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Murrr4er [49]
3 years ago
8

Holly has $125 she spends $35 on gas for her car which model represents how much money holly has left idk if the answer is right

Mathematics
1 answer:
inn [45]3 years ago
5 0

Answer:

The second one would be the answer.

Step-by-step explanation:

If you look at the model, it shows it on 125 going down to 90.

If she has $125 and she spends $35 on gas, you would subtract that from how much money she has.

125 - 35 = 90.

Feel free to let me know if you need more help. :)

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What is the ratio cos C?
OLga [1]

Answer:

Roughly, sinC = 15/17

=> AB = 15, BC = 17 => AC = sqrt(17^2-15^2) = 8

=> cos C = AC/BC = 8/17

7 0
4 years ago
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1. The numbers ______ and _____ are respectively the additive and multiplicative identities of rational numbers
miss Akunina [59]

Answer:  0 and 1, in that order

The numbers <u>   0   </u> and <u>   1   </u> are respectively the additive and multiplicative identities of rational numbers.

===========================================================

Explanation:

The additive identity is 0 because adding 0 to any number leads to the original number. For instance, 7+0 = 7. In general we can say x+0 = x or we could also say 0+x = x.

The multiplicative identity is 1 because multiplying 1 with anything leads to that original number. Example: 1*5 = 5 or 9*1 = 1. The general template is x*1 = x which is the same as saying 1*x = x.

These ideas not only apply to rational numbers, but to real numbers as well.

8 0
3 years ago
A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean l
tankabanditka [31]

Answer:

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the mean of the Population = 95

Given that the standard deviation of the Population = 5

Let 'X' be the random variable in a normal distribution

Let X⁻ = 96.3

Given that the size 'n' = 84 monitors

<u><em>Step(ii):-</em></u>

<u><em>The Empirical rule</em></u>

         Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }

        Z = \frac{96.3 - 95}{\frac{5}{\sqrt{84} } }

       Z = 2.383

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

    P(X⁻ ≥ 96.3) = P(Z≥2.383)

                      =  1- P( Z<2.383)

                      =  1-( 0.5 -+A(2.38))

                      = 0.5 - A(2.38)

                     = 0.5 -0.4913

                    = 0.0087

<u><em>Final answer:-</em></u>

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

3 0
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Answer:

Step-by-step explanation:

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