I need help with my work

Solve the equation x2 = 10. x = 3/10 X = ✓10 x = +5 X = 5

Norma-Jean [14]

Answer:

the answer is 5, because 5x2=10.

6 0

4 years ago

How much higher is 500m than 400m

zaharov [31]

Answer:

100m

Step-by-step explanation:

just subtract them from each other

500m

- 400m

-----------

100m

7 0

3 years ago

Read 2 more answers

The owner of a farm plants apple trees and mango trees in a ratio of 16:6. How many oak trees are planted if 198 mango trees are

BARSIC [14]

Answer:528 apple trees

Step-by-step explanation:Listen 198/6=33 so then just multiply 33x16 that equals 528 hope this helps

3 0

2 years ago

Exercise 5.2. Suppose that X has moment generating function

soldi70 [24.7K]

Answer:

a) Mean, E(X) = - 0.5

Variance = = 9.25

b) $M_{X}(t)=E(e^{tX})$

or

⇒ $=\sum e^{tx}p(x)$

⇒ $=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}$

Step-by-step explanation:

Given:

moment generating function of X as:

MX(t) = $\frac{1}{2} + \frac{1}{3}e^{-4t} + \frac{1}{6} e^{5t}$

a) Now

Mean, E(X) = $M_{X}'(t=0)$

Thus,

$M_{X}'(t)=\frac{1}{3}(-4)e^{-4t}+\frac{1}{6}(5)e^{5t}$

or

$M_{X}'(t)=\frac{-4}{3}e^{-4t}+\frac{5}{6}e^{5t}$

also,

$E(X^{2})=M_{X}''(t=0)$

Thus,

$M_{X}''(t)=\frac{-4}{3}(-4)e^{-4t}+\frac{5}{6}(5)e^{5t}$

or

$M_{X}''(t)=\frac{16}{3}e^{-4t}+\frac{25}{6}e^{5t}$

Therefore,

Mean, E(X) = $M_{X}'(t=0)=\frac{-4}{3}e^{-4(0)}+\frac{5}{6}e^{5(0)}$

or

Mean, E(X) = - 0.5

and

$E(X^{2})=M_{X}''(t=0)=\frac{16}{3}e^{-4(0)}+\frac{25}{6}e^{5(0)}$

or

$E(X^{2})$ = 9.5

also,

Variance(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

b) Now,

Let f(x) be the PMF of X

Thus,

$M_{X}(t)=E(e^{tX})$

or

⇒ $=\sum e^{tx}p(x)$

⇒ $=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}$

Therefore,

at x = 0, P(x) = $\frac{1}{2}$

at x= - 4 ,P(x) = $\frac{1}{3}$

at x = 5, P(x) = $\frac{1}{6}$

Thus,

E(X) = $\sum xP(x)=0(\frac{1}{2})+(-4)(\frac{1}{3})+5(\frac{1}{6})$

or

E(X) = - 0.5

also, $E(X^{2})=\sum x^{2}P(x)=0^{2}(\frac{1}{2})+(-4)^{2}(\frac{1}{3})+5^{2}(\frac{1}{6})$

$E(X^{2})$ = 9.5

Hence,

Var(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

4 0

4 years ago

TO

ipn [44]

Answer:

50 soldiers must be transferred elsewhere.

Step-by-step explanation:

We solve this question by proportions, using a rule of three.

As the number of soldiers decrease, the provisions last for more time. This means that the measures are inversely proportional, and we have an inverse rule of three, using line multiplication, instead of cross.

30 days of provisions - 200 soldiers

40 days of provisions - x soldiers

$40x = 200*30$

Simplifying by 40

$x = 5*30 = 150$

The provisions will last for 40 days with 150 soldiers, which means that 200 - 150 = 50 soldiers must be transferred elsewhere.

4 0

3 years ago

I need help with my work​

I need help with my work