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Vadim26 [7]
3 years ago
10

Math scores on the SAT exam are normally distributed with a mean of 514 and a standard deviation of 118. If a recent test-taker

is selected at random, what is the probability the student scored 691 or greater on the exam ?
Mathematics
1 answer:
suter [353]3 years ago
7 0

Answer:

The probability is 0.06681

Step-by-step explanation:

To calculate this, we need to calculate the standard score or z-score

Mathematically, the standard score can be calculated using the formula;

z-score = (x - mean)/SD

from the question, the mean is 514 and the standard deviation is 118

The z-score is thus = (691-514)/118 = 177/118 = 1.5

The probability we are trying to calculate is thus;

P(x ≥ 691) or P(z ≥ 1.5)

Using standard score table or calculator,

Recall, P( x < 691) = 1 - P( x ≥ 691)

Hence, P( x ≥ 691) = 1 - P( x < 691)

P( x ≥ 691) = 1 - 0.93319

= 0.06681

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Find the distance between the two points in simplest radical form.
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Answer:

d=\sqrt{58}\approx7.6

Step-by-step explanation:

To find the distance between any two points, we can use the distance formula:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2

Our first point, A, is at (1, 1) and our second point, B, is at (-2, 8).

Let's let A(1, 1) be (x₁, y₁) and B(-2, 8) be (x₂, y₂). Substitute this into the distance formula:

d=\sqrt{(-2-1)^2+(8-1)^2

Subtract:

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Square:

d=\sqrt{9+49}

Add:

d=\sqrt{58}

This cannot be simplified.

So, the distance between the two points is √58 or about 7.6 units.

And we're done!

So... where's my cookie :)?

6 0
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Step-by-step explanation:

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Step 2: Factor out variable x.

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Step 3: Divide both sides by f-4.

x(f−4)/ f−4 = −96/ f−4

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