As I increases by 18, y decreases by 12. From the last value in the table, I needs to increase by 2·18 for it to become zero. Hence the y-intercept will be 2·12 less than the last value in the table:

1 - 2·12 = -23 . . . the y-intercept

6 0

4 years ago

Use the normal distribution of SAT critical reading scores for which the mean is 502 and the standard deviation is 116. Assume t

PSYCHO15rus [73]

Answer:

(a) 93.19%

(b) 267.3

Step-by-step explanation:

The population mean and standard deviation are given as 502 and 116 respectively.

Consider, X be the random variable that shows the SAT critical reading score is normally distributed.

(a) The percent of the SAT verbal scores are less than 675 can be calculated as:

$P(X$

Thus, the required percentage is 93.19%

(b)

The number of SAT verbal scores that are expected to be greater than 575 can be calculated as:

$P(X>575)=P(\frac{x-502}{116}>\frac{575-502}{116}\\P(X>575)=P(Z>\frac{575-502}{116})\\P(X>575)=P(Z>0.6293)\\P(X>575)=0.2673$

So,

Out of 1000 randomly selected SAT verbal scores, 1000(0.2673) = 267.3 are expected to have greater than 575.

8 0

3 years ago

Select the equivalent expression.

Vanyuwa [196]

For this case we must find an expression equivalent to: $(y ^ 3 * 2 ^ 5) ^ 2$

By definition of power properties we have to meet:

$(a ^ n) ^ m = a ^ {n * m}$

So, rewriting the expression we have:

$y^{3 * 2} * 2^ {5 * 2} =\\y^ 6 * 2^{10}$

Answer:

Option B

$y^6*2^{10}$

4 0

4 years ago