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2 years ago

Question

Mathematics

2 answers:

mars1129 [50]2 years ago

8 0

Answer:

x=11

Step-by-step explanation:

2x+7=36
2x=36-7
2x=22
x=11

liq [111]2 years ago

7 0

Answer:

Here you go

Step-by-step explanation:

Hope this helps :)

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A lotion is made from an oil blend costing $1.50 per ounce and glycerin costing $1.00 per ounce. Four ounces of lotion costs $5.

vovikov84 [41]

Answer:

A=(4-g) 1.5

Step-by-step explanation:

Let g be the amount of glycerin in the lotion. We know that there are 4 ounces total of lotion; this means that there are 4-g ounces of the oil blend.

The oil blend costs $1.50 per ounce; this gives us the expression (4-g)1.5 for the oil blend.

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Prime numbers up to 30

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2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Step-by-step explanation:

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2 years ago

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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 $3p(1-p)(p^{2} + (1 - p)^{2})$

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3 years ago

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Answer:

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2 years ago

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Hi pls help me out asap 1

faltersainse [42]

Answer:

B: 112.5

Step-by-step explanation:

first, we split up the rhombus into two triangles, with the height of 15 and the length of 7.5. when you multiply these you will get 112.5, and normally if you are just getting the area of the triangle, then you would divide it by two, but seeing that we have 2 triangles here, we would multiply it by two afterwards so an easier way to do it is just to not divide.

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