Guinea pig (p) = 4 years old
Giraffe (g) = 10 years old
Tiger (t) = 16 years old

P = g-6
T = 4p

30 = g+g-6+(4(g-6))
30 = 2g-6+4g-24
30 = 6g-30
+30 +30
60 = 6g
6 6
10 = g

P = 10-6
P = 4

T = 4(4)
T = 16

8 0

2 years ago

A random sample of math majors taking an introductory statistics course were surveyed after completing the final exam. They were

quester [9]

Answer:

$E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8$

Now we can find the second moment with this formula:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4$

The variance would be given by:

$Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96$

And the deviation would be:

$Sd(X) =\sqrt{6.96}= 2.638$

Step-by-step explanation:

For this case we have the following distribution given:

X 1 2 7

P(X) 1/5 2/5 2/5

We need to begin finding the mean with this formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8$

Now we can find the second moment with this formula:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4$

The variance would be given by:

$Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96$

And the deviation would be:

$Sd(X) =\sqrt{6.96}= 2.638$

7 0

3 years ago

4)) Solve for w.<br> 100 = w + 2<br> w=[

Anettt [7]

Answer: w= 98

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

What is 1+7/30(4x)=12

ahrayia [7]

165 X= ——— 14 1 + 7/30 (4x) = 20

8 0

3 years ago