u4774748 answer

7 0

3 years ago

If an outcome measure is normally distributed, this means that

Inessa [10]

A normally distributed result means that when the data is plotted on a graph, we can actually see a bell shaped curve. That highest frequency or peak is centered on the average value. Because it is shaped like this, we can say that:

most scores are around the average with some scores being lower or higher

5 0

3 years ago

13. Find the equations of the straight lines which passes

ipn [44]

9514 1404 393

Answer:

y = 1/2x +2
y = -2x +7

Step-by-step explanation:

The slope of a line is the tangent of the angle it makes with the x-axis. The given line has a slope of -1/3, so the lines we want will have slopes of ...

m1 = tan(arctan(-1/3) +45°) = 0.5 . . . . . using a calculator

m2 = tan(arctan(-1/3) -45°) = -2

Of course, these two lines are perpendicular to each other, so their slopes will have a product of -1: (0.5)(-2) = -1.

__

We can use the point-slope form of the equation for a line to write the desired equations:

y = m(x -h) +k . . . . . line with slope m through point (h, k)

<u>Line 1</u>:

y = 1/2(x -2) +3

y = 1/2x +2

<u>Line 2</u>:

y = -2(x -2) +3

y = -2x +7

3 0

3 years ago

Natalie uses a 15% off coupon when she buys a camera. The original price of the camera is $45.00. How much money does Natalie sa

Contact [7]

Answer:

6.75

Step-by-step explanation:

First, I just gave you the answer.

Second, just use it.

4 0

3 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

3 years ago