Shannon test-drives a car in Germany and drives 95 kilometers per hour. What is her speed in the miles per hour ?

X^3+3x^5=x5 what's the missing expression

shutvik [7]

x3+3x5=x53x5+x3=x5Subtract x^5 from both sides.3x5+x3−x5=x5−x52x5+x3=0Factor left side of equation.x3(2x2+1)=0Set factors equal to 0.x3=0‌‌ or ‌‌2x2+1=0x=0

3 0

3 years ago

Read 2 more answers

Subtract:

choli [55]

Answer:

it's answer is D -92

As -57-19-16= -92

3 0

3 years ago

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Please help me... I really need it.

MAXImum [283]

Answer:

If you are writing a seesya let me help you

Step-by-step explanation:

Make sure to hilight key info

Also use grammarly to correct past tense and sentences

Make sure to make sense

8 0

2 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

A study is designed to examine the effect of doing synchronized movements (such as marching in step or doing synchronized dance

Vlada [557]

Answer:

The complete question in the graphic

A.

After synchronized exercise, it was observed that the average proximity rate has increased, which shows that the mean of the variable is positive and there is high synchronization in the groups.

.

B.

If groups with low synchronization and high effort are compared, the result of the average will be positive, so the average proximity rate will also increase after the synchronization exercise is performed.

On the contrary, if we observe the groups that have low synchronization and low effort, the mean of the variable is negative, thus the average proximity rate after performing the exercise will decrease.

(C)

considering that the total degrees of freedom is 259, it will show 260 students in the analysis.

D.

With a significance level of 5% or 0.05, the P value is less than the significance level, which leads to differentiating the mean of the scores before and after performing the combinations.

E.

With a significance level of 1% or 0.01, the p-value is higher than the significance level. With this data we can conclude that there is no difference between the results before and after the synchronized exercise

3 0

3 years ago