$V=\dfrac{1}{3}\pi r^2H\\\\2r=22m\to r=11m;\ H=12m\\\\V=\dfrac{1}{3}\pi\cdot11^2\cdot12=\pi\cdot121\cdot4=484\pi\ (m^3)\\\\V\approx484\cdot3.14=1,519.76\approx1,519.8\ (m^3)\\\\Answer:1,520.5\ cm^3$

8 0

3 years ago

A chocolate factory produces mints that weigh 10 grams apiece. The standard deviation of the weight of a box of 10 mints is 3 gr

frozen [14]

The confidence that the box you bought did not come from the factory is 90 %.

<u>Step-by-step explanation:</u>

Since we have given that

Population Mean weight (\mu)= 10 grams a piece

Standard deviation of the weight of a box = 3 grams

Number of mints = 10

We need to buy a box of mints that weighs 95 grams.

Sample mean is given by

x = $\frac{95}{10}=9.5$ grams .

First we find out the standard error which is given by

$s=\frac{\sigma}{\sqrt{n}}\\\\=\frac{3}{\sqrt{10}}\\\\$ = 0.94868

Since it is normal distribution, so, we will find z-score.

$z=\frac{x-\mu}{s}\\\\z=\frac{9.5-10}{0.94868}\\\\z=-0.527\\\\$

z = - 0.53

The area to the left of a z-score of -0.53 = 0.29805.

So, it may be 90% or 95 % confidence.

For 95% confidence level,

$\alpha=\frac{1-0.95}{2}=0.025$

Similarly,

For 90% confidence level,

$\alpha=\frac{1-0.90}{2}=0.05$

The value is much smaller than 0.05.

So, we will get 90% confidence and the critical value = 1.645

Margin of error is given by

(Standard deviation) $\times$ (critical\ value) = 0.94868 $\times$ 1.645

=1.56

So, confidence interval will be

(10-1.56,10+1.56)

=(8.44,11.56)

4 0

3 years ago

How to solve this I am very confused and have a test tomorrow

Anastaziya [24]

4(x+3) = 16

First divide by 4 on both sides

x+3 = 4

Then subtract 3 from both sides

x = 1

Then check your work

4(1+3) = 16
4(4) = 16
16 = 16

Hope this helped! :)

3 0

3 years ago

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