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dexar [7]
3 years ago
8

Solve for a. a+|-23+6|=-32 Plz help

Mathematics
2 answers:
prisoha [69]3 years ago
5 0

Answer:

a = -49

Step-by-step explanation:

a+|-23+6|=-32

Simplify the absolute value term

a+|-17|=-32

a+17=-32

Subtract 17 from each side

a +17-17 = -32-17

a = -49

Serga [27]3 years ago
3 0

<u>Answer:</u>

a = −49

<u>Step-by-step explanation:</u>

a + ∣ − 17∣ = −32

a + 17 = −32

a = −32 − 17

<u>Simplify.</u>

-32 - 17 to -49.

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

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(a)

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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

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