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Anon25 [30]
3 years ago
5

The diameter of a cylindrical water tank is 10 ft, and its height is 9 ft. What is the volume of the tank?

Mathematics
2 answers:
andreev551 [17]3 years ago
4 0

Answer:

225w

Step-by-step explanation:

volume formula: w*r²*9

w*5²*9

25*9w

225w

Ludmilka [50]3 years ago
4 0

Answer: V=707

Step-by-step explanation:

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How do you round the area of a circle with 3.14?
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Answer:

(This is a example since you don’t have anything to round to)

Example Answer if the radius is 4 - 50.24

Step-by-step explanation:

Example Radius : 4

Area : 3.14

We have the mulitiply 3.14 times The Example Radius 2 times because The diameter of a circle is 2 times its radius.

( 3.14 x 4 x 4) 50.24

If you know the diameter, it’s a half or 1/2 as large.

Therefore, if the radius was 4, it would be 50.24.

Youre welcome :)

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3 years ago
Samantha can run one mile in 8 minutes. At this rate how long will it take her to run in 5 miles
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It would take Samantha 40 minutes to run 5 miles!!!
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8 0
3 years ago
A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours a
natita [175]

Answer:

the standard deviation increases

Step-by-step explanation:

Let x₁ , x₂, .   .   .  , x₂₃ be the actual data observed by the student

The sample means  = x₁  +  x₂  +  .   .   .  , x₂₃ / 23

= \frac{x_1 +x_2 +...x_2_3}{23}

= 2.3hr

⇒\sum xi =2.3 \times 23 = 52.9hrs

let x₁ , x₂, .   .   .  , x₂₃  arranged in ascending order

Then x₂₃ was 10  and has been changed to 14

i.e x₂₃ increase to 4

Sample mean  = \frac{x_1 +x_2 +...x_2_3}{23}

\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47

therefore, the new sample mean is 2.47

2) For the old data set

the median is x_1_2(th) values

[\frac{n +1}{2} ]^t^h value

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = [{x_1_2]^t^h value remain unchange

sample median = 2.5hrs

The Standard deviation of old data set is calculated

=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)

The new sample standard sample deviation is calculated as

= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)

Now, when we compare (1) and (2)  the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)²  ∑ (xi - 2.47)²

Therefore , the standard deviation increases

6 0
3 years ago
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