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777dan777 [17]
3 years ago
7

Find the common factorsa. 2a² and 10ab​

Mathematics
2 answers:
tensa zangetsu [6.8K]3 years ago
8 0

Answer:

Step-by-step explanation:

common factor of 2a² and 10ab= 2a

lukranit [14]3 years ago
8 0
2 is a factor of 10
a is a factor of a and a2
only 10ab has b, so b is not a factor
hence the answer is 2a
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Which of the following events below will have an equally likely chance of happening?
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Answer:

B

Step-by-step explanation:

Think of it like a quarter you got a half and half shot if getting heads or tails. meaning a 50/50 chance.

5 0
3 years ago
There were 400 sweets in a pack. The principal of a school bought 25 such packs of sweets for 2000 children o Children's Day. If
AlexFokin [52]

Answer:

10 more packs

Step-by-step explanation:

400 sweets bought 25 such packs = 400*25=10000

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How do i find the intervals on which a function is continuous?
VikaD [51]
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8 0
4 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
PLEASE HELP!!!! I don't understand!
cricket20 [7]
I think the answer is D
7 0
4 years ago
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