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LuckyWell [14K]
3 years ago
9

A psychology professor assigns letter grades on a test according to the following scheme. A: Top 11% of scores B: Scores below t

he top 11% and above the bottom 61% C: Scores below the top 39% and above the bottom 16% D: Scores below the top 84% and above the bottom 6% F: Bottom 6% of scores Scores on the test are normally distributed with a mean of 81.8 and a standard deviation of 7.8. Find the minimum score required for an A grade. Round your answer to the nearest whole number, if necessary.
Mathematics
1 answer:
GaryK [48]3 years ago
7 0

Answer: the minimum score required for an A grade is 91

Step-by-step explanation:

Since the scores on the test are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = scores on the test.

µ = mean score

σ = standard deviation

From the information given,

µ = 81.8

σ = 7.8

The probability value for the scores in the top 11% would be (1 - 11/100) = (1 - 0.11) = 0.89

Looking at the normal distribution table, the z score corresponding to the probability value is 1.23

Therefore,

1.23 = (x - 81.8)/7.8

Cross multiplying by 114, it becomes

1.23 × 7.8 = x - 81.8

9.594 = x - 81.8

x = 9.594 + 81.8

x = 91 rounded to the nearest whole number.

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4 0
4 years ago
The x- intercepts of a parabola are (0,-6) and (0,4). The parabola crosses the y- axis at -120. Lucas said that an equation for
astraxan [27]

Given:

The x- intercepts of a parabola are (0,-6) and (0,4).

The parabola crosses the y- axis at -120.

Lucas said that an equation for the parabola is y=5x^2+10x-120 and that the coordinates of the vertex are (-1, -125).

To find:

Whether Lucas is correct or not.

Solution:

The x- intercepts of a parabola are (0,-6) and (0,4). It means (x+6) and (x-4) are the factors of the equation of the parabola.

y=a(x+6)(x-4)             ...(i)

The parabola crosses the y- axis at -120. It means the equation of the parabola must be true for (0,-120).

-120=a(0+6)(0-4)

-120=a(6)(-4)

-120=-24a

Divide both sides by -24.

\dfrac{-120}{-24}=a

5=a

Substituting a=5 in (i), we get

y=5(x+6)(x-4)

y=5(x^2+6x-4x-24)

y=5(x^2+2x-24)

y=5x^2+10x-120

So, the equation of the parabola is y=5x^2+10x-120.

The vertex of a parabola f(x)=ax^2+bx+c is:

Vertex=\left(-\dfrac{b}{2a},f(-\dfrac{b}{2a})\right)

In the equation of the parabola, a=5,b=10,c=-120.

-\dfrac{b}{2a}=-\dfrac{10}{2(5)}

-\dfrac{b}{2a}=-\dfrac{10}{10}

-\dfrac{b}{2a}=-1

Putting x=-1 in the equation of the parabola, we get

y=5(-1)^2+10(-1)-120

y=5-10-120

y=-125

So, the vertex of the parabola is at point (-1,-125).

Therefore, Lucas is correct.

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Answer:

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