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3 years ago

) Simplify the expression.

Mathematics

1 answer:

alexira [117]3 years ago

4 0

Answer:

-15/4 or -3.75

Step-by-step explanation:

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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

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.

3 0

3 years ago

Arte-miy333 [17]

Answer:

Yes, it is a function because domain ( x value) is not repeated

Domain: {7, 1, -1, 0}

Range: { 1, 1, 7, -1}

6 0

3 years ago

A survey of 46 college athletes found that 24 played volleyball, while 22 played basketball. a) If we pick one athlete survey pa

PSYCHO15rus [73]

Answer:

A) $\dfrac{11}{23}$

B) $\dfrac{88}{345}$

Step-by-step explanation:

A survey of 46 college athletes found that

24 played volleyball,
22 played basketball.

A) If we pick one athlete survey participant at random, the probability they play basketball is

$P_1=\dfrac{22}{46}=\dfrac{11}{23}$

B) If we pick 2 athletes at random (without replacement),

the probability we get one volleyball player is $\dfrac{24}{46}=\dfrac{12}{23};$
the probability we get another basketball player is $\dfrac{22}{45}$ (only 45 athletes left).