What is 300cm in meters

Y = - 2x + 4 missing value

zhenek [66]

Your trash Hope this helped

3 0

4 years ago

Read 2 more answers

Paddy and Anna each invest $2000 for 5 years.

Serga [27]

Answer: r= 1.22

Step-by-step explanation:

Formula for amount with simple interest = $P(1+rt)$

, where

P= principal value , r= rate of interest , t = time.

Given: P= $2000, t= 5 years, r= 1.25% = 0.0125

$A=2000(1+0.0125\times5)\\\\=2000(1.0625)=2125$

Formula to compute compound amount : $P(1+r)^t$

$=2000(1+r)^5$

When both have same worth then

$2000(1+r)^5=2125\\\\\\ (1+r)^5=\dfrac{2125}{2000}\\\\\\ (1+r)^5=1.0625$

taking log on both sides , we get

$5\ln (1+r)=\ln 1.0625\\\\\\ 5\ln (1+r)=0.0606246\\\\\\ \ln (1+r)=0.012125\\\\\\ 1+r=e^{0.005266}\\\\\\ 1+r=1.0122\\\\ r=0.0122\\\\ \\ r=1.22\%$

Hence, Value of r= 1.22

6 0

3 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

There is a loss of Rs 5 when a watch is sold at the discount of 10%. IF the watch is sold at the discount of 5%, there is profit

Colt1911 [192]

At disc of 10% there was 5rupees loss and at disc of 5 % dis there was 22.5 profit
so 5% *x= 22.5+5=27.5
x=27.5 *20
x=550

8 0

3 years ago

A2=11, a12 = 51<br><br> can someone please help me ??

Nimfa-mama [501]

could you specify your question a bit more?

3 0

3 years ago