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k0ka [10]
3 years ago
7

Which is the best estimate for the mass of a pair of eyeglasses? A. 25 grams B. 250 grams C. 25 kilograms D. 250 kilograms

Mathematics
1 answer:
Yuliya22 [10]3 years ago
4 0
About A. 25 grams because glasses are really not heavy at all and the other options are way too illogical.
Hope this helps!
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ASAP PLEASE ANSWER
pickupchik [31]

Answer:

B

Step-by-step explanation:

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7 0
3 years ago
What is the slope of the line passing through the points (1, 2) and (5, 4)?​
bulgar [2K]
<h2>SOLVING</h2>

\Large\maltese\underline{\textsf{A. What is Asked}}

What is the slope of the line passing through the point (1,2) and (5,4)

\Large\maltese\underline{\textsf{This problem has been solved!}}

Formula used, here  \bf{\dfrac{y2-y1}{x2-x1}

_______________________________________________________

\bf{\dfrac{4-2}{5-1} | simplify

\bf{\dfrac{2}{4} | reduce

\bf{\dfrac{1}{2}

\rule{300}{1.7}

\bf{Result:}

         \bf{=Slope:\dfrac{1}{2}

\boxed{\bf{aesthetic\not101}}

3 0
2 years ago
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What is the area of the composite figure?( 20 points)​
kicyunya [14]
The answer is B. 20.285
5 0
3 years ago
What is 1.9 million plus 8.7 million?<br><br> Thanks so much!!!!!
Tju [1.3M]
The answer is ten million six hundred thousand.
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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
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