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RideAnS [48]
3 years ago
5

Check all the statements that are true:

Mathematics
1 answer:
Dahasolnce [82]3 years ago
4 0

Answer:

The true statements are;

A, B, G, H, I, J

Step-by-step explanation:

To answer the question, we test each option as follows

A. If a and b both divide c, then ab divides c².

The above statement is true as c/a exists,

c/b exits therefore c²/ab = c/a×c/b

B. If p and q are distinct primes, then p2q2 has exactly 11 positive divisors.

The above statement is true as p² and q² each have 3 positive divisors, therefore, p²q² will also have pq and p²q² as possible divisors, therefore, true

C. If p and q are distinct primes, then p+q is prime as well.

The above statement is not correct as 5 + 7 = 12 an even number

D. If a divides b and c divides d, then a+c divides b+d.

The above statement is not correct as

8 is divisible by 2 and

9 is divisible by 3

but 17 is not divisible by 5

E. If p is prime, then so is p+2.

The above statement is not correct as 7 + 2 = 9 which is divisible by 3

G. If a and b both divide c, and a and b are relatively prime, then ab divides c.

The above statement is true as both a and b are factors of c

H. There are infinitely many prime numbers.

The above statement is true as there are infinitely many numbers

I. If p is prime, then p2 has exactly 3 positive divisors.

The above statement is true

1, p and p²

J. There are three consecutive odd numbers that are prime.

The above statement is true

3, 5, 7.

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Answer:

The margin of error is 6.45.

Step-by-step explanation:

The complete question is:

As an early intervention effort, a school psychologist wants to estimate the average score on the Stanford-Binet Intelligence Scale for all students with a specific type of learning disorder using a simple random sample of 36 students with the disorder.

Determine the margin of error, of a 99% confidence interval for the mean IQ score of all students with the disorder. Assume that the standard deviation IQ score among the population of all students with the disorder is the same as the standard deviation of IQ score for the general population, σ = 15 points.

The (1 - <em>α</em>)% confidence interval for population mean <em>μ</em> is:

CI=\bar x\pm z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}

The margin of error for this interval is:

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Given:

<em>n</em> = 36

σ = 15

(1 - <em>α</em>)% = 99%

Compute the critical value of <em>z</em> for 99% confidence level as follows:

z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58

*Use a <em>z</em>-table.

Compute the value of MOE as follows:

MOE= z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}

          =2.58\times\frac{15}{\sqrt{36}}

          =2.58\times 2.5\\=6.45

Thus, the margin of error is 6.45.

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