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Bezzdna [24]
4 years ago
9

What is ​ 7/10 ​ expressed as a decimal? Enter your answer in the box.

Mathematics
1 answer:
ki77a [65]4 years ago
6 0
Dividing a number by 10 moves the decimal over one place to the left.  Knowing this:

7 ÷ 10 = 0.7
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Help mee!!! I am so confused. I need help ASAP
Hatshy [7]

x=5, y=0. (x,y)=(5,0).

3 0
3 years ago
What is an example of parallel or perpendicular lines in every day life​
Olegator [25]

Answer: Road lines

Step-by-step explanation:

Road lines are parallel lines as they are always the same distance from each other, and never intersect.

6 0
3 years ago
The hawks lost eighty three games they won 14 more than they lost. how many games did they win?
Daniel [21]
The hawks won 97 games 
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3 years ago
Read 2 more answers
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
I need help with this equation -11(-8×-15) my teacher said that I need to show my work pls help me.​
Zielflug [23.3K]

First calculate (-8x-15)=120

Then multiply - 11x120

=-1320

3 0
3 years ago
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