34 baseball cards. 2 times more baseball cards than his friend thomas.how many baseball cards does thomas have?

A new bookcase costs $76 if you buy it already painted the color you want. If you buy it unfinished (not painted at all) then it

lapo4ka [179]

23.68421053%

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8 0

3 years ago

Help please it is due today

Svet_ta [14]

Answer:

Jane's situation is the one which represents a porportional relationship.

Step-by-step explanation:

To verify the proportionality betwen a ratio you have to divide all the ratios and all them have to get the same constant factor.

So, if you divide the matt's ratios you aren't gonna find the same constant factor

12/1=12

20/2=10

31/2=15,5

But if you try with Jane's ones:

12/1=12

24/2=12

36/3=12

This 12 is the known constant factor, which show us that these ratios are proportional.

8 0

3 years ago

PLEASE HELP!! Write the equation of the graph.

Jet001 [13]

When you first look at thius graph, you should notice two things immedietly.

1. This graph will have a positive slope, because the line is going up.

2. the y-intercept is -2 because that's where it is on the y-axis.

Now, you have been given two points, (2, 1) and (0, -2). You can put these two points into the slope formula, which is: m = $\frac{y2 - y1}{x2 - x1}$ $\frac{(-2) - 1}{0 - 2}$ = $\frac{3}{2}$ = 3/2.

So, now you have your slope and your y-intercept. Now, just put them into y = mx + b form!

y = 3/2x - 2 (it is ubtracting two instead of adding two because it is a negative two.)

Hope this helps!!
~Kiwi

8 0

2 years ago

Find the x and y intercepts for the equation 2x + 7y = 42.

o-na [289]

Answer:

see explanation

Step-by-step explanation:

Given

2x + 7y = 42

To find the x- intercept let y = 0 in the equation and solve for x

2x + 7(0) = 42, that is

2x = 42 ( divide both sides by 2 )

x = 21 ← x- intercept or (21, 0 )

To find the y- intercept let x = 0 in the equation and solve for y

2(0) + 7y = 42 , that is

7y = 42 ( divide both sides by 7 )

y = 6 ← y- intercept or (0, 6 )

6 0

3 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