Need help...with this graph

help in need help !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Aleks [24]

answer is 2+ 1/4x + 1 the simplest form is x= -3/4 or -0.75

5 0

3 years ago

Find the solution of the system of equations. <br>7x -- 8y = -23<br>7x – 7y = -14<br><br>(?,?)

gtnhenbr [62]

Answer:

(7,9)

Step-by-step explanation:

6 0

3 years ago

Pascale leaves the office to go home. She walks 12 blocks due north and then 9 blocks due west. How far is she from the office?

Stells [14]

Answer:

15 blocks

Step-by-step explanation:

We can use the Pythagorean theorem to solve the right triangle

a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse

12^2 + 9^2 = h^2

144+81 = h^2

225= h^2

Taking the square root of each side

sqrt(225) = sqrt(h^2)

15 = h

7 0

4 years ago

Read 2 more answers

PLEASE HELP 25 POINTSS I WILL GIVE BRAILIEST

kykrilka [37]

The solution to the system of equation is (1, 4).

In order to find this, we can first just see where the graphs intersect each other. This will give us the solution set.

As for what it represents, the x value in the increase in temperature and the y value is the increase in customers.

Therefore, we know that we want the temperature to go up by 1 (although we don't know the units) and that would result in the amount of people coming, and staying longer by 4 (again, we don't know the units of measure).

6 0

4 years ago

A 95% confidence interval was computed using a sample of 16 lithium batteries, which had a sample mean life of 645 hours. The co

polet [3.4K]

Answer:

Step-by-step explanation:

Hello!

The mean life of 16 lithium batteries was estimated with a 95% CI:

(628.5, 661.5) hours

Assuming that the variable "X: Duration time (life) of a lithium battery(hours)" has a normal distribution and the statistic used to estimate the population mean was s Student's t, the formula for the interval is:

[X[bar]± $t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }$ ]

The amplitude of the interval is calculated as:

a= Upper bond - Lower bond

a= [X[bar]+ $t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }$ ] -[X[bar]- $t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }$ ]

and the semiamplitude (d) is half the amplitude

d=(Upper bond - Lower bond)/2

d=([X[bar]+ $t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }$ ] -[X[bar]- $t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }$ ] )/2

d= $t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }$

The sample mean marks where the center of the calculated interval will be. The terms of the formula that affect the width or amplitude of the interval is the value of the statistic, the sample standard deviation and the sample size.

Using the semiamplitude of the interval I'll analyze each one of the posibilities to see wich one will result in an increase of its amplitude.

Original interval:

Amplitude: a= 661.5 - 628.5= 33

semiamplitude d=a/2= 33/2= 16.5

1) Having a sample with a larger standard deviation.

The standard deviation has a direct relationship with the semiamplitude of the interval, if you increase the standard deviation, it will increase the semiamplitude of the CI

↑d= $t_{n-1;1-\alpha /2}$ * ↑S/√n

2) Using a 99% confidence level instead of 95%.

d= $t_{n_1;1-\alpha /2}$ * S/√n

Increasing the confidence level increases the value of t you will use for the interval and therefore increases the semiamplitude:

95% ⇒ $t_{15;0.975}= 2.131$

99% ⇒ $t_{15;0.995}= 2.947$

The confidence level and the semiamplitude have a direct relationship:

↑d= ↑ $t_{n_1;1-\alpha /2}$ * S/√n

3) Removing an outlier from the data.

Removing one outlier has two different effects:

1) the sample size is reduced in one (from 16 batteries to 15 batteries)

2) especially if the outlier is far away from the rest of the sample, the standard deviation will decrease when you take it out.

In this particular case, the modification of the standard deviation will have a higher impact in the semiamplitude of the interval than the modification of the sample size (just one unit change is negligible)

↓d= $t_{n_1;1-\alpha /2}$ * ↓S/√n

Since the standard deviation and the semiamplitude have a direct relationship, decreasing S will cause d to decrease.

4) Using a 90% confidence level instead of 95%.

↓d= ↓ $t_{n_1;1-\alpha /2}$ * S/√n

Using a lower confidence level will decrease the value of t used to calculate the interval and thus decrease the semiamplitude.

5) Testing 10 batteries instead of 16. and 6) Testing 24 batteries instead of 16.

The sample size has an indirect relationship with the semiamplitude if the interval, meaning that if you increase n, the semiamplitude will decrease but if you decrease n then the semiamplitude will increase:

From 16 batteries to 10 batteries: ↑d= $t_{n_1;1-\alpha /2}$ * S/√↓n

From 16 batteries to 24 batteries: ↓d= $t_{n_1;1-\alpha /2}$ * S/√↑n

I hope this helps!

4 0

3 years ago