Add (-x^2-2x+6)+(-9x^2-7x+6)

If f(x) = 22 – 4x – 5, find the following:<br> a. f(-2)<br> b. f (10)<br> .c. f(0)

coldgirl [10]

Answer:

a.25

b.-23

c.17

Step-by-step explanation:

Plug each value in for x and solve

4 0

3 years ago

Miss smith’s pay is directly proportional to the number of hours she work at the airport. Her pay for 22 hours is $297.How much

Troyanec [42]

Answer:

She should earn $540 in a 40 hour week

Step-by-step explanation:

we know that

Her pay for 22 hours is $297

so

using proportion

Find out how much should she earn in a 40 hours week

$\frac{22}{297}=\frac{40}{x}\\\\x=297*40/22\\\\x= \$540$

7 0

3 years ago

Can you help me with my math

romanna [79]

I can help if you want me to

5 0

2 years ago

Find the difference between the points (6, -3) And (-5,-3)

Varvara68 [4.7K]

We can use the distance formula to find the distance between these points.

Distance formula: $\sqrt{(x.2 - x.1)^{2} + (y.2 - y.1)^{2}}$

√(-5 - 6)² + (-3 - (-3))²

√121 = 11

The difference between the points is 11.

5 0

2 years ago

Based on the Nielsen ratings, the local CBS affiliate claims its 11:00 PM newscast reaches 41 % of the viewing audience in the a

ZanzabumX [31]

Answer:

1) Null hypothesis: $p\geq 0.41$

Alternative hypothesis: $p < 0.41$

2) $\hat p=0.36$ estimated proportion of people indicated that they watch the late evening news on this local CBS station

3) $z_{crit}=-2.33$

And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"

4) $z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{1000}}}=-1.017$

5) $z_{crit}=-1.28$

And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"

6) We see that $|t_{calculated}|$ so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.

7) Null hypothesis: $p\geq 0.41$

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

X represent the people indicated that they watch the late evening news on this local CBS station

$\hat p=0.36$ estimated proportion of people indicated that they watch the late evening news on this local CBS station

$p_o=0.41$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) Part 1We need to conduct a hypothesis in order to test the claim that 11:00 PM newscast reaches 41 % of the viewing audience in the area: Null hypothesis:[tex]p\geq 0.41$

Alternative hypothesis: $p < 0.41$

Part 2

$\hat p=0.36$ estimated proportion of people indicated that they watch the late evening news on this local CBS station

Part 3

Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.01 of the area on the left and on this case this value is :

$z_{crit}=-2.33$

And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"

Part 4

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{1000}}}=-1.017$

Part 5

Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.1 of the area on the left and on this case this value is :

$z_{crit}=-1.28$

And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"

Part 6

We see that $|t_{calculated}|$ so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.

Part 7

Null hypothesis: $p\geq 0.41$

4 0

2 years ago