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igomit [66]
3 years ago
10

Need help with this

Mathematics
2 answers:
cupoosta [38]3 years ago
7 0

a²+b² = c² where a and b are legs and c is the hypotenuse

So...

a²+4²=13²

a²+16=169

a²=153

a=√153

answer: approx. 12.37 units

WITCHER [35]3 years ago
5 0

The equation would be a^ + 4^ = 13^

First we need to bring the numbers to the second power, 4 to the second would be 16, 13 to the second power is 169.

Then we subtract 16 from 169, which equals to 153.

Then we "square root" the 153 because of the 2nd power on top of the a

The correct answer would be 12.4

I hope this helps :)

STAY SAFE!! :D

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Answer:

17576

Step-by-step explanation:

Each of the three rotors contained the letters A through Z.

For the first rotor: There are 26 Possible Initial Settings

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For the second rotor: There are 26 possible initial combination with the first rotor likewise.

For the third rotor:There are also 26 possible combinations with the first and second rotors.

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Use a protractor to draw a quadrilateral so that three of its four sides measure 3 centimeters each. The angles between these si
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The length of the fourth side of the quadrilateral is 2.2 cm. Then the correct option is B.

<h3>What is a quadrilateral?</h3>

It is a polygon that has four sides. The sum of the internal angle is 360 degrees.

A quadrilateral so that three of its four sides measure 3 centimeters each.

The angles between these sides have measures of 80° and 85°.

The following procedures are to draw the diagram will be

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The length of the fourth side of the quadrilateral is 2.2 cm

Thus, the correct option is B.

More about the quadrilateral link is given below.

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Answer:

72.92% probability that more than half (that is, 6 or more) carry just one person.

Step-by-step explanation:

For each vehicle, there are only two possible outcomes. Either they carry only one person, or they carry more than one person. The vehicles are chosen at random, which means that the probability of a vehicle carrying just one driver is independent from other vehicles. 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.

64% of vehicles on the roads are occupied by just the driver.

This means that p = 0.64

(a) If you choose 10 vehicles at random, what is the probability that more than half (that is, 6 or more) carry just one person?

This is P(X \geq 6) when n = 10

We have that

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10).

In which

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

P(X = 6) = C_{10,6}.(0.64)^{6}.(0.36)^{4} = 0.2424

P(X = 7) = C_{10,7}.(0.64)^{7}.(0.36)^{3} = 0.2462

P(X = 8) = C_{10,8}.(0.64)^{8}.(0.36)^{2} = 0.1642

P(X = 9) = C_{10,9}.(0.64)^{9}.(0.36)^{1} = 0.0649

P(X = 10) = C_{10,10}.(0.64)^{10}.(0.36)^{0} = 0.0115

So

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Answer:

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