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AnnyKZ [126]
3 years ago
8

HELP OVER DUEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

Mathematics
2 answers:
jarptica [38.1K]3 years ago
6 0
1. 70
2. 22
3. 54.5
4. 38.5
5. 65.5
Fudgin [204]3 years ago
5 0
Answers
1)70
2)22
3)54.5
4)38.5
5)65.5
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Ken is thinking of a number. nine more than the product of 4 and the number is 73. Find Kens number.
rewona [7]
I got 16 for this one
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Define z_alpha to be a z-score with an area of alpha to the right. For Example: z_0.10 means P(Z > z_0.10) = 0.10. We would a
Reptile [31]

Answer:

a) P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

"=NORM.INV(0.025,0,1)"

And for this case the two values are :z_{crit}= \pm 1.96

b) P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates \alpha/2 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

"=NORM.INV(alpha/2,0,1)"

c) For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got z_{\alpha/2}=1.64

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Part a

P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

"=NORM.INV(0.025,0,1)"

And for this case the two values are :z_{crit}= \pm 1.96

Part b

P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates \alpha/2 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

"=NORM.INV(alpha/2,0,1)"

Part c

For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got z_{\alpha/2}=1.64

6 0
3 years ago
Chapter 2 Review
Ivan
It’s A
Because I just did it
4 0
3 years ago
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STatiana [176]

r= 9 because 45/5 is 9 and it works the same way for the rest of the problems.

5 0
3 years ago
At the beginning of the school year, 70 students signed up for music lessons. Students could choose drums, guitar, piano, or vio
Irina18 [472]

Answer:

D.

Step-by-step explanation:

1. Answer A:

15 students signed up for violin lessons (4 boys + 11 girls)

15 / 70 = approx. 21% of the class.

Answer A says 28%, so this is incorrect.

2. Answer B:

17 students signed up for drum lessons (11 boys + 6 girls)

17/70 = approx. 24%

B says 17% signed up for drums, so this is incorrect.

3. Answer C.

Wrong because more girls signed up for piano lessons (15 girls, whereas only 9 boys signed up for piano)

4. D. The same percent of boys and girls are taking guitar lessons.

Get the total number of boys and girls in the class (just add the results for each separately for all instruments)

Boys: 11 + 6 + 9 + 4 = 30

Girls: 6+8+15+11 = 40

6 out of 30 boys signed up for guitar

6/30= 20%

8 out of 40 girls signed up for guitar

8/40 = 20%

Answer is D

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3 years ago
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