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uranmaximum [27]
3 years ago
6

I NEED HELP FAST!!-i chose a random answer.

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
8 0

Answer:

yes, 17 is correct

Step-by-step explanation:

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PLS HELP WILL MARK U BRAINLIEST
Olenka [21]
They are identical lines so they have infinity pairs

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3 years ago
Calculate the missing measure of the line?
labwork [276]

Your answer would be 18.1 cm.

To find this length you have to notice that AC is the hypotenuse of the right-angled triangle ACD. This means that we can find it by using Pythagoras' Theorem (a² + b² = c²), however we need to find the length CD.

We can also find the length CD using Pythagoras' Theorem, because the length CD is the same as the base of the triangle at the top if the shape was split into a rectangle and a triangle.

The height of this triangle would be 11 - 4 = 7 cm, which means we can find CD by doing:

16² - 7² = 256 - 49 = 207 = CD²

Now we can find AC by doing √(CD² + 11²) = √(207 + 121) = √368 = 18.1

I hope this helps! Let me know if you have any questions :)

7 0
3 years ago
Read 2 more answers
Two corporate baseball teams are scheduled to play a game together. They agree that if both teams attend or if neither team atte
sp2606 [1]

Answer:

2.99% probability that the cost will be paid by only one team

Step-by-step explanation:

The binomial probability distribution is used to solve this question.

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.

Probability that a team plays:

A team plays if it has at most 2 injured players out of 11.

11 players, so n = 11

Each player with a 5% probability of injury, so p = 0.05

Then

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

In which

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

P(X = 0) = C_{11,0}.(0.05)^{0}.(0.95)^{11} = 0.5688

P(X = 1) = C_{11,1}.(0.05)^{1}.(0.95)^{10} = 0.3293

P(X = 2) = C_{11,2}.(0.05)^{2}.(0.95)^{9} = 0.0867

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5688 + 0.3293 + 0.0867 = 0.9848

Each team has a 0.9848 probability of showing up to play.

What is the probability that the cost will be paid by only one team?

This happens if one team shows up and the other do not.

2 teams, so n = 2

Each team has a 0.9848 probability of showing up to play, so p = 0.9848.

This probability is P(X = 1).

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

P(X = 1) = C_{2,1}.(0.9848)^{1}.(0.0152)^{1} = 0.0299

2.99% probability that the cost will be paid by only one team

7 0
4 years ago
Given the formula k = Lmn what is the formula for m
Luden [163]
\sf Solve \ for \ m \\  \\ k = lmn \\  \\ Flip \ equation  \\  \\ lmn = k \\  \\ Divide \ both \ sides \ by \ ln \\  \\  \dfrac{lmn}{ln} =  \dfrac{k}{ln}  \\  \\ m =  \dfrac{k}{ln}
8 0
4 years ago
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There are 8 brooms and 6 mops in a janitors closet. What is the ratio of the number of mops to the number of brooms?
STatiana [176]
6/8 Mops first then brooms
simplify ; 3/4

C.
5 0
3 years ago
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