E[Y] = 0.4·1 + 0.3·2 + 0.2·3 + 0.1·4 = 2 E[1/Y] =0.4·1/1 + 0.3·1/2 + 0.2·1/3 + 0.1·1/4 = 0.4 + 0.15 + 0.0666 + 0.025?0.64 V[Y] =E[Y2]-E[Y]2= (0.4)·12+(0.3)·22+(0.2)·32+(0.1)·42-22= 0.4+1.2+1.8+1.6-4= 5-4 = 1

8 0

4 years ago

Jenny and Natalie are selling cheesecakes for a school fundraiser. Customers can buy chocolate cakes and vanilla cakes. Jenny so

IrinaVladis [17]

The cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

Solution:

Let "c" be the cost of 1 chocolate cake

Let "v" be the cost of 1 vanilla cake

Jenny sold 14 chocolate cakes and 5 vanilla cakes for 119 dollars

Therefore, we can frame a equation as:

14 x cost of 1 chocolate cake + 5 x cost of 1 vanilla cake = 119

$14 \times c + 5 \times v=119$

14c + 5v = 119 ------- eqn 1

Natalie sold 10 chocolate cakes and 10 vanilla cakes for 130 dollars

Therefore, we can frame a equation as:

10 x cost of 1 chocolate cake + 10 x cost of 1 vanilla cake = 130

$10 \times c + 10 \times v = 130$

10c + 10v = 130 -------- eqn 2

Let us solve eqn 1 and eqn 2

Multiply eqn 1 by 2

28c + 10v = 238 ------ eqn 3

Subtract eqn 2 from eqn 3

28c + 10v = 238

10c + 10v = 130

( - ) --------------------------

18c = 108

c = 6

Substitute c = 6 in eqn 1

14(6) + 5v = 119

84 + 5v = 119

5v = 119 - 84

5v = 35

v = 7

Thus cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

8 0

3 years ago

Worth 12 Points, i spent 24

rewona [7]

Answer:

$\displaystyle m^{\frac{35}{12}}n^{-\frac{28}{15}}=\frac{m^{\frac{35}{12}}}{n^{\frac{28}{15}}}$

Step-by-step explanation:

The relevant rule of exponents is ...

(a^b·c^d)^e = a^(be)·c^(de)

Then ...

(m^(5/4)·n^(-4/5))^(7/3) = m^(5/4·7/3)·n^(-4/5·7/3)

= m^(35/12)·n^(-28/15)

Since you want positive rational exponents, you can write this as ...

= m^(35/12)/n^(28/15)

6 0

3 years ago