What is the orgins of a line if the slope is -2

IQ scores on the WAIS test approximate a normal curve with a mean of 100 and a standard deviation of 15. What IQ score is identi

riadik2000 [5.3K]

Answer:

IQ scores of at least 130.81 are identified with the upper 2%.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 100 and a standard deviation of 15.

This means that $\mu = 100, \sigma = 15$

What IQ score is identified with the upper 2%?

IQ scores of at least the 100 - 2 = 98th percentile, which is X when Z has a p-value of 0.98, so X when Z = 2.054.

$Z = \frac{X - \mu}{\sigma}$

$2.054 = \frac{X - 100}{15}$

$X - 100 = 15*2.054$

$X = 130.81$

IQ scores of at least 130.81 are identified with the upper 2%.

7 0

2 years ago

(8.99x10^15) in standard form

Hoochie [10]

8990000000000000 i think

5 0

3 years ago

If you launch the cannonball from the ground instead (hi=0), what does that change in the original equation?

ikadub [295]

Answer:

the value of hi changes from 58 to 0:

h(t) = -4.9t^2 +19.8t

Step-by-step explanation:

If the initial height changes, then the term in the equation that represents that height will change. The initial height term is hi, so it changes from 58 to 0.

$h(t)=-4.9t^2+19.8t+0\\\\\boxed{h(t)=-4.9t^2+19.8t}$

7 0

2 years ago

A firm has the marginal-demand function where D(x)=-1400x÷√25-x² is the number of units sold at x dollars per unit. Find the dem

borishaifa [10]

The equation of the demand function is D(x) = 1400√(25-x²) + 11400

<h3>How to determine the demand function?</h3>

From the question, we have the following parameters that can be used in our computation:

Marginal demand function, D'(x) = -1400x÷√25-x²

Also, we have

D = 17000, when the value of x = 3

To start with, we need to integrate the marginal demand function, D'(x)

So, we have the following representation

D(x) = 1400√(25-x²) + C

Recall that

D = 17000 at x = 3

So, we have

17000 = 1400√(25-3²) + C

Evaluate

17000 = 5600 + C

Solve for C

C = 17000 - 5600

So, we have

C = 11400

Substitute C = 11400 in D(x) = 1400√(25-x²) + C

D(x) = 1400√(25-x²) + 11400

Hence, the function is D(x) = 1400√(25-x²) + 11400