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zysi [14]
3 years ago
12

HELPPPPPPPPP A cylinder has a radius of 24 m and a height of 9 m. What is the exact volume of the cylinder? Question 1 options:

972π m3 1296π m3 108π m3 216π m3
Mathematics
2 answers:
7nadin3 [17]3 years ago
5 0

Answer:

VOLUME =πR^2H

π x 24 x 24 x 9

5184π= 5184 x 22/7=16292.5714

your options are wrong!

s2008m [1.1K]3 years ago
3 0

Answer: 1,296π if that radius is actually the diameter? Otherwise the answer is 5,184π which isn't listed.

Step-by-step explanation: Volume of a cylinder is πr^2h. This is because we're solving for the area of the circle, then extending it upwards. So we square 24 and multiply by 9 then by pi. This gives us an answer that isn't an option so I'm assuming radius is incorrect. Dividing by 2 gives us 144π9 = 1296pi.

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Help will give 20 points
siniylev [52]

Answer:1/3 divided by 6 equals 18

Step-by-step explanation: If you have a cloth that is 1/3 long then divide it by 6 it would look like 1/3 divided by 1/6 which would then equal 1/18.

6 0
3 years ago
Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
tigry1 [53]

Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

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3 years ago
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elena-s [515]
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3 0
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