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Anni [7]
3 years ago
5

A teacher gives a test in which the average of the test scores is 68 and the standard deviation is 14. Realizing that the test w

as probably too difficult, the teacher decides to scale the tests by adding 10 points to all scores. What is the mean and standard deviation of the scaled test scores?
Mathematics
1 answer:
Alika [10]3 years ago
6 0

Answer:

Mean = 78 ;

Standard deviation = 14

Step-by-step explanation:

Given that:

Mean score = 68

Standard deviation = 14

When a constant value is added to the individual value of a dataset, the new mean value will become the sum of the initial mean value and the constant term added to the values.

Constant Added score = 10

The mean of scaled test score will be:

Mean score + constant added score

68 + 10 = 78

The standard deviation remains unchanged ; Hence, standard deviation = 14

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4k^3 + k^2 - 2k + 7

Step-by-step explanation:

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Answer:

6

Step-by-step explanation:

Hayden uses 3 cups of water to make a broth for soup

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A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station. Find the ra
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Answer:

The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

Step-by-step explanation:

Given information:

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.

z=1

\frac{dx}{dt}=430

We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

y=2

According to Pythagoras

hypotenuse^2=base^2+perpendicular^2

y^2=x^2+1^2

y^2=x^2+1               .... (1)

Put z=1 and y=2, to find the value of x.

2^2=x^2+1^2

4=x^2+1

4-1=x^2

3=x^2

Taking square root both sides.

\sqrt{3}=x

Differentiate equation (1) with respect to t.

2y\frac{dy}{dt}=2x\frac{dx}{dt}+0

Divide both sides by 2.

y\frac{dy}{dt}=x\frac{dx}{dt}

Put x=\sqrt{3}, y=2, \frac{dx}{dt}=430 in the above equation.

2\frac{dy}{dt}=\sqrt{3}(430)

Divide both sides by 2.

\frac{dy}{dt}=\frac{\sqrt{3}(430)}{2}

\frac{dy}{dt}=372.390923627

\frac{dy}{dt}\approx 372

Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

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3 years ago
Given the following expression, what is the rate of decay (write in decimal form, not as a percentage)? 15000 (.87)^7​
34kurt

Answer:

0.87

Step-by-step explanation:

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Answer: 0.87

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