I have to solve each problem and round to the nearest tenth.

Tony is taking inventory in a warehouse. He's counting tennis shoes. There are 43 pairs of women's tennis shoes and 39 pairs of

n200080 [17]

There would be 82 pairs of shoes in all

5 0

3 years ago

Read 2 more answers

Quick algebra 1 question for 10 points!

Maslowich

The range of the <em>quadratic</em> function y = (2 / 3) · x² - 6 is {- 6, 0, 18, 48}.

<h3>What is the range of a quadratic equation?</h3>

In this case we have a <em>quadratic</em> equation whose domain is stated. The domain of a function is the set of x-values associated to only an element of the range of the function, that is, the set of y-values of the function. We proceed to evaluate the function at each element of the domain and check if the results are in the choices available.

x = - 9

y = (2 / 3) · (- 9)² - 6

y = 48

x = - 6

y = (2 / 3) · (- 6)² - 6

y = 18

x = - 3

y = (2 / 3) · (- 3)² - 6

y = 0

x = 0

y = (2 / 3) · 0² - 6

y = - 6

x = 3

y = (2 / 3) · 3² - 6

y = 0

x = 6

y = (2 / 3) · 6² - 6

y = 18

x = 9

y = (2 / 3) · 9² - 6

y = 48

The range of the <em>quadratic</em> function y = (2 / 3) · x² - 6 is {- 6, 0, 18, 48}.

To learn more on functions: brainly.com/question/12431044

#SPJ1

7 0

2 years ago

You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob

koban [17]

Answer:

a) No, because the probabilities of winning the 2nd game are dependant on the result of the 1st game. The probabilities are different if you win or lose the first game.

b) P=0.42

c) P=0.08

d)

X | P(X)

------------------

0 | 0.42

1 | 0.50

2 | 0.08

e) E(x)=0.66

s.d.=0.62

Step-by-step explanation:

a) No, because the probabilities of winning the 2nd game are dependant on the result of the 1st game. The probabilities are different if you win or lose the first game.

b) The probability of losing the first game is

$P(G_1=L)=1-P(G_1=W)=1-0.4=0.6$

The probability of losing the second game, given that the first game was lost is:

$P(G_2=L|G_1=L)=1-P(G_2=W|G_1=L)=1-0.3=0.7$

So the probability of losing both games is:

$P(G_2=L\&G_1=L)=P(G_1=L)*P(G_2=L|G_1=L)=0.6*0.7=0.42$

c) The probability of winning both games is:

$P(G_2=W\&G_1=W)=P(G_1=W)*P(G_2=W|G_1=W)=0.4*0.2=0.08$

d) The variable X can take values 0, 1 and 2.

X=0 is when both games are lost. This happens with probability P=0.42.

X=2 is when both games are won. This happens with probability P=0.08.

X=1 is when one game is won and the other is lost. This happens with probability P=1-0.42-0.08=0.50.

Then the table of probabilities become:

X | P(X)

------------------

0 | 0.42

1 | 0.50

2 | 0.08

e) The expected value is:

$E(x)=\sum p_ix_i=0.42*0+0.50*1+0.08*2=0.00+0.50+0.16=0.66$

The variance and standard deviation of x are:

$V(x)=\sum p_i(x_i-E(x))^2\\\\V(x)=0.42(0-0.66)^2+0.50(1-0.66)^2+0.08(2-0.66)^2\\\\V(x)=0.42*0.4356+0.50*0.1156+0.08*1.7956\\\\V(x)=0.183+0.058+0.144=0.385\\\\\\\sigma=\sqrt{V(x)}=\sqrt{0.385}=0.62$