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Natasha_Volkova [10]
2 years ago
9

Khan Academy Volume of Sphere Answers!!!

Mathematics
1 answer:
agasfer [191]2 years ago
5 0

Explanation:

The N stands for Pi so when you type in the answers to Khan Academy use the Pi symbol when theres an N! These are the common questions and common radius Khan Academy gives when they ask to find the radius of the circle.

Answers:

For 4 - 267.95  

For 1/4 - 1/48n  

For 1 - 4/3n    

For 9 - 972n

For 7 - 1372/3n

For 6 - 288n

For 3 - 36n

For 10 - 4000/3n

For 8 - 2048/3n

For 1/2 - 1/6n

For 5 - 500/3n

For 2 - 33.49

Step-by-step explanation:

<u><em>I did the go math so all of these rights and the majority of the answers are the questions Khan Academy will ask you to find the radius of the circle.</em></u>

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Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):
Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, \mu_X represent the population mean for Brand B and let \mu_Y represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

        P.Q. = \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } ~ t_n__1+n_2-2

where, Xbar = Sample mean for Brand B data = 36.9

            Ybar = Sample mean for Brand A data = 32.9

              n_1  = Sample size for Brand B data = 13

              n_2 = Sample size for Brand A data = 9

              s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2}  }{n_1+n_2-2} } = \sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} } = 3.013

Here, s^{2}_X and s^{2} _Y are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < t_2_0 < 2.528) = 0.98

P(-2.528 < \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } < 2.528) = 0.98

P(-2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (Xbar -Ybar) -(\mu_X-\mu_Y) < 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

P( (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (\mu_X-\mu_Y) < (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} , (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ]

[ (36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} , (36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

                     

4 0
3 years ago
What is the average of 17/15
blondinia [14]
The average of 17 and 15 is 16 because if you add 17 and 15 together, and you divide them by 2, you get 16.
4 0
3 years ago
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HELP ASAP THIS IS DUE IN AN HOUR ILL MARK U BRIANLY
AlexFokin [52]

Answer:

5.8 plus 5.1 plus 2

final answer is 11.9

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Alex wants to paint the four walls of her new office. The room has dimensions 18 feet by
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Answer:

58

Step-by-step explanation:

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Solve the equation and check the solution4.4x-4=3.4x
PolarNik [594]
4.4x-4=3.4x

Substract '-3.4x' at LHS nad the RHS of the above expression.

\begin{gathered} 4.4x-4-3.4x=3.4x-3.4x \\ 4.4x-3.4x-4=0 \\ x-4=0 \end{gathered}

Add '4' on both LHS and RHs of the above expression.

undefined

4 0
1 year ago
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