Hello! Please help! (directions on the image) :D

Select all of the statements that are true about the following expression: 5x3+x4−8

Lapatulllka [165]

I think the correct answer is c

6 0

3 years ago

There are 64 pretzels in a 16 ounce bag of chocolate covered pretzels.

FromTheMoon [43]

16:64
5:x (Cross Multiply)
16x = 64*5
16x = 320 (Then Divide both sides by 16 to isolate x)
x= 20

5 0

3 years ago

Can anyone help me? SOOOOO hard!!

emmasim [6.3K]

Answer:

i think its 16/34

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Calculate the value of the sample variance. Round your answer to one decimal place. 9_5,9_5,2,9_5

Ugo [173]

Answer:

$s^2 = 0.01$

Step-by-step explanation:

Given

Values: 9/5, 9/5, 2, 9/5

Required

Calculate the sample variance

Sample variance is calculated using:

$s^2 = \frac{\sum (x_i - \overline x)^2}{n - 1}$

First, we calculate the mean

$\overline x = \frac{\sum x}{n}$

$\overline x = \frac{9/5 + 9/5 + 2 + 9/5}{4}$

$\overline x = \frac{7.4}{4}$

$\overline x = 1.85$

$s^2 = \frac{\sum (x_i - \overline x)^2}{n - 1}$ becomes

$s^2 = \frac{(9/5 - 1.85)^2+(9/5 - 1.85)^2+(2 - 1.85)^2+(9/5 - 1.85)^2}{4 - 1}$

$s^2 = \frac{(-0.05)^2+(-0.05)^2+(0.15)^2+(-0.05)^2}{4 - 1}$

$s^2 = \frac{0.0025+0.0025+0.0225+0.0025}{3}$

$s^2 = \frac{0.03}{3}$

$s^2 = 0.01$

<em>Hence, the variance is 0.01</em>

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