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Musya8 [376]
3 years ago
10

SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You are interested in estimating the average SA

T score of first year students at your college. If you would like to limit the margin of error of your 98% confidence interval to 40 points, at least how many students should you sample?
Mathematics
1 answer:
kotegsom [21]3 years ago
3 0

Answer:

I should use at least 304 students

Step-by-step explanation:

Margin error (E) = t × sd/√n

E = 40

sd = 300

confidence level (C) = 98% = 0.98

significance level = 1 - C = 1 - 0.98 = 0.02 = 2%

t-value corresponding to 2% significance level and infinity degree of freedom is 2.326

n = (t×sd/E)^2 = (2.326×300/40)^2 = 17.445^2 = 304 (to the nearest integer)

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Which comparison is correct?<br><br> 7 -7<br> -9 &gt; 4<br> -6 &gt; -5
ArbitrLikvidat [17]

Answer:

7 > -7

Step-by-step explanation:

-9 is not greater than -4

-6 is not greater than -5

I am not sure what 7 -7 means in your first option but if it is consistent, then yes, 7 is greater than -7.

8 0
2 years ago
Mrs.Benite earns $3,400 every month.She spends 25% of her drawings on rent and 30% of the remainder on transportation.How much d
lutik1710 [3]
25% of 3,400 is 3,400/4=$850 spent rent
3,400-850= 2,550 remaining
30% of 2,550= 2550•0.3= $765 spent on transportation
4 0
3 years ago
A​ half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of​
Ainat [17]

Answer:

99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of​ today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of​ today's women in their 20's have been taken.

<u><em /></u>

<u><em>Let </em></u>\bar X<u><em> = sample mean heights</em></u>

The z-score probability distribution for sample mean is given by;

                          Z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean height of women = 64.7 inches

            \sigma = standard deviation = 2.07 inches

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the sample of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches is given by = P(\bar X \geq 65.86 inches)

  P(\bar X \geq 65.86 inches) = P( \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } } ) = P(Z \geq -2.57) = P(Z \leq 2.57)

                                                                        = <u>0.99492  or  99.5%</u>

<em>The above probability is calculated by looking at the value of x = 2.57 in the z table which has an area of 0.99492.</em>

<em />

Therefore, 99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

3 0
3 years ago
Factor.<br> 9y2 - 12y - 12 Please put it as (x+b) (x+b) form
Step2247 [10]

Answer:

3(3y+2)(y-2)

hope this helps

4 0
3 years ago
Read 2 more answers
If someone could help me figure out #7 ASAP I’d really appreciate it
IceJOKER [234]

Answer: Angle N

Explanation:

For triangle ABC, we refer to the middle angle, Angle B.

For triangle MNP, we refer to its middle angle, Angle N.

Good luck!

8 0
2 years ago
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