Identify the domain and range of the following graph.

Evaluate the following functions at x = 10<br>g(x)=√(x+1)

V125BC [204]

g(x)=√(x+1)

Let's plug 10 in for x

g(10) = √(10+1)

Simplify the exponents

g(10) = √(11)

This is the exact form but it can also be expressed as a decimal.

√(11) = 3.31662479

You can simplify as needed

6 0

3 years ago

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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

3 years ago

The balance of a loan is $2,570 in January, and the monthly payment is $125.50. The relationship between the loan balance, y, an

Hoochie [10]

1600<2570-125.5x<2000 subtract 2570 from all terms...

-970<-125.5x<-570 divide all terms by -125.5 (and reverse signs because of division by a negative!)

7.73>x>4.54 and x is months since January, and since months can only be integers...

x=[5,7]

So January + 5, 6, and 7 respectively are the three months that satisfy the equation...

June, July, and August.

4 0

3 years ago

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Write 2 7/10% as a fraction.

Andre45 [30]

Step-by-step explanation:

Can you please reword your problem

6 0

3 years ago

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A triangle has a base length given by the equation b = 2x + 4 and a height given by the equation h

Anna11 [10]