What is the equation of the line passing through the point (1,2) and (-2,5)

Need help will give brainlest!!

Evgesh-ka [11]

Answer:

14/20

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Which line segment is a diameter of circle L A.GK B.HK C.IL D.JL

serious [3.7K]

B is the answer because diameter goes all the way through a circle

4 0

2 years ago

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Assign two variables for each problem, and write the equations. Do not solve.

GalinKa [24]

Call the amount of pineapple "P', and the amount of gingerale "G".

We know that a mix of pineapple and gingerale makes 200 cups. So:
$P+G=200$

We also know that double the pineapple (i.e. 2P) and triple the gingerale (i.e. 3G) makes 420 cups. So:
$2P+3G=420$

We can now solve by substitution:
$2P+3(200-P)=420$
$2P+600-3P=420$
$-P=-180$
$P=180$

$180+G=200$
$G=20$

For 420 cups we need 2P=360 cups of pineapple and 3G=60 cups of gingerale]

3 0

2 years ago

Find an antiderivative F(x) with F′(x) = f(x) = 6 + 24x^3 + 18x^5 and F(1)=0.

7nadin3 [17]

Answer:

The antiderivative is $F(X) = 6x + 6x^4 + 3x^6 - 15$ .

Step-by-step explanation:

Antiderivative F(x)

This is the integral of $F^{\prime}(x)$

So

F′(x) = f(x) = 6 + 24x^3 + 18x^5

Then:

$F(x) = \int (6 + 24x^3 + 18x^5) dx$

$F(x) = 6x + \frac{24x^4}{4} + \frac{18x^6}{6} + K$

$F(x) = 6x + 6x^4 + 3x^6 + K$

F(1)=0

$F(X) = 0$ when $x = 1$ . We use this to find K.

$F(x) = 6x + 6x^4 + 3x^6 + K$

$0 = 6 + 6 + 3 + K$

$K = -15$

Thus

The antiderivative is $F(X) = 6x + 6x^4 + 3x^6 - 15$ .

7 0

2 years ago

Every day your friend commutes to school on the subway at 9 AM. If the subway is on time, she will stop for a $3 coffee on the w

Shtirlitz [24]

Answer:

1.02% probability of spending 0 dollars on coffee over the course of a five day week

7.68% probability of spending 3 dollars on coffee over the course of a five day week

23.04% probability of spending 6 dollars on coffee over the course of a five day week

34.56% probability of spending 9 dollars on coffee over the course of a five day week

25.92% probability of spending 12 dollars on coffee over the course of a five day week

7.78% probability of spending 12 dollars on coffee over the course of a five day week

Step-by-step explanation:

For each day, there are only two possible outcomes. Either the subway is on time, or it is not. Each day, the probability of the train being on time is independent from other days. So we use the binomial probability distribution to solve this problem.

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.

In this problem we have that:

The probability that the subway is delayed is 40%. 100-40 = 60% of the train being on time, so $p = 0.6$

The week has 5 days, so $n = 5$

She spends 3 dollars on coffee each day the train is on time.

Probabability that she spends 0 dollars on coffee:

This is the probability of the train being late all 5 days, so it is P(X = 0).

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

$P(X = 0) = C_{5,0}.(0.6)^{0}.(0.4)^{5} = 0.0102$

1.02% probability of spending 0 dollars on coffee over the course of a five day week

Probabability that she spends 3 dollars on coffee:

This is the probability of the train being late for 4 days and on time for 1, so it is P(X = 1).

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

$P(X = 1) = C_{5,1}.(0.6)^{1}.(0.4)^{4} = 0.0768$

7.68% probability of spending 3 dollars on coffee over the course of a five day week

Probabability that she spends 6 dollars on coffee:

This is the probability of the train being late for 3 days and on time for 2, so it is P(X = 2).

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

$P(X = 2) = C_{5,2}.(0.6)^{2}.(0.4)^{3} = 0.2304$

23.04% probability of spending 6 dollars on coffee over the course of a five day week

Probabability that she spends 9 dollars on coffee:

This is the probability of the train being late for 2 days and on time for 3, so it is P(X = 3).

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

$P(X = 3) = C_{5,3}.(0.6)^{3}.(0.4)^{2} = 0.3456$

34.56% probability of spending 9 dollars on coffee over the course of a five day week

Probabability that she spends 12 dollars on coffee:

This is the probability of the train being late for 1 day and on time for 4, so it is P(X = 4).

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

$P(X = 4) = C_{5,4}.(0.6)^{4}.(0.4)^{1} = 0.2592$

25.92% probability of spending 12 dollars on coffee over the course of a five day week

Probabability that she spends 15 dollars on coffee:

Probability that the subway is on time all days of the week, so $P(X = 5)$ .

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

$P(X = 5) = C_{5,5}.(0.6)^{5}.(0.4)^{0} = 0.0778$

7.78% probability of spending 12 dollars on coffee over the course of a five day week

8 0

3 years ago