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2 years ago

What is the lateral surface area of this cylinder? r = 7ft h = 16ft

Mathematics

1 answer:

Stolb23 [73]2 years ago

7 0

Answer:

703.36 sq ft

Step-by-step explanation:

The lateral area is the product of the circumference and the height.

LA = 2π·r·h = 2·3.14·(7 ft)(16 ft) = 703.36 ft²

_____

Comment on the answer choices

Please note this is not the actual area, which rounds to 703.72 ft². The above value is obtained only by using an inappropriate value for π. The 5-significant-digit answer is not supported by a 3-significant-digit value for π. More appropriate would be π≈3.1416.

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Three guinea pigs living in the same cage need at least 12 square feet of living spàce.Is a cage that measure 3 feet by 5 feet b

Arte-miy333 [17]

Yes, there is enough space . If the guinea pigs need 12 feet, and 5*3= 15 feet then they have enough.
( idk if means 12 feet of living space for each or all 3 , but here i hope it helps :) )

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2 years ago

What is the perimeter of the quadrilateral?

olasank [31]

Answer:

24.6

Step-by-step explanation:

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2 years ago

Read 2 more answers

You borrowed $59,000 for 2 years at 11% which was compounded annually. What

gayaneshka [121]

Answer:

$72693.9

Step-by-step explanation:

To get this answer you need to use the compound interest formula, which will be A=P(1+r/n)^n(t). P=59,000 r=11%=0.11 n=1 (annually) t=2 years. From there you should be able to figure the rest out and get the answer. Hope this helps!

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2 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

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1 year ago

Find the median of the data

kicyunya [14]

The median is the middle number of the data plot. Since there 30 numbers, there isn't a exact middle. To find the median you must add the two middle number then divide by 2.
In this case you must add 87 and 88 which gives 175.
Then divide 175 by 2 which gives 87.5
87.5 is the median

5 0

2 years ago