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3 years ago

Suppose that XNY are in rational numbers such that x>y. Which of the following inequalities must be true

Mathematics

1 answer:

igomit [66]3 years ago

8 0

Answer:

D. 2x > 2y

Step-by-step explanation:

This question illustrates some of the difficulties you can run into with inequalities.

Consider each of the expressions when x = 1 and y = -2. We know that

x > y

1 > -2

A. x^2 = 1, y^2 = 4 . . . . x^2 > y^2 is not true

B. -x = -1, -y = 2 . . . . -x > -y is not true

C. 1/x = 1/1 = 1, 1/y = 1/-2 = -1/2 . . . . 1/x > 1/y is true

Now consider x=2, y=1. x > y because 2 > 1, but ...

1/x = 1/2, 1/y = 1 . . . . 1/x > 1/y is not true

D. We know that multiplying the numbers by 2 only stretches them apart on the number line. It does not change their order. For our example, ...

2x = 2, 2y = -4 . . . . 2 > -4 is true

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Simplify: -2x+5y-6x-4y

Westkost [7]

-2x+5y-6x-4y = -8x + y

5 0

3 years ago

Solve each porportion 10/k =8/4 and tell me how you git your answer

Anettt [7]

It is only 1 proportion, and it is 1 problem.

10 8
---- = ---
X 4

1). It is 8x=40
2). Divide both sides
by 8
3). X=5

7 0

3 years ago

A report on consumer financial literacy summarized data from a representative sample of 1,664 adult Americans. Based on data fro

daser333 [38]

Answer:

a. The 95% confidence interval for the proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is (0.54, 0.588). This means that we are 95% sure that the true proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is between these two bounds.

b. Because this confidence interval is entirely above 0.5, the interval is consistent with the statement that a majority of adult Americans would give themselves a grade of A or B.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Sample of 1,664 adult Americans, 939 people in the sample would have given themselves a grade of A or B in personal finance.

This means that $n = 1664, \pi = \frac{939}{1664} = 0.5643$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5643 - 1.96\sqrt{\frac{0.5643*0.4357}{1644}} = 0.54$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5643 + 1.96\sqrt{\frac{0.5643*0.4357}{1644}} = 0.588$

The 95% confidence interval for the proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is (0.54, 0.588). This means that we are 95% sure that the true proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is between these two bounds.

(b) Is the confidence interval from part (a) consistent with the statement that a majority of adult Americans would give themselves a grade of A or B?

Yes, because the confidence interval is entirely above 0.5.

Because this confidence interval is entirely above 0.5, the interval is consistent with the statement that a majority of adult Americans would give themselves a grade of A or B.

8 0

3 years ago

Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is the following. F(x) =

Troyanec [42]

Answer:

a) P (x <= 3 ) = 0.36

b) P ( 2.5 <= x <= 3 ) = 0.11

c) P (x > 3.5 ) = 1 - 0.49 = 0.51

d) x = 3.5355

e) f(x) = x / 12.5

f) E(X) = 3.3333

g) Var (X) = 13.8891 , s.d (X) = 3.7268

h) E[h(X)] = 2500

Step-by-step explanation:

Given:

The cdf is as follows:

F(x) = 0 x < 0

F(x) = (x^2 / 25) 0 < x < 5

F(x) = 1 x > 5

Find:

(a) Calculate P(X ≤ 3).

(b) Calculate P(2.5 ≤ X ≤ 3).

(d) What is the median checkout duration ? [solve 0.5 = F()].

(e) Obtain the density function f(x). f(x) = F '(x) =

(f) Calculate E(X).

(g) Calculate V(X) and σx. V(X) = σx =

(h) If the borrower is charged an amount h(X) = X2 when checkout duration is X, compute the expected charge E[h(X)].

Solution:

a) Evaluate the cdf given with the limits 0 < x < 3.

So, P (x <= 3 ) = (x^2 / 25) | 0 to 3

P (x <= 3 ) = (3^2 / 25) - 0

P (x <= 3 ) = 0.36

b) Evaluate the cdf given with the limits 2.5 < x < 3.

So, P ( 2.5 <= x <= 3 ) = (x^2 / 25) | 2.5 to 3

P ( 2.5 <= x <= 3 ) = (3^2 / 25) - (2.5^2 / 25)

P ( 2.5 <= x <= 3 ) = 0.36 - 0.25 = 0.11

c) Evaluate the cdf given with the limits x > 3.5

So, P (x > 3.5 ) = 1 - P (x <= 3.5 )

P (x > 3.5 ) = 1 - (3.5^2 / 25) - 0

P (x > 3.5 ) = 1 - 0.49 = 0.51

d) The median checkout for the duration that is 50% of the probability:

So, P( x < a ) = 0.5

(x^2 / 25) = 0.5

x^2 = 12.5

x = 3.5355

e) The probability density function can be evaluated by taking the derivative of the cdf as follows:

pdf f(x) = d(F(x)) / dx = x / 12.5

f) The expected value of X can be evaluated by the following formula from limits - ∞ to +∞:

E(X) = integral ( x . f(x)).dx limits: - ∞ to +∞

E(X) = integral ( x^2 / 12.5)

E(X) = x^3 / 37.5 limits: 0 to 5

E(X) = 5^3 / 37.5 = 3.3333

g) The variance of X can be evaluated by the following formula from limits - ∞ to +∞:

Var(X) = integral ( x^2 . f(x)).dx - (E(X))^2 limits: - ∞ to +∞

Var(X) = integral ( x^3 / 12.5).dx - (E(X))^2

Var(X) = x^4 / 50 | - (3.3333)^2 limits: 0 to 5

Var(X) = 5^4 / 50 - (3.3333)^2 = 13.8891

s.d(X) = sqrt (Var(X)) = sqrt (13.8891) = 3.7268

h) Find the expected charge E[h(X)] , where h(X) is given by:

h(x) = (f(x))^2 = x^2 / 156.25

The expected value of h(X) can be evaluated by the following formula from limits - ∞ to +∞:

E(h(X))) = integral ( x . h(x) ).dx limits: - ∞ to +∞

E(h(X))) = integral ( x^3 / 156.25)

E(h(X))) = x^4 / 156.25 limits: 0 to 25

E(h(X))) = 25^4 / 156.25 = 2500

8 0

3 years ago

In triangle LMN, LM = 6, MN = 9, and LN = 7. Find m∠M to the nearest tenth of a degree.

kozerog [31]

7^2 = 6^2 + 9^2 - 2(6)(9)cos(m)
-68 = 2(6)(9)cos(m)
cos(m) = -0.6296
m = arcos ( -0.6296)
m = 129.0 degrees

6 0

3 years ago