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3 years ago

Help Needed Please!!! First the volume of the sphere

Mathematics

2 answers:

Kisachek [45]3 years ago

7 0

Answer:

$V=1767$

Step-by-step explanation:

Step 1: Find the volume of the sphere

Volume of a Sphere: $V=\frac{4}{3}\pi r^{3}$

Find the radius

$Radius = Diameter / 2$

$Radius = 15 / 2$

Plug in the values and solve

$V=\frac{4}{3}\pi * \frac{15}{2}^{3}$

$V=\frac{4}{3}\pi * \frac{3375}{8}$

$V=\frac{1125}{2}\pi$

$V=1767$

Answer: $V=1767$

satela [25.4K]3 years ago

5 0

Answer:

1766 using 3.14 for pi

1767 using the pi button

Step-by-step explanation:

The volume of a sphere is given by

V = 4/3 pi r^3

we need to know the radius

r =d/2 = 15/2 = 7.5

V = 4/3 (pi)( 7.5)^3

V =562.5 pi

Approximating pi by 3.14

V = 562.5 (3.14)

V =1766.25

To the nearest whole number

V = 1766

If you use the pi button

1767.14586

To the nearest whole number

1767

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Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

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2 years ago

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andrey2020 [161]

Answer:

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2 years ago

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Furkat [3]

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3 years ago

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3 years ago

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Harlamova29_29 [7]

Answer:

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