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3 years ago

PLEASE HELP WITH THISS.......

Mathematics

1 answer:

Mashcka [7]3 years ago

5 0

GF = ½ RT

Step-by-step explanation:

SG=GR (given)

SF=FT (given)

line GF is a midpoint in triangle RST, so it is therefore ½ the angle parallel to it(viz. RT)

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(a/c)²+(b/c)²=1

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3 years ago

A basketball player's free throw percentage is 82.5%, or 0.825.

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Answer:

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825/1000

Step-by-step explanation:

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1 year ago

Ben has ran for president of his class if there are 400 and 10 students and he received 72% of the vote. How many students were

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Answer:

287

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3 years ago

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IceJOKER [234]

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3 years ago

Read 2 more answers

The random variable x takes on the values 1, 2, or 3 with probabilities (1 + 3k)/3, (1 + 2k)/3, and (0.5 +5k)/3, respectively.(a

givi [52]

Answer:

a) $\frac{1+3k}{3} + \frac{1+2k}{3} +\frac{0.5+5k}{3}= 1$

We can multiply both sides by 3 and we got:

$1+3k+ 1+2k + 0.5+5k = 3$

$10 k = 3-1-0.5-1=0.5$

$k = 0.05$

And if we replace we got:

X 1 2 3

P(X) 0.383 0.367 0.25

b) $E(X)= \sum_{i=1}^n X_i P(X_i) = 1*0.383 + 2*0.367 +3*0.25= 1.867$

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)=1^2*0.383 + 2^2*0.367 +3^2*0.25= 4.101$

$Var(X) = E(X^2) -[E(X)]^2 = 4.101- [1.867]^2 =0.615$

c) X 1 2 3

F(X) 0.383 0.367+0.383=0.75 0.25+0.75 = 1

Step-by-step explanation:

For this case we have the following probability mass function given:

X 1 2 3

P(X) (1+3k)/3 (1+2k)/3 (0.5+5k)/3

Part a

In order to satisfy the definition of probability distribution the sum of all the probabilities needs to be 1 and each of the individual probabilities needs to be higher or equal than 0, using this we can do that:

$\frac{1+3k}{3} + \frac{1+2k}{3} +\frac{0.5+5k}{3}= 1$

We can multiply both sides by 3 and we got:

$1+3k+ 1+2k + 0.5+5k = 3$

$10 k = 3-1-0.5-1=0.5$

$k = 0.05$

And if we replace we got:

X 1 2 3

P(X) 0.383 0.367 0.25

And we satisfy all the conditions.

Part b

For this case we can find the mean using this formula:

$E(X)= \sum_{i=1}^n X_i P(X_i) = 1*0.383 + 2*0.367 +3*0.25= 1.867$

For the variance we need to find the second moment first like this:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)=1^2*0.383 + 2^2*0.367 +3^2*0.25= 4.101$

And we can find the variance with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 4.101- [1.867]^2 =0.615$

Part c

The cumulative distribution on this case is given by:

X 1 2 3

F(X) 0.383 0.367+0.383=0.75 0.25+0.75 = 1

4 0

3 years ago