Assume that z scores are normally distributed with a mean of 0 and a standard deviation of 1. if p(z >

in a circle graph above you can see that 47 percent of the companys monthly income is used to pay the owners personal salary if

Marizza181 [45]

The first thing you have to do is to divide the personal income by the percen it takes up

2,176.10/47=46.3

that means that 46.3 is 1% of the whole, so if we multiply it by 100 you will get the whole

46.3*100=4630

so the answer is

4,630

6 0

2 years ago

Deal with these relations on the set of real numbers: R₁ = {(a, b) ∈ R² | a > b}, the "greater than" relation, R₂ = {(a, b) ∈

uranmaximum [27]

Answer:

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 = $\{ (a,b) \in R^2 \}$

d) R1ºR4 = $\{ (a,b) \in R^2 \}$

e) R1ºR5 = R1

f) R1ºR6 = $\{ (a,b) \in R^2 \}$

g) R2ºR3 = $\{ (a,b) \in R^2 \}$

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(a,c) is in R1ºR1 if there exists b such that (a,b) is in R1 and (b,c) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2

This case is similar to the previous one. (a,c) is in R1ºR2 if there exists b such that (a,b) is in R2 and (b,c) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that ac. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

5 0

2 years ago

The steps to derive the quadratic formula are shown below: Step 1 ax2 + bx + c = 0 Step 2 ax2 + bx = − c Step 3 x2 + b over a ti

Nostrana [21]

Answer:

$x + \frac{b}{2a} = \frac{+/ - \sqrt{b^2-4ac} }{2a}$

Step-by-step explanation:

Step 1:

$ax^2+bx+c = 0$

Step 2:

$ax^2+bx = -c$

Step 3:

$\frac{ax^2+bx}{a} = \frac{-c}{a}$

Step 4:

Adding $\frac{b^2}{4a^2}$ to both sides to complete the square

$x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} = \frac{-c}{a} + \frac{b^2}{4a^2}$

Step 5:

$x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} = \frac{-4ac+b^2}{4a^2}$

Step 6:

Taking square root on both sides

$x + \frac{b}{2a} = \frac{+/ - \sqrt{b^2-4ac} }{2a}$

3 0

2 years ago

sophia says that you can solve the problem in the Example by multiplying both quantities in the ratio 60 : 36 by 1 . Is Sophia c

guapka [62]

Answer:

No multiplying both quantities in the ratio 60 : 36 by 1 would not give the answer .

But multiplying both quantities in the ratio 60 : 36 by 1/6 would solve it.

Step-by-step explanation:

Multiplying any number or ratio by 1 gives the same answer no matter what the expression or how much bigger the number is.This will not give the solution .

If it is multiplied with a fraction like 1/6 that would give the correct answer.

Multiplying it with 1/6 reduces the ratio and gives a better estimate.

A:B

60:36

60*1/6: 36*1/6

10: 6

Meaning that for a total of 16 there are 10 of category A and 6 of category B.

But if we do the same procedure with the answer is same which does not solve the question.

A:B

60:36

60*1: 36*1

60: 36

4 0

2 years ago

Andrea's family stopped at the gas station to get gas. At gas stations, the price of gas per gallon is given to the nearest thou

Evgen [1.6K]

Answer:

$2.498

Step-by-step explanation:

5 0

2 years ago