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2 years ago

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scores on the act college entrance exam follow a bell-shaped distribution with mean 18 and standard deviation 6. Wayne's standar

dized score on the ACT was -0.5. what was Wayne's actual ACT score?

1 answer:

WITCHER [35]2 years ago

7 0

Answer:

His actual score is 15

Step-by-step explanation:

Here, we are interested in calculating Wayne’s actual score on the ACT

when we say the scores have being standardized, it means the score was reported in terms of the z-score and not the initial raw scores.

Now, mathematically, for the scores to have a negative z-score, it means it is actually below the mean.

The formula for the z-score or standard score is given below;

z-score = (x - mean)/SD

where in this case, x = ? which is the score we are looking for , z-score = -0.5 , mean score = 18 and standard deviation of the scores = 6

So, substituting these values into the z-score equation, we have;

-0.5 = (x-18)/6

x-18 = 6(-0.5)

x -18 = -3

x = -3 + 18

x = 15

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3 years ago

A random sample of 20 purchases showed the amounts in the table (in $). The mean is $49.57 and the standard deviation is $20.28.

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Answer:

The 98% confidence interval for the mean purchases of all customers is ($37.40, $61.74).

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

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2 years ago

Read 2 more answers

Janeel has a 10-inch by 12-inch photograph. She wants to scan the photograph, then reduce the result

Vaselesa [24]

<u>Complete Question:</u>

Janeel has a 10 inch by 12 inch photograph. She wants to scan the photograph, then reduce the results by the same amount in each dimension to post on her Web site. Janeel wants the area of the image to be one eight of the original photograph. Write an equation to represent the area of the reduced image. Find the dimensions of the reduced image.

<u>Correct Answer:</u>

A) $(10-x)(12-x)=15$

B) Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch

<u>Step-by-step explanation:</u>

a. Write an equation to represent the area of the reduced image.

Let the reduced dimensions is by x , So the new dimensions are

$length=10-x\\breadth=12-x$

According to question , Area of new image is :

⇒ $Area = \frac{1}{8}Length(breadth)$

⇒ $Area = \frac{1}{8}(10)(12)$

⇒ $Area = 15$

So the equation will be :

⇒ $(10-x)(12-x)=15$

b. Find the dimensions of the reduced image

Let's solve : $(10-x)(12-x)=15$

⇒ $120-10x-12x+x^2=15$

⇒ $120-22x+x^2=15$

⇒ $x^2-22x+105=0$

By Quadratic formula :

⇒ $x = \frac{-b \pm \sqrt{b^2-4ac} }{2a}$

⇒ $x = \frac{22 \pm8 }{2}$

⇒ $x = \frac{22 +8 }{2} , x = \frac{22 -8 }{2}$

⇒ $x = 15 , x =7$

x = 15 is rejected ! as 15 > 10 ! Side can't be negative

⇒ $x =7$

Therefore, Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch

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2 years ago