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2 years ago

Need help with #13 please help anyone

Mathematics

1 answer:

Anuta_ua [19.1K]2 years ago

7 0

No sorry I can't I hope you find someone who cab tho

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Which statement comparing these measures is true? A microgram is 1,000 times greater than a milligram.

ololo11 [35]

The statement is false.

The statement below is true.
A milligram is 1000 times greater than a microgram.

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2 years ago

David just accepted a job at a new company where he will make an annual salary of $69000. David was told that for each year he s

Nataliya [291]

Answer:

After 4 years working for the company, he would make $79,000 of salary.

His salary after t years will be:

[te]S(t) = 69000 + 2500t[/tex]

Step-by-step explanation:

David just accepted a job at a new company where he will make an annual salary of $69000. David was told that for each year he stays with the company, he will be given a salary raise of $2500.

This means that after t years, his salary will be given by:

[te]S(t) = 69000 + 2500t[/tex]

How much would David make as a salary after 4 years working for the company?

[te]S(4) = 69000 + 2500*4 = 69000 + 10000 = 79000[/tex]

After 4 years working for the company, he would make $79,000 of salary.

5 0

2 years ago

a store buys a jacket for$20.00 and then sells it for 80% more what is the selling price for the jacket

victus00 [196]

Answer: 100$.

100% - 80% = 20%

20$ / 0.20(20%) =100$

7 0

2 years ago

Answer for brainlist & extra points <3

DENIUS [597]

Answer:

Equation:500÷8×2

Answer:125

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2 years ago

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73

faust18 [17]

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 = $(Mean - \frac{t\times\sigma}{\sqrt{n}})$

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

Lower limit of CI = 98.11

and,

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

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

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

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

7 0

2 years ago