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soldi70 [24.7K]
3 years ago
15

If 45.78 ÷ 1.7 is written in long division form as long division set up where 17 is on the outside of the division symbol and 45

780 is on the inside so that the divisor is written as a whole number, where should the decimal point be placed in the dividend?
Mathematics
1 answer:
morpeh [17]3 years ago
5 0

Answer:

the divided will be 457.80 and divisor will be 17

Step-by-step explanation:

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The absolute value parent function.
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Which of the following are not congruence theorems for right triangles? Check all that apply.
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Answer:

c) HH E)AA

Step-by-step explanation:

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There were 800 math instructors at a mathematics convention. Forty instructors were randomly selected and given an IQ test. The
Leya [2.2K]

Answer:

95% confidence interval for the mean of the 800 instructors is [126.80 , 133.20].

Step-by-step explanation:

We are given that there were 800 math instructors at a mathematics convention.

Forty instructors were randomly selected and given an IQ test. The scores produced a mean of 130 with a standard deviation of 10.

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 mean score = 130

             s = sample standard deviation = 10

             n = sample of instructors = 40

             \mu = population mean of 800 instructors

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

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

P(-2.0225 < t_3_9 < 2.0225) = 0.95  {As the critical value of t at 39 degree of

                                         freedom are -2.0225 & 2.0225 with P = 2.5%}  

P(-2.0225 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.0225) = 0.95

P( -2.0225 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.0225 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.0225 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.0225 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u><em /></u>

<u><em>95% confidence interval for</em></u> \mu = [ \bar X-2.0225 \times {\frac{s}{\sqrt{n} } } , \bar X+2.0225 \times {\frac{s}{\sqrt{n} } } ]

                   = [ 130-2.0225 \times {\frac{10}{\sqrt{40} } } , 130+2.0225 \times {\frac{10}{\sqrt{40} } } ]

                   = [126.80 , 133.20]

Therefore, 95% confidence interval for the mean of the 800 instructors is [126.80 , 133.20].

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What are the factors for 24, 64 and 88
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<span>
The common factors of 24, 64, and 88 are: 1, 2, 4, and 8.</span>
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The answer is D, 8, that's what I got
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