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3 years ago

Please help hdhdhhdhdhdhhdhdhdhhdhdhdhdhhdhehrhhrhrhrhrhrhrhhehe QUSTION IS IN THE PICTURE

Mathematics

2 answers:

Lana71 [14]3 years ago

7 0

Answer:

No solution

Step-by-step explanation:

6x+3(-2x+4)=-12

6x-6x+12=-12

12≠-12

Mars2501 [29]3 years ago

5 0

Answer:

<em>no solution</em>

Step-by-step explanation:

6x+ 3(-2x+4)=-12

6x-6x+12=-12

12 is not equal to -12.

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(3.6 × 10^-5) ÷ (1.8 × 10^2)

kumpel [21]

Step-by-step explanation:

(3.6 × 10^-5) ÷ (1.8 × 10^2)

2 × 10 ^-7

Hope it helps ya

3 0

3 years ago

Read 2 more answers

(THIS IS ACUALLY ART) PLEASE PLEASE HELP

koban [17]

I think it is d or c idk if I’m right or not tho

5 0

3 years ago

7.12 Equations quiz answers for k12

Darya [45]

Answer:

7.12 equations? K12? whats that

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3 years ago

Waiting on the platform, a commuter hears an announcement that the train is running five minutes late. He assumes the arrival ti

natima [27]

Answer:

D. 91%

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Less than 15 minutes.

Event B: Less than 10 minutes.

We are given the following probability distribution:

$f(T = t) = \frac{3}{5}(\frac{5}{t})^4, t \geq 5$

Simplifying:

$f(T = t) = \frac{3*5^4}{5t^4} = \frac{375}{t^4}$

Probability of arriving in less than 15 minutes:

Integral of the distribution from 5 to 15. So

$P(A) = \int_{5}^{15} = \frac{375}{t^4}$

Integral of $\frac{1}{t^4} = t^{-4}$ is $\frac{t^{-3}}{-3} = -\frac{1}{3t^3}$

Then

$\int \frac{375}{t^4} dt = -\frac{125}{t^3}$

Applying the limits, by the Fundamental Theorem of Calculus:

At $t = 15$ , $f(15) = -\frac{125}{15^3} = -\frac{1}{27}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A) = -\frac{1}{27} + 1 = -\frac{1}{27} + \frac{27}{27} = \frac{26}{27}$

Probability of arriving in less than 15 minutes and less than 10 minutes.

The intersection of these events is less than 10 minutes, so:

$P(B) = \int_{5}^{10} = \frac{375}{t^4}$

We already have the integral, so just apply the limits:

At $t = 10$ , $f(10) = -\frac{125}{10^3} = -\frac{1}{8}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A \cap B) = -\frac{1}{8} + 1 = -\frac{1}{8} + \frac{8}{8} = \frac{7}{8}$

If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{7}{8}}{\frac{26}{27}} = 0.9087$

Thus 90.87%, approximately 91%, and the correct answer is given by option D.

3 0

3 years ago

Draw a picture in the space in the right to show how you would find 1/2 of 8 marbles

777dan777 [17]

Simply present that you would split them up as a factor of two. ( make two lines going either vertically or horizontally. place in varriations, a marble in one line and a marble in the other and then count the leftover

7 0

4 years ago