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3 years ago

(-5z)^3 using only positove exponents

Mathematics

1 answer:

Elena L [17]3 years ago

8 0

Answer:

Shown below

Step-by-step explanation:

First allocate the power of three to every number

$-5^{3} * z^{3}$ = $-125 * z^{3}$

not sure what you mean by positive exponents, the exponents already are positive. Maybe you mean positive exponents and integers?

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3 years ago

Mathematics achievement test scores for 300 students were found to have a mean and a variance equal to 600 and 3600, respectivel

Zina [86]

Answer:

(a) Approximately 205 students scored between 540 and 660.

(b) Approximately 287 students scored between 480 and 720.

Step-by-step explanation:

A mound-shaped distribution is a normal distribution since the shape of a normal curve is mound-shaped.

Let <em>X</em> = test score of a student.

It is provided that $X\sim N(\mu = 600, \sigma^{2} = 3600)$ .

(a)

The probability of scores between 540 and 660 as follows:

$P(540\leq X\leq 660)=P(\frac{540-600}{\sqrt{3600} }\leq \frac{X-600}{\sqrt{3600} }\leq \frac{660-600}{\sqrt{3600} })\\=P(-1 \leq Z\leq 1)\\= P(Z\leq 1)-P(Z\leq -1)\\=0.8413-0.1587\\=0.6826$

Use the standard normal table for the probabilities.

The number of students who scored between 540 and 660 is:

300 × 0.6826 = 204.78 ≈ 205

Thus, approximately 205 students scored between 540 and 660.

(b)

The probability of scores between 480 and 720 as follows:

$P(480\leq X\leq 720)=P(\frac{480-600}{\sqrt{3600} }\leq \frac{X-600}{\sqrt{3600} }\leq \frac{720-600}{\sqrt{3600} })\\=P(-2 \leq Z\leq 2)\\= P(Z\leq 2)-P(Z\leq -2)\\=0.9772-0.0228\\=0.9544$

Use the standard normal table for the probabilities.

The number of students who scored between 480 and 720 is:

300 × 0.9544 = 286.32 ≈ 287

Thus, approximately 287 students scored between 480 and 720.

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3 years ago