If the measure of ZDEF is 65°, what is the measure of DE?

How do you do this problem 2/5+4a=-6/5 what does A = to?

a_sh-v [17]

Answer: A = -2/5

Step-by-step explanation:

- 6/5 - 2/5 = - 8/5 aka - 1 3/5

-1 3/5 ÷ 4 = - 2/5

4 0

3 years ago

James rolls a number cube, with sides labeled 1 through 6, two times. What is the probability James will roll an even number the

charle [14.2K]

Answer:

1/6

Step-by-step explanation:

4 0

3 years ago

Angie and kenny plays online games. Angie buys 1 software package and 1 month of game play. Kenny buys 1 software package 3 mont

Arte-miy333 [17]

kenny 10.32

angie 3.44

5 0

3 years ago

Jason can peel 15 potatoes in 25 minutes. Janette can peel 8 potatoes in 1/10 hour. If they start peeling at the same time how m

xenn [34]

9514 1404 393

Answer:

210 minutes

Step-by-step explanation:

Jason's rate of peeling is ...

(15 potatoes)/(25 minutes) = 15/25 potatoes/minute = 3/5 potatoes/minute

Janette's rate of peeing potatoes is ...

(8 potatoes)/(6 minutes) = 4/3 potatoes/minute

Their combined rate is ...

(3/5 +4/3) = (9 +20)/15 = 29/15 . . . . potatotes/minute

Then the time required for 406 potatoes is ...

(406 potatoes)/(29/15 potatoes/minute) = (406×15/29) minutes

= 210 minutes

6 0

2 years ago

Two teams A and B play a series of games until one team wins three games. We assume that the games are played independently and

Olenka [21]

Answer:

The probability that the series lasts exactly four games is $3p(1-p)(p^{2} + (1 - p)^{2})$

Step-by-step explanation:

For each game, there are only two possible outcomes. Either team A wins, or team A loses. Games are played independently. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

We also need to know a small concept of independent events.

Independent events:

If two events, A and B, are independent, we have that:

$P(A \cap B) = P(A)*P(B)$

What is the probability that the series lasts exactly four games?

This happens if A wins in 4 games of B wins in 4 games.

Probability of A winning in exactly four games:

In the first two games, A must win 2 of them. Also, A must win the fourth game. So, two independent events:

Event A: A wins two of the first three games.

Event B: A wins the fourth game.

P(A):

A wins any game with probability p. 3 games, so n = 3. We have to find P(A) = P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(A) = P(X = 2) = C_{3,2}.p^{2}.(1-p)^{1} = 3p^{2}(1-p)$

P(B):

The probability that A wins any game is p, so P(B) = p.

Probability that A wins in 4:

A and B are independent, so:

$P(A4) = P(A)*P(B) = 3p^{2}(1-p)*p = 3p^{3}(1-p)$

Probability of B winning in exactly four games:

In the first three games, A must win one and B must win 2. The fourth game must be won by 2. So

Event A: A wins one of the first three.

Event B: B wins the fourth game.

P(A)

P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(A) = P(X = 1) = C_{3,1}.p^{1}.(1-p)^{2} = 3p(1-p)^{2}$

P(B)

B wins each game with probability 1 - p, do P(B) = 1 - p.

Probability that B wins in 4:

A and B are independent, so:

$P(B4) = P(A)*P(B) = 3p(1-p)^{2}*(1-p) = 3p(1-p)^{3}$

Probability that the series lasts exactly four games:

$p = P(A4) + P(B4) = 3p^{3}(1-p) + 3p(1-p)^{3} = 3p(1-p)(p^{2} + (1 - p)^{2})$

The probability that the series lasts exactly four games is $3p(1-p)(p^{2} + (1 - p)^{2})$

8 0

3 years ago