What is the value of x? triangle has a side of 16 and 34

When Mischa filled her car's gas tank before beginning a trip, the car's odometer read 44,587 miles. When she stopped for gas ne

pantera1 [17]

Answer:

18.11 miles per gallon

Step-by-step explanation:

Get the miles traveled from the trip odometer, or subtract the original odometer reading from the new one.

Divide the miles traveled by the amount of gallons it took to refill the tank. The result will be your car’s average miles per gallon yield for that driving period.

44,895-44,587 = 308 miles

308/17 = 18.11

6 0

3 years ago

An easier question all wrong or unhelpful or no-work answers will be reported.

ipn [44]

Elimination:

3x - 9y = 3
6x - 3y = -24

3x - 9y = 3
18x - 9y = -72
(subtract)
-15x = 75
÷ -15
x = -5

(3 × -5) - 9y = 3
-15 - 9y = 3
+ 15
-9y = 18
÷ -9
y = -2

Substitution:

6x - 3y = -24
+ 3y
6x = -24 + 3y

÷ 6
x = 4 + 0.5y

3(4 + 0.5y) - 9y = 3
12 + 1.5y - 9y = 3
12 - 7.5y = 3
- 12
-7.5y = -9
÷ -7.5
y = 1.2

x = 4 + (0.5 × 1.2)
x = 4 + 0.6
x = 4.6

So this one didn't fail as much, but I got different numbers. If you have to give in values, I'd give in the values from the elimination because I don't trust myself when it comes to the substitution

4 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).

(c) Calculate P(X > 3.5).

(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

Given the following trigonometric ratio, enumerate the meaning ratio

Likurg_2 [28]

Answer:

The trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

Step-by-step explanation:

From Trigonometry we know the following definitions for each trigonometric ratio:

Sine

$\sin \theta = \frac{y}{h}$ (1)

Cosine

$\cos \theta = \frac{x}{h}$ (2)

Tangent

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}$ (3)

Cotangent

$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{y}$ (4)

Secant

$\sec \theta = \frac{1}{\cos \theta} = \frac{h}{x}$ (5)

Cosecant

$\csc \theta = \frac{1}{\sin \theta} = \frac{h}{y}$ (6)

Where:

$x$ - Adjacent leg.

$y$ - Opposite leg.

$h$ - Hypotenuse.

The length of the hypotenuse is determined by the Pythagorean Theorem:

$h = \sqrt{x^{2}+y^{2}}$

If $y = AC$ and $x = BC$ , then the trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

6 0

3 years ago

The meaning of significance. Asked to explain the meaning of "statistically significant at the α=0.05 level," a student says: "T

Irina18 [472]

Answer:

This explanation is not correct

Explanation:

Significance level is the probability that the null hypothesis rejected is in fact true. If we say there is a significance level of 0.10 or 10%(α=0.10 level) then there is a 10% chance of incorrectly rejecting the null hypothesis when there is actually no difference.

If we wish to reach a conclusion on rejection or acceptance of the null hypothesis, we could use the p value(the probability that observed results or more extremee results are true given null hypothesis)which we compare with the significance level to conclude. If the significance level is 0.10 and p value is 0.07(since p value is less than significance level) , we reject the null hypothesis and vice versa.

3 0

3 years ago