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ValentinkaMS [17]
3 years ago
6

The Two questions below is that I need help with. I need to break apart the numbers into tens and ones

Mathematics
1 answer:
Aleks [24]3 years ago
7 0
First Problem:

44 + 57 = 101

10 + 10 + 10 + 14 

10 + 10 + 10 + 10 + 17 = 101

Second Problem: 

25 + 64 = 89

10 + 10 + 5

10 + 10 + 10 + 10 + 10 + 14 = 89

Hope this helps. :)

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Answer:

$7,056.83

Step-by-step explanation:

Rewrite given data in the following table:

\begin{array}{ccc}\text{Operation}&\text{Amount}&\text{Mathematical operation}\\\text{Beginning balance}&\$9,145.87$\\\text{Made deposit}&\$2,783.71&"+"\\\text{Wrote checks}&\$4,871.90&"-"\\\text{Paid a maintenance fee}&\$12&"-"\\\text{Earned interest}&\$11.15&"+"\end{array}

To find the balance at the end of the month, complete all mathematical operations:

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t the Fidelity Credit Union, a mean of 5.8 customers arrive hourly at the drive-through window. What is the probability that, in
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Answer:

0.5217

Step-by-step explanation:

Given that:

mean \mu = 5.8

Using Poisson Probability formula:

P(X=x) = \frac{   (e^{- \mu} * {\mu^x) }}{x!}

P(X>5.8)=1-P(X \leq5)

= 1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)]

= 1 -[\frac{(e^{-5.8}*5.8^0)    }{ 0!   } + \frac{(e^{-5.8}*5.8^1)    }{ 1!   } +\frac{(e^{-5.8}*5.8^2)    }{ 2!   } +\frac{(e^{-5.8}*5.8^3)    }{ 3!   } +\frac{(e^{-5.8}*5.8^4)    }{ 4!   } +\frac{(e^{-5.8}*5.8^5)    }{ 5!   } ]

=1-(0.003+0.0176+0.0509+0.0985+0.1428+0.1656)

=1-0.4783

=0.5217

∴ the probability that, in any hour, more than 5 customers will arrive =0.5217

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