Suppose a normal distribution has a mean of 50 and a standard deviation of

What are the real zeros of the function g(x) = x3 + 2x2 − x − 2?

Nat2105 [25]

Upon a slight rearrangement this problem gets a lot simpler to see.

x^3-x+2x^2-2=0 now factor 1st and 2nd pair of terms...

x(x^2-1)+2(x^2-1)=0

(x+2)(x^2-1)=0 now the second factor is a "difference of square" of the form:

(a^2-b^2) which always factors to (a+b)(a-b), in this case:

(x+2)(x+1)(x-1)=0

So g(x) has three real zero when x={-2, -1, 1}

4 0

3 years ago

Need help ASAP. Last two problems on my study guide, can't seem to solve them?

Dimas [21]

Answer:

9) 4

10) p¹⁵/q⁹

Step-by-step explanation:

9)

As per the law of indices, (xᵃ)ᵇ=xᵃᵇ.

So divide 8 by 2 to get 4

10)

p(p⁻⁷q³)⁻²q⁻³

p(p¹⁴q⁻⁶)q⁻³ (because (xᵃ)ᵇ=xᵃᵇ)

pq⁻³(p¹⁴q⁻⁶)

p¹⁵q⁻⁹

p¹⁵/q⁹ (because x⁻ᵃ =1 /xᵃ)

4 0

3 years ago

A college job placement office collected data about students’ GPAs and the salaries they earned in their first jobs after gradua

frez [133]

Answer:

X is the GPA

Y is the Salary

Standard deviation of X is 0.4

Standard deviation of Y is 8500

E(X)=2.9

E(Y)=47200

We are given that The correlation between the two variables was r = 0.72

a) $y = a+bx$

$b = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2} = \frac{r \times \sqrt{var(X) \times Var(Y)}}{Var(X)} = \frac{0.72 \times \sqrt{0.4^2 \times 8500^2}}{0.4^2} = 15300$

$a=y-bx = 47200-(15300 \times 29) = 2830$

So, slope = 15300

Intercept = 2830

So, equation : $y = 2830+15300x$

b) Your brother just graduated from that college with a GPA of 3.30. He tells you that based on this model the residual for his pay is -$1880. What salary is he earning?

$y = 2830+15300 \times 3.3 = 53320$

Observed salary = Residual + predicted = -1860+53320 = 51440

c)) What proportion of the variation in salaries is explained by variation in GPA?

The proportion of the variation in salaries is explained by variation in GPA = $r^2 = (0.72)^2 =0.5184$

8 0

3 years ago

Plotting these two equations on a graph. p=6-2q p=4+q

Mekhanik [1.2K]

Put into slope intercept form, y=mx+b where m=slope
p=-2q+6
slope is -2
subsitute values for q and get values for p
q=0, p=6
(q,p)
(0,6)
(1,4)

p=q+4
slope is 1
subsitutte
q=0,p=4
(q,p)
(0,4)
(1,5)

below are the graphs

7 0

3 years ago

Consider the two data sets below:

alina1380 [7]

Answer:

Option c) The data sets will have the same values of their interquartile range.

Explanation:

1. The values are in order: they are in increasing oder, from lowest to highest value.

2. Calculate the interquartile range.

Interquartile range, IQR, is the third quartile, Q3, less the first quartile Q1:

IQR = Q3 - Q1

To find the first and the third quartile, first find the median:

Data Set 1: 19, 25, 35, 38, 41, 49, 50, 52, 59

[19, 25, 35, 38], 41, [49, 50, 52, 59]

↑

median = 41

Data Set 2: 19, 25, 35, 38, 41, 49, 50, 52, 99

[19, 25, 35, 38] , 41, [49, 50, 52, 99]

↑

median = 41

Now find the median of each subset: the values below the median and the values above the median.

Data set 1: First quartile

[19, 25, 35, 38],

↑

Q1 = [25 + 35] / 2 = 30

Third quartile

[49, 50, 52, 59]

↑

Q3 = [50 + 52] / 2 = 51

IQR = Q3 - Q1 = 51 - 30 = 21

Data set 2: First quartile

[19, 25, 35, 38]

↑

Q1 = [25 + 35] / 2= 30

Third quartile

[49, 50, 52, 99]

↑

Q3 = [52 + 50]/2 = 51

IQR = 51 - 30 = 21

Thus, it is shown that the data sets have will have the same values for the interquartile range: IQR = 21. (option c)

This happens because replacing one extreme value (in this case the maximum value) by other extreme value does not affect the median.

An outlier will change the range because the range is the maximum value less the minimum value.

5 0

3 years ago