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Vikentia [17]
3 years ago
13

Giving Brainliest correct answer

Mathematics
1 answer:
sammy [17]3 years ago
8 0

Answer:

The answer is C

Step-by-step explanation:

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Anchara drops a penny from a height of 60 feet above the ground. The equation h=-16t^2 + 60 models the penbys height h i feet as
weeeeeb [17]

Answer:

t = 1.94 seconds

Step-by-step explanation:

we have to solve the following polynomial equation:

-16t² + 60 = 0

16t² = 60

t² = 60/16 = 3.75

t = √3.75 = 1.936491673 seconds

the equation also has another possible answer = -1.936491673 seconds, but since time cannot be negative, we have to dismiss that option

7 0
3 years ago
The measure of an angles suppliment is 53° more than twice that of its compliment. find the angle.
Anettt [7]
(180-x)-2(90-x)= 53
Remember that complimentary angles add to 90 and supplementary angles add to 180.

180-x-180+2x=53
x=53

Final answer: The angle is 53 degrees.
3 0
3 years ago
Math please help i have 8 more
Tanzania [10]

Answer:

Step-by-step explanation:

he bout 6 red rose

4 0
3 years ago
Read 2 more answers
Answer the question in the picture. If its right I’ll mark brainliest
Kisachek [45]

Answer:

The answer is c!

Step-by-step explanation:

5 0
3 years ago
Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n 2 if heads comes up
Artyom0805 [142]

Answer:

In the long run, ou expect to  lose $4 per game

Step-by-step explanation:

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.

Assuming X be the toss on which the first head appears.

then the geometric distribution of X is:

X \sim geom(p = 1/2)

the probability function P can be computed as:

P (X = n) = p(1-p)^{n-1}

where

n = 1,2,3 ...

If I agree to pay you $n^2 if heads comes up first on the nth toss.

this implies that , you need to be paid \sum \limits ^{n}_{i=1} n^2 P(X=n)

\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2        ∵  X \sim geom(p = 1/2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =6

Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6

= $4

∴

In the long run, you expect to  lose $4 per game

3 0
4 years ago
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