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3 years ago

Some one plz help me with this math qustion

Mathematics

2 answers:

madreJ [45]3 years ago

5 0

The answer to your question is A. 40%

Valentin [98]3 years ago

4 0

Answer is A 40 hope this helps

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At a certain university, the average attendance at basketball games has been 3125. Due to the dismal showing of the team this ye

Neporo4naja [7]

Answer:

The test statistic needed to evaluate the claim is t = -1.08.

Step-by-step explanation:

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the expected value of the mean, s is the standard deviation of the sample and n is the size of the sample.

At a certain university, the average attendance at basketball games has been 3125. The athletic director claims that the attendance is the same as last year.

This means that $\mu = 3125$

Due to the dismal showing of the team this year, the attendance for the first 9 games has averaged only 2915 with a standard deviation of 585.

This means that $n = 9, X = 2915, s = 585$

What is the test statistic needed to evaluate the claim?

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{2915 - 3125}{\frac{585}{\sqrt{9}}}$

$t = -1.08$

The test statistic needed to evaluate the claim is t = -1.08.

6 0

3 years ago

Assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. Based on this assumption,

kompoz [17]

If the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder, then its volume is

$V_{flask}=V_{sphere}+V_{cylinder}.$

Use following formulas to determine volumes of sphere and cylinder:

$V_{sphere}=\dfrac{4}{3}\pi R^3,\\ \\V_{cylinder}=\pi r^2h,$

wher R is sphere's radius, r - radius of cylinder's base and h - height of cylinder.

Then

$V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;$
$V_{cylinder}=\pi r^2h=\pi \cdot \left(\dfrac{1}{2}\right)^2\cdot 3=\dfrac{3\pi}{4}\approx 2.36;$
$V_{flask}=V_{sphere}+V_{cylinder}\approx 47.71+2.36=50.07.$

Answer 1: correct choice is C.

If both the sphere and the cylinder are dilated by a scale factor of 2, then all dimensions of the sphere and the cylinder are dilated by a scale factor of 2. So

R'=2R, r'=2r, h'=2h.

Write the new fask volume:

$V_{\text{new flask}}=V_{\text{new sphere}}+V_{\text{new cylinder}}=\dfrac{4}{3}\pi R'^3+\pi r'^2h'=\dfrac{4}{3}\pi (2R)^3+\pi (2r)^2\cdot 2h=\dfrac{4}{3}\pi 8R^3+\pi \cdot 4r^2\cdot 2h=8\left(\dfrac{4}{3}\pi R^3+\pi r^2h\right)=8V_{flask}.$