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arlik [135]
4 years ago
8

A cotton candy holder is shaped like a cone. The height is 12 in., and the diameter is 12 in. What is the volume of the holder?

Use 3.14 to approximate pi, and express your final answer to the nearest tenth.. ______in3
Mathematics
2 answers:
babymother [125]4 years ago
6 0
The formula for the volume of a cone is
V=\frac{1}3\pi r^2h
(you can remember it like this: it's literally just 1/3 the volume of a cylinder, just as a pyramid is 1/3 the volume of a prism)

We know that the diameter is 12...the radius is always half the diameter, so 6.
The height is 12.
Let's plug these into our volume formula.

V=\frac{1}3\pi \times6^2\times12
V=\frac{1}3\pi \times36\times12
V=\frac{1}3\pi \times432
V=144\times \pi
V\approx144\times3.14
\boxed{V\approx452.16\ in^3}
alexgriva [62]4 years ago
6 0

that answer was close but I just took the test and its 452.2

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Keith_Richards [23]
1/2 = 8/16 . . . so there are (<u><em>8</em></u>) - 1/16 pounds servings of chocolate in a 1/2 pound chocolate bar
6 0
3 years ago
A class consisting of 4 graduate and 12 undergraduate students is random divided into 4 groups of 4. what is the probability tha
MrRissso [65]
Total = 16
Graduates = 4

\frac{4}{16}  =  \frac{1}{4}
\frac{1}{4}  \div 4  \\ =  \frac{1}{4}  \times  \frac{1}{4}  \\  =  \frac{1}{16}
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Rounding off to 2 decimal places,
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Hope this helps. - M
5 0
3 years ago
An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberr
seraphim [82]

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean compression strength from the extra carbonation of strawberry drink and μ2 be the mean compression strength from the extra carbonation of cola.

The random variable is μ1 - μ2 = difference in the mean compression strength from the extra carbonation of strawberry drink and the mean compression strength from the extra carbonation of cola.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 < μ2 H1 : μ1 - μ2 < 0

This is a left tailed test.

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

x1 = 530

x2 = 553

s1 = 23

s2 = 16

n1 = 10

n2 = 10

t = (530 - 533)/√(23²/10 + 16²/10)

t = - 2.6

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [23²/10 + 16²/10]²/[(1/10 - 1)(23²/10)² + (1/10 - 1)(16²/10)²] = 6162.25/383.752

df = 16

We would determine the probability value from the t test calculator. It becomes

p value = 0.0097

Since alpha, 0.05 > than the p value, 0.0097, then we would reject the null hypothesis. Therefore, at 5% significance level, the data suggests that the extra carbonation of cola results in a higher average compression strength.

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