10 pecebt in simplest form

MZJNM = 103°. Find mZJNK 2 * 24 32 K M L

just olya [345]

Answer:

m<ZJNK=47°

Step-by-step explanation:

m<ZJNK=103°-24°-32°=47°

8 0

3 years ago

Explain how knowing that 5divided by eight equals 0.625 help you write the decimal for 4 and five eights

Black_prince [1.1K]

4 and five eights is equal to 4 + 5/8.
You know that the decimal representation of 5/8 is 0.625, so the decimal representations of 4 and five eights will be 4.625.

7 0

3 years ago

Solve for y. 40 = 25y Simplify your answer as much as possible.

Romashka [77]

Mark me as the brainliest

6 0

2 years ago

Read 2 more answers

Robert Masterson bought 99 shares of stock Eddie price of $55 per share and $150 shares at $52 per share. The annual dividend pe

Luda [366]

Answer:

2.39%

Step-by-step explanation:

Given that

99 shares at $55 per share = $5445

150 share at $52 per share = $7800

Hence, total share cost => 5445 + 7800 => $13,245

Therefore, given that the average annual yield is the summation of all the income gotten on investment and dividing the amount by the number of years the money was invested.

With a $1.27 dividend per share, and a total share of 99 + 150 = 249 shares

=> 1.27 * 249 = $316.23

Hence, average annual yield = $316.23 ÷ 13,245

=> 0.0239

=> 2.39%

Therefore, Robert Mastersons' average annual yield is 2.39%

8 0

3 years ago

Based upon a long period of record keeping the following represents the probability distribution of the number of times the John

Nesterboy [21]

Given a table representing the probability distribution of the number of times the John Jay wifi network is slow during a week. We call the random variable x.

$\begin{tabular}
{|c|c|c|c|c|c|c|c|}
x&0&1&2&3&4&5&6\\[1ex]
p(x)&.08&.17& .21& k& .21& k& .13
\end{tabular}$

Part A:

The total value of p(x) = 1.

Thus, 

.08 + .17 + .21 + k + .21 + k + .13 = 1

0.8 + 2k = 1

2k = 1 - 0.8 = 0.2

k = 0.2 / 2 = 0.1

Therefore, the value of k is 0.1

Part B:

The expected value of x is given by

$E(x)=\Sigma
 xp(x) \\ \\ =0(0.08)+1(0.17)+2(0.21)+3(0.1)+4(0.21)+5(0.1)+6(0.13) \\ 
 \\ =0+0.17+0.42+0.3+0.84+0.5+0.78=3.01$

Therefore, the expected value of x is 3.01

Part C:

The expected value of $x^2$ is given by

$E(x^2)=\Sigma x^2p(x) 
\\ \\ =0^2(0.08)+1^2(0.17)+2^2(0.21)+3^2(0.1)+4^2(0.21)+5^2(0.1)+6^2(0.13) \\ \\ 
=0(0.08)+1(0.17)+4(0.21)+9(0.1)+16(0.21)+25(0.1)+36(0.13) \\ \\ =0+0.17+0.84+0.9+3.36+2.5+4.68=12.45$

Therefore, the expected value of $\bold{x^2}$ is 12.45


Part D:

The variance of x is given by

$Var(x)=E(x^2)-(E(x))^2 \\ \\ =12.45 - (3.01)^2=12.45-9.06 \\ \\ =3.39$

Therefore, the variance of x is 3.39.

Part E

The standard deviation of x is given by

$\sqrt{Var(x)} = \sqrt{3.39} =1.84$

Therefore, the standard deviation of x is 1.84.

Part F:

The variance of ax, where a is a constant is given by

$Var(aX)=a^2Var(X)$

Thus, the variance of 3x is given by

$Var(3X)=3^2Var(X)=9(3.39)=30.51$

Therefore, the variance of 3x is 30.51.

Part G:

The probability that the network has no more that 4 slow times in one week is given by

$P(x\leq4)=P(0)+P(1)+P(2)+P(3)+P(4) \\ \\ =0.08+0.17+0.21+0.1+0.21=0.77$

Since, the network slowness is independent from week to week, the probability that if we look at 5 separate weeks, the network has no more than 4 slow times in any of those weeks is given by

$(0.77)^5=0.27$

Therefore, the probability that if we look at 5 separate weeks, the network has no more than 4 slow times in any of those weeks is 0.27

Part H:

The variance of x^2 is given by

$Var(x^2)=E((x^2)^2)-(E(x^2))^2=E(x^4)-(E(x^2))^2$

$E(x^4)=\Sigma
 x^4p(x) \\ \\ 
=0^4(0.08)+1^4(0.17)+2^4(0.21)+3^4(0.1)+4^4(0.21)+5^4(0.1) \\ +6^4(0.13)
 \\ \\ 
=0(0.08)+1(0.17)+16(0.21)+81(0.1)+256(0.21)+625(0.1)\\+1,296(0.13) \\ \\
 =0+0.17+3.36+8.1+53.76+62.5+168.48=296.37$

Thus,

$Var(x^2)=296.37-(12.45)^2=296.37-155.00=141.37$

Therefore, the variance of the random variable $\bold{x^2}$ is 141.37

5 0

3 years ago