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3 years ago

Can anyone help finish this assignment I need help on 3456and 8 pleaseee

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

4 0

Answer:

Below, are the answers.

Step-by-step explanation:

1. x < -2 or x > 0

2. x < 0 and x > -2 transforms to ==> -2 < x < 0

3. x ≤ -1 or x ≥ 3

4. -3 ≤ x ≤ 1

5. x ≤ 0 or x ≥ 4

6. -5 ≤ x ≤ 1

7. 3 < x < 5

8. x ≤ -1 or x ≥ 2

9. x < -2 or x > $\frac{2}{3}$

Hope that helps :)

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3 years ago

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A cylinder has a height of 16 and a radius of . A sphere has a radius of 2. The volume of the cylinder and the sphere are equal.

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2 years ago

The mean MCAT score 29.5. Suppose that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.

Paul [167]

Answer:

We conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

Step-by-step explanation:

We are given that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.2 with a standard deviation of 4.2.

Let $\mu$ = population mean score

So, Null Hypothesis, $H_0$ : $\mu \leq$ 29.5 {means that the students that took the Kaplan tutoring have a mean score less than or equal to 29.5}

Alternate Hypothesis, $H_A$ : $\mu$ > 29.5 {means that the students that took the Kaplan tutoring have a mean score greater than 29.5}

The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean MCAT score = 32.2

s = sample standard deviation = 4.2

n = sample of students = 40

So, the test statistics = $\frac{32.2-29.5}{\frac{4.2}{\sqrt{40} } }$ ~ $t_3_9$

= 4.066

The value of t-test statistics is 4.066.

Now, at 0.05 level of significance, the t table gives a critical value of 1.685 at 39 degrees of freedom for the right-tailed test.

Since the value of our test statistics is more than the critical value of t as 4.066 > 1.685, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

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3 years ago

The radius of a circle is 8 feet. What is the area of sector bounded by a 90° arc?

Lera25 [3.4K]

Since 90 degree is 1/4 of a circle, the area will be 1/4 th area of the full circle.

The radius is given by, r = 8 ft, therefore, the area of the full circle is,

$\begin{gathered} A=\pi\times r^2 \\ A=3.14\times8^2=200.96ft^2 \end{gathered}$

Therefore, the area of the sector is,

$A^{\prime}=\frac{1}{4}\times200.96ft^2=50.24ft^2$ $\begin{gathered} A^{\prime}=\frac{\theta}{360}\times\pi\times r^2 \\ \text{ =}\frac{\text{90}}{360}\times\pi\times8^2 \\ \text{ =}\frac{1}{4}\pi\times64=16\pi \end{gathered}$

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2 years ago