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ololo11 [35]
3 years ago
13

Employee B got a one time bonus of $90. He makes $13 an hour. If he was paid $415, how

Mathematics
1 answer:
Marina CMI [18]3 years ago
3 0

Answer:

y=mx+b format... 8th grade math class memories...ahhh

Step-by-step explanation:

Employee B-

one time bonus of 90 gets plugged into the b spot

13 an hour goes into the m space

415 total paid goes into the y spot

415=13(x)+90

subtract 90 from 415 and 90

325=13(x)

divide 325 by 13 and 13

25=(x)

Now plug x into the y=mx+b format with all your other numbers to check

y=13(25)+90

y=415!!

Hope this wasn't too confusing, good luck!

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18cm

Step-by-step explanation:

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The number at the end of the equation shifts the graph up or down.

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On the checkerboard, 4/10 of the pieces left are red. What fraction of the pieces left are black? Write the equation.​
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Answer:

The fraction of black pieces left is \frac{6}{10}

Step-by-step explanation:

Given as :

On the checkerboard , The fraction of red pieces left = \frac{4}{10}

Let The fraction of black pieces left                              = x

So, The fraction of black pieces left = x = 1 - \frac{4}{10}

Or,                                                          x = \frac{10-4}{10}

Or,                                                          x = \frac{6}{10}

Hence The fraction of black pieces left is \frac{6}{10}  Answer

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Half of a class took Form A of a test, and half took Form B. Of the students who took Form B, 39% passed. What is the probabilit
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A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73
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Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62      [from the standard t table]

Therefore,

CI = (Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})

Thus,

Lower limit of CI =  (Mean - \frac{t\times\sigma}{\sqrt{n}})

or

Lower limit of CI =  (98.3 - \frac{2.62\times0.73}{\sqrt{103}})

or

Lower limit of CI = 98.11

and,

Upper limit of CI =  (Mean + \frac{t\times\sigma}{\sqrt{n}})

or

Upper limit of CI =  (98.3 + \frac{2.62\times0.73}{\sqrt{103}})

or

Upper limit of CI = 98.49

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The value of 98.6°F suggests that this is significantly higher and  the mean temperature could very possibly be 98.6°F

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