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Olin [163]
3 years ago
10

You invest $300 in an account at 7.5% per year simple interest. How much will you have in the account at the beginning of the 11

th year
Mathematics
2 answers:
padilas [110]3 years ago
5 0

Answer:

The answer is 525........ on apex

Step-by-step explanation:

Dahasolnce [82]3 years ago
4 0
By the beginning of the 11th year, you'll have $580.22
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-10 +log3 (n+3)= -10
Rufina [12.5K]

Answer:

n= -3 or in exact form: N= - log27/log3

Step-by-step explanation:

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aksik [14]

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Step-by-step explanation:

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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
2 years ago
6. The distance from Tory's home to the mall is 14 km. The distance from her home
Minchanka [31]

Answer: 6 km to 22 km !

Step-by-step explanation:

the distances from the school mall to bus station could be, in the worst case, 14+8=22 km apart, and in the best case 14-8=6 km apart

so the range is 6 km to 22 km !

8 0
2 years ago
Mary has saved 17.50 in the past three weeks at this rate how much will she save in 15 weeks
Kryger [21]
Mary will have saved $87.50

$17.50 per three weeks
x per 15 weeks

17.50 x 5 = $87.50
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3 years ago
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