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Scrat [10]
3 years ago
11

Which ratio forms a proportion with 25/35

Mathematics
1 answer:
Art [367]3 years ago
4 0
5/7 is your answer

25/5 = 5
35/5 = 35

5:7 is proportionate to 25:35

hope this helps
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Please help me with this question below
BabaBlast [244]

Answer:

381 cubic inches

Step-by-step explanation:

To find the volume of the cube, use the volume formula for a rectangular prism V = l*w*h and for a triangular prism V = 1/2 l*w*h.

<u>Rectangular prism</u>

Substitute l  = 16, w = 7, and h = 3.

V = 16*7*3 =  336 cubic inches

<u>Triangular prism</u>

Substitute l = 5, w = 3 and h = 6.

V = 1/2*5*3*6 = 45 cubic inches

Together the volume will be 336 + 45 = 381.

4 0
4 years ago
The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this information, find a 99% confide
kondor19780726 [428]

Answer:

99% confidence interval for the population standard deviation = (74.97 , 635.20).

Step-by-step explanation:

We are given that the ages of a random sample of five university professors are 39, 54, 61, 72 and 59. Also, it is provided that the ages of university professors are normally distributed.

So, firstly the pivotal quantity for 99% confidence interval for the population standard deviation is given by;

         P.Q. = \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2} __n_-_1

where, s = sample standard deviation

            \sigma = population standard deviation

            n = sample of university professors = 5

Also, s^{2} = \frac{\sum (X-\bar X)^{2} }{n-1} = 144.5

So, 99% confidence interval for population standard deviation,\sigma is;

P(0.2070 < \chi^{2} __5_-_1 < 14.86) = 0.99 {As the table of \chi^{2} at 4 degree of freedom

                                                      gives critical values of 0.2070 & 14.86}

P(0.2070 < \frac{(n-1)s^{2} }{\sigma^{2} } < 14.86) = 0.99

P( \frac{ 0.2070}{(n-1)s^{2} } < \frac{1 }{\sigma^{2} } < \frac{ 14.86}{(n-1)s^{2} } ) = 0.99

P(\frac{ (n-1)s^{2}}{14.86 } < \sigma^{2} < \frac{ (n-1)s^{2}}{0.2070 } ) = 0.99

99% confidence interval for \sigma^{2} = ( \frac{ (n-1)s^{2}}{14.86 } , \frac{ (n-1)s^{2}}{0.2070 } )

                                                   = ( \frac{ (5-1) \times 144.5^{2}}{14.86 } , \frac{ (5-1) \times 144.5^{2}}{0.2070 } )

                                                   = (5620.525 , 403483.092)

99% confidence interval for \sigma = ( \sqrt{5620.525} , \sqrt{403483.092} )

                                                  = (74.97 , 635.20)

Therefore, 99% confidence interval for the population standard deviation of the ages of all professors at the university is (74.97 , 635.20).

8 0
4 years ago
Quien quiere puntos:3
ratelena [41]

Answer:

?

Step-by-step explanation:

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4 years ago
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I believe it to be 1 over 3 as 1 as the numerator and 3 as the denominator
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