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sveta [45]
3 years ago
11

The sphere at the right fits snugly inside a cube with 8​-in. edges. What is the approximate volume of the space between the sph

ere and​ cube?
Mathematics
1 answer:
boyakko [2]3 years ago
5 0

Answer:the answer is

Find the vol of the sphere:

radius of the sphere = 3 in

V = %284%2F3%29%2Api%2A3%5E3

V = 97.858

:

Find the vol between the sphere and cube:

216 - 97.858 = 118.142 cu/in

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The length of the window is 2m longer than its width and the area is 32m^
eimsori [14]

Answer:

X¹= -2 + 2√33

X²= -2 - 2√33

Step-by-step explanation:

X × (X+2) = 32

X² + 2X = 32

X² + 2X - 32= 0

Δ=132

X= -2 ± 2√33

X¹= -2 + 2√33

X²= -2 - 2√33

(You can change the 2√33 to 11,5 if you want to)

4 0
3 years ago
50 pts!!!! Luciana's laptop has 3,000 pictures. The size of the pictures is skewed to the right, with a mean of 3.7MB and a stan
skelet666 [1.2K]

Answer:

<u></u>

  • <u>Part A: No, you cannot.</u>

<u></u>

  • <u>Part B: 0.4491</u>

Explanation:

<u>Part A:</u>

The mean of samples of symmetrical (bell shaped) distributions follow a normal distribution pattern.

Thus, for symmetrical distributions you can use the z-score tables to make calculations that permit calculate probabilities for particular values.

For <em>skewed </em>distributions, in general, the samples do not follow a normal distribution pattern, except that the samples are large.

Since <em>a sample of 20 pictures</em> is not large enough, the answer to this question is negative: <em>you cannot accurately calculate the probability that the mean picture size is more than 3.8MB for an SRS (skewed right sample) of 20 pictures.</em>

<u>Part B.</u>

For large samples, the<em> Central Limit Theorem</em> will let you work with samples from skewed distributions.

Although the distribution of a population is skewed, the <em>Central Limit Theorem</em> states that  large samples follow a normal distribution shape.

It is accepted that samples of 30 data is large enough to use the <em>Central Limit Theorem.</em>

Hence, you can use the z-core tables for standard normal distributions to calculate the probabilities for a <em>random sample of 60 pictures</em> instead of 20.

The z-score is calculated with the formula:

  • z-score = (value - mean/ (standard deviation).
  • z=(x-\mu )/\sigma

Here, mean = 3.7MB, value = 3.8MB, and standard deviation = 0.78MB.

Thus:

         z=(3.7M-3.8MB)/(0.78MB)=-0.128

Then, you must use the z-score table to find the probability that the z-score is greater than - 0.128.

There are tables that show the cumulative probability in the right end and tables that show the cumulative probability in the left end of a normal standard distribution .

The probability that a z-score is greater than -0.128 is taken directly from a table with the cumulative probability in the left end. It is 0.4491.

4 0
3 years ago
1/3 of 9 is 3<br> How can you figure out these problems
gtnhenbr [62]
I don’t see the problems you want me to solve. Sorry
6 0
3 years ago
Savannah runs a day care center. Of the 12 children at the day care center, 3 of them are
sertanlavr [38]

Answer:

\frac{1}{4}

Step-by-step explanation:

3 out of 12 children at the day care center are 5-year-olds. So, the probability of a randomly selected child being a five-year old is \frac{3}{12}, or \frac{1}{4}.

6 0
3 years ago
Help please!! How do I solve this
egoroff_w [7]
This is an exponential equation. We will solve in the following way. I do not have special symbols, functions and factors, so I work in this way
 2 on (2x) - 5 2 on x + 4=0 =>. (2 on x)2 - 5 2 on x + 4=0  We will replace expression ( 2 on x) with variable t => 2 on x=t  =. t2-5t+4=0 => This is quadratic equation and I solve this in the folowing way => t2-4t-t+4=0 =>     t(t-4) - (t-4)=0 => (t-4) (t-1)=0 => we conclude t-4=0 or t-1=0 => t'=4 and t"=1 now we will return t' => 2 on x' = 4 => 2 on x' = 2 on 2 => x'=2 we do the same with t" => 2 on x" = 1 => 2 on x' = 2 on 0 => x" = 0 ( we know that every number on 0 gives 1). Check 1: 2 on (2*2)-5*2 on 2 +4=0 =>                   2 on 4 - 5 * 4+4=0 => 16-20+4=0 =. 0=0 Identity proving solution. 
Check 2:   2 on (2*0) - 5* 2 on 0 + 4=0 => 2 on 0 - 5 * 1 + 4=0 => 
1-5+4=0 => 0=0  Identity provin solution.
5 0
3 years ago
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