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Oksana_A [137]
3 years ago
5

Giving brainlist to whoever answers

Mathematics
1 answer:
just olya [345]3 years ago
6 0

Answer:

False

Step-by-step explanation:

V=bhw\\V=7.5*13*2\\V=195

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It would be 295 square feet.
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The box plots below show student grades on the most recent exam compared to overall grades in the class:
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<span>he box plots below show attendance at a local movie theater and high school basketball games: two box plots shown. The top one is labeled Movies. Minimum at 60, Q1 at 65, median at 95, Q3 at 125, maximum at 150. The bottom box plot is labeled Basketball games. Minimum at 90, Q1 at 95, median at 125, Q3 at 145, maximum at 150. Which of the following best describes how to measure the spread of the data? The IQR is a better measure of spread for movies than it is for basketball games. The standard deviation is a better measure of spread for movies than it is for basketball games. The IQR is the best measurement of spread for games and movies. The standard deviation is the best measurement of spread for games and movies.</span>

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How do you know if a whole number is a perfect square
kirill [66]

Answer:

you simply think of a number that when multiplied by itself it will give you that whole number

Step-by-step explanation:

let's take for instance the number 4 it's a whole number and still a perfect square because when you think of a number from 1to 10, which number when you multiply by itself will give you four apart from 2

4 0
3 years ago
What’s the value of x? and what’s the measure of angel JHK?
enot [183]

Answer:

x = 14

JHK =  21

Step-by-step explanation:

The angles are vertical angles and vertical angles are equal

3x-21 = x+7

Subtract x from each side

3x-x -21 = x+7-x

2x-21 = 7

Add 21 to each side

2x-21+21 = 7+21

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x = 14

JHK = 3x-21 = 3(14) -21 = 42-21 = 21

3 0
2 years ago
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For a quality control test, a company checks the miles per gallon (mpg) for a simple random sample of 100 minivans drawn from it
amm1812

Answer:

95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].

Step-by-step explanation:

We are given that a company checks the miles per gallon (mpg) for a simple random sample of 100 minivans drawn from its current inventory.

The average mpg of these 100 minivans is 15.6 with a standard deviation of 1.9 mpg.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

                         P.Q. = \frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average mpg = 15.6 mpg

             s = sample standard deviation = 1.9 mpg

            n = sample of minivans = 100

            \mu = population average mpg

<em>Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.</em>

So, 95% confidence interval for the population mean, \mu is ;

P(-1.987 < t_9_9 < 1.987) = 0.95  {As the critical value of t at 99 degree

                                         of freedom are -1.987 & 1.987 with P = 2.5%}  

P(-1.987 < \frac{\bar X -\mu}{\frac{s}{\sqrt{n} } } < 1.987) = 0.95

P( -1.987 \times {\frac{s}{\sqrt{n} } } < {\bar X -\mu} < 1.987 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-1.987 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.987 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-1.987 \times {\frac{s}{\sqrt{n} } } , \bar X+1.987 \times {\frac{s}{\sqrt{n} } } ]

                                      = [ 15.6-1.987 \times {\frac{1.9}{\sqrt{100} } } , 15.6+1.987 \times {\frac{1.9}{\sqrt{100} } } ]

                                      = [15.22 mpg , 15.98 mpg]

Therefore, 95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].

It is appropriate to compute a confidence interval for this problem using the Normal curve as t test statistics is used when data should follow normal distribution.

7 0
3 years ago
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