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3 years ago

15

the mean on a advanced algebra test was 78 with a standard deviation of 8. if the test scores are normal distributed, find the i

nterval aboitthe mean that contains 99.7% of the scores.

1 answer:

Maksim231197 [3]3 years ago

6 0

Maybe 54-102 is the correct interval

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I will give brainlest if the answer is right Enter the missing numbers in the boxes to complete the table of equivalent ratios.

rjkz [21]

Answer:

2 , 10 , and 15 in order of missing spaces, up to down

Step-by-step explanation:

You find the ration 8/20 and find out if the ration is simplifiable and turn it to 2/5 which you multiply the rest of the answers too. Hope this answers your question :)

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4 years ago

Arthur is treating his friends to a meal at Chez Rich. The meals added up to $152.30. If he tips the waiter 15%, what is the amo

True [87]

152.30(0.15)= 22.845

22.845 rounded to the nearest cent is $22.85.

Therefore, the answer is C. Hope this helps!

8 0

3 years ago

Read 2 more answers

The minute hand of a clock is 9.5cm long. How far will the hand travel in one day, to the nearest meter

Luba_88 [7]

136.8 meter, or 137 meters

Work:

60 minutes x 24 hours =
1,440 minutes in a day

1,440 minutes x 9.5 cm =
13,680 centimeters

13,680 cm = 136.8 meters // 137

5 0

3 years ago

3.In 1995, the Educational Testing Service in Princeton, New Jersey (which administers SAT exam) re-centered the scores so that

Travka [436]

Answer:

38.22% probability that of 10 randomly selected students, lessthan four will be between 900 and 1,100

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they score between 900 and 1,100, or they score outside this range. The probability of a student scoring between 900 and 1,100 is independent from other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

About 40 percent of these students’ scores were between 900 and 1,100.

This means that $p = 0.4$ .

a)Based on this estimate, what is the probability that of 10 randomly selected students, lessthan four will be between 900 and 1,100?

This is $P(X < 4)$ when $n = 10$ . So

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0060$

$P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403$

$P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209$

$P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0060 + 0.0403 + 0.1209 + 0.2150 = 0.3822$

38.22% probability that of 10 randomly selected students, lessthan four will be between 900 and 1,100

8 0

4 years ago

In testing the difference between two population means using two independent samples, the sampling distribution of the sample me

Neko [114]

Answer:

C. populations are non normal and the sample sizes are large.

Step-by-step explanation:

To test hypotheses about the difference between two populations means we deal with the following three cases.

1) both the populations are normal with known standard deviations.

2)both the populations are normal with unknown standard deviations.

3) both the populations are non normal in which case both the sample sizes are necessarily large.

So option C is the correct answer.

4 0

4 years ago