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Firlakuza [10]
2 years ago
14

What’s the answer to this question

Mathematics
1 answer:
Semenov [28]2 years ago
4 0

Area Formula:

a + b  \times .5 \times h

7+10.4 x .5 x 6.7

Answer: 58.29

I hope this is right!

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the length of a varies string varies inversely as the frequency of its vibrations. A violin string 10 inches long vibrates at a
kolbaska11 [484]

Answer: <u><em>512</em></u>

Step-by-step explanation: <u><em>the length of a varies string varies inversely as the frequency of its vibrations. A violin string 10 inches long vibrates at a frequency of 512 cycles per second. Find the frequency of an 8-inch string.​</em></u>

<u><em /></u>

4 0
3 years ago
Read 2 more answers
What is (x-2)-(x-5)? Plz I will mark u brainliest.
Olenka [21]

Answer: 3

This is the simplified way

7 0
3 years ago
A car insurance company has determined that 8% of all drivers were involved in a car accident last year. Among the 15 drivers li
svetlana [45]

Answer:

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

Step-by-step explanation:

For each driver surveyed, there are only two possible outcomes. Either they were involved in a car accident last year, or they were not. This means that we solve this problem using binomial probability concepts.

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}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

In this problem

15 drivers are randomly selected, so n = 15.

A success consists in finding a driver that was involved in an accident. A car insurance company has determined that 8% of all drivers were involved in a car accident last year.  This means that \pi = 0.08.

What is the probability of getting 3 or more who were involved in a car accident last year?

This is P(X \geq 3).

Either less than 3 were involved in a car accident, or 3 or more were. Each one has it's probabilities. The sum of these probabilities is decimal 1. So:

P(X < 3) + P(X \geq 3) = 1

P(X \geq 3) = 1 - P(X < 3)

In which

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{15,0}.(0.08)^{0}.(0.92)^{15} = 0.2863

P(X = 1) = C_{15,1}.(0.08)^{1}.(0.92)^{14} = 0.3734

P(X = 2) = C_{15,2}.(0.08)^{2}.(0.92)^{13} = 0.2273

So

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2863 + 0.3734 + 0.2273 = 0.887

Finally

P(X \geq 3) = 1 - P(X < 3) = 1 - 0.887 = 0.113

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

5 0
3 years ago
Round 1.530 to the nearest hundred​
Bumek [7]
1.53

Since it is 0 in the thousands place it, doesn’t change.
4 0
2 years ago
Can somone really help me with this! Thank You! ( Giving out Brainly)
prohojiy [21]

Answer:

U= (1,-6)

V= (4,-6)

W= (1,-9)

3 0
2 years ago
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