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Anestetic [448]
3 years ago
10

Find the length of the third side to the nearest 10th

Mathematics
2 answers:
Serggg [28]3 years ago
7 0

Answer: 2.8

Step-by-step explanation:

VARVARA [1.3K]3 years ago
5 0
The other side is 15
You might be interested in
Solve −2(n+3)−4=8 . The solution is n= .
Anton [14]

Answer:

n= -9

Step-by-step explanation:

Simplifying

-2(n + 3) + -4 = 8

Reorder the terms:

-2(3 + n) + -4 = 8

(3 * -2 + n * -2) + -4 = 8

(-6 + -2n) + -4 = 8

Reorder the terms:

-6 + -4 + -2n = 8

Combine like terms: -6 + -4 = -10

-10 + -2n = 8

Solving

-10 + -2n = 8

Solving for variable 'n'.

Move all terms containing n to the left, all other terms to the right.

Add '10' to each side of the equation.

-10 + 10 + -2n = 8 + 10

Combine like terms: -10 + 10 = 0

0 + -2n = 8 + 10

-2n = 8 + 10

Combine like terms: 8 + 10 = 18

-2n = 18

Divide each side by '-2'.

n = -9

Simplifying

n = -9

3 0
3 years ago
Worth 15 points please answer asap!!!!
ExtremeBDS [4]
So do the opposite of the answer like for example 4 +3\2 and your answer is y
the you do the rest
6 0
3 years ago
Find the range of the parent function below y=x^2
marusya05 [52]

you don't show the graph, but graphing y=x^2 it is a U shaped line that starts at (0,0) and the lines go upwards on both sides of the Y axis,

 This means all real numbers above 0 and 0 will solve it

 so the range is y≥0

6 0
3 years ago
Suppose a basketball player has made 231 out of 361 free throws. If the player makes the next 2 free throws, I will pay you $31.
statuscvo [17]

Answer:

The expected value of the proposition is of -0.29.

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either the player will make it, or he will miss it. The probability of a player making a free throw is independent of any other free throw, which means that 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.

Suppose a basketball player has made 231 out of 361 free throws.

This means that p = \frac{231}{361} = 0.6399

Probability of the player making the next 2 free throws:

This is P(X = 2) when n = 2. So

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

P(X = 2) = C_{2,2}.(0.6399)^{2}.(0.3601)^{0} = 0.4095

Find the expected value of the proposition:

0.4095 probability of you paying $31(losing $31), which is when the player makes the next 2 free throws.

1 - 0.4059 = 0.5905 probability of you being paid $21(earning $21), which is when the player does not make the next 2 free throws. So

E = -31*0.4095 + 21*0.5905 = -0.29

The expected value of the proposition is of -0.29.

3 0
2 years ago
A. -3/2<br><br> b. 3/2<br><br> c. 2/3 <br><br> d. -2/3
zubka84 [21]
D. -2/3 because when you do (y2-y1)/(x2-x1) you get -4/6 which simplifies to -2/3
7 0
3 years ago
Read 2 more answers
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