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3 years ago

Create a model to show that 2(2x+3y)=4x+6y

1 answer:

4 0

You would first do 2 multiplied by 2x and that would give you 4x, then you would do 2 multiplied by 3y which would give you 6y, you would put 4x under 2x and put 6y under 3y and bring down your addition symbol

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A physical education teacher wanted to test whether her county’s students are falling below the national proportion. It is known

Ipatiy [6.2K]

Answer:

(a) The population parameter is students who ran the mile in less than 10 minutes.

(b) The hypothesis being tested is:

H0: p = 0.70

Ha: p < 0.70

(d) In order to conduct a one-sample proportion z-test, the following conditions should be met:

The data are a simple random sample from the population of interest.

The population is at least 10 times as large as the sample.

n⋅p ≥ 10 and n⋅(1−p) ≥ 10, where n is the sample size and p is the true population proportion.

n⋅p = 756 * 0.70 ≥ 10

n⋅(1−p) = 756 * (1 - 0.70) ≥ 10

All the conditions are met.

(e) p = 504/756 = 0.67

The test statistic, z = (p - p)/√p(1-p)/n = (0.67 - 0.70)/√0.70(1-0.70)/756 = -2.00

(f) The p-value is 0.0228.

(g) Since the p-value (0.0228) is less than the significance level (0.05), we can reject the null hypothesis.

(h) Therefore, we can conclude that less than 70% of students run one mile in less than 10 minutes.

Step-by-step explanation:

7 0

3 years ago

32 small bags of soil are needed to fill three clay pots how many small bags of soil should be purchased to fill 8 clay pots

lapo4ka [179]

Answer:

i think

Step-by-step explanation:

4 0

3 years ago

Pls only put the right answer

Makovka662 [10]

Answer:

100°

Step-by-step explanation:

No explanation, by your request.

3 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

Hi, so i have a question and i'm giving away free points so here's the question! If you answer ill put out another question and

Shtirlitz [24]

Answer:

1. Saturday

2(a) 10,944

2(b) 32700

3. (16,24)

Step-by-step explanation:

8 0

3 years ago