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3 years ago

Does anyone know this?

Mathematics

1 answer:

andreyandreev [35.5K]3 years ago

7 0

I’m pretty sure it’s 38, when you divide you don’t get an answer shown, same for subtraction, and multiplication. when you add you get 38 which is a choice; unless there is a formula i’m missing then this should be the answer

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Suppose that salaries for recent graduates of one university have a mean of $25,700 with a standard deviation of $1350. Using Ch

sergeinik [125]

Answer:

75% of the data will reside in the range 23000 to 28400.

Step-by-step explanation :

To find the range of values :

We need to find the values that deviate from the mean. Since we want at least 75% of the data to reside between the range therefore we have,

$1 - \frac{1}{k^2} =\frac{75}{100}$

Solving this, we would get k = 2 which shows the value one needs to find lies outside the range.

Range is given by : mean +/- (z score) × (value of a standard deviation)

⇒ Range : 25700 +/- 2 × 1350

⇒ Range : (25700 - 2700) to (25700 + 2700)

Hence, 75% of the data will reside in the range 23000 to 28400.

3 0

3 years ago

How do you move the decimal point to find 10% of a number?

drek231 [11]

You divide the number by 10 like this: x/10=0.x

3 0

3 years ago

Solve the following equation. * x2 – 13x – 30 = 0 What’s the answer

Basile [38]

Sorry for the late response if u still don’t got it the answer it is X=-30

5 0

4 years ago

What is the constant of proportionality of the following equation? B = 1.25C

vladimir1956 [14]

Answer:

The answer is 122

Step-by-step explanation: 0.25 is the constant proportionality in this equation .

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6 0

3 years ago

Read 2 more answers

A simple random sample of size nequals17 is drawn from a population that is normally distributed. The sample mean is found to be

IRINA_888 [86]

Answer:

95% confidence intervals about the population mean is

(51.7656 , 60.2344)

Step-by-step explanation:

Step(i):-

Given random sample of size 'n' =17

Given mean of the sample 'x⁻' = 56

Given standard deviation of sample 's' = 10

95% confidence intervals about the population mean is determined by

$(x^{-} - t_{0.05} \frac{s}{\sqrt{n} } ,x^{-} + t_{0.05} \frac{s}{\sqrt{n} } )$

Degrees of freedom

ν = n-1 = 17-1 =16

t₀.₀₅ = 1.7459 (from t-table)

Step(ii):-

95% confidence intervals about the population mean is determined by

$(x^{-} - t_{0.05} \frac{s}{\sqrt{n} } ,x^{-} + t_{0.05} \frac{s}{\sqrt{n} } )$

$(56 - 1.7459 \frac{10}{\sqrt{17} } ,56 + 1.7459 \frac{10}{\sqrt{17} } )$

( 56 - 4.2344 , 56 + 4.2344)

(51.7656 , 60.2344)

Conclusion:-

95% confidence intervals about the population mean is

(51.7656 , 60.2344)

4 0

3 years ago