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Nadusha1986 [10]
4 years ago
11

Sam typed 570 words and 15 minutes if he continues to type at this rate how many words will he have typed in two hours

Mathematics
2 answers:
leonid [27]4 years ago
7 0
120 minutes in 2 hours
120/15=8
570*8=4560 words
Hope this helps!
Svetllana [295]4 years ago
7 0
So first, you will have to find out how many words he types in 1 minute, so you divide 570 by 15 and then you'll get 38, so Sam types 38 words per minute. 
Then, we know that 2 hours is equal to 120 minutes. So you multiply 38 by 120 because you want to calculate how many words Sam types in 120 minutes. 
So your answer should be 4,560 words in 2 hours. 

I hope this helped (:
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You’ve bought a half-dozen (six) eggs from the store but you forgot to check them first! The probability that no eggs are broken
GREYUIT [131]

Answer:

a)

P(X = 0) = C_{6,0}.(0.1416)^{0}.(0.8584)^{6} = 0.4

P(X = 1) = C_{6,1}.(0.1416)^{1}.(8584)^{5} = 0.3960

P(X = 2) = C_{6,2}.(0.1416)^{2}.(8584)^{4} = 0.1633

P(X = 3) = C_{6,3}.(0.1416)^{3}.(8584)^{3} = 0.0359

P(X = 4) = C_{6,4}.(0.1416)^{4}.(8584)^{2} = 0.0044

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P(X = 6) = C_{6,6}.(0.1416)^{6}.(8584)^{0} = 0.00001

b) 56.77% probability that an even number of eggs is broken.

c)

Expectation: 0.8496

Variance: 0.7293

Step-by-step explanation:

For each egg, there are only two possible outcomes. Either it is broken, or it is not. The probability of an egg being broken is independent from other eggs. So we use the binomial probability distribution 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.

The expected value of the binomial distribution is:

E(X) = np

The variance of the binomial distribution is:

V(X) = np(1-p)

There are 6 eggs

So n = 6

The probability that no eggs are broken is 0.4.

This means that P(X = 0) = 0.4. So

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

P(X = 0) = C_{6,0}.p^{0}.(1-p)^{6}

(1 - p)^{6} = 0.4

Taking the sixth root from both sides of the equality

(1 - p) = 0.8584

p = 0.1416

Each egg has a 0.1416 probability of being broken

(a) Write out the pmf of X.

Probability of each value, from 0 to 6

P(X = 0) = C_{6,0}.(0.1416)^{0}.(0.8584)^{6} = 0.4

P(X = 1) = C_{6,1}.(0.1416)^{1}.(8584)^{5} = 0.3960

P(X = 2) = C_{6,2}.(0.1416)^{2}.(8584)^{4} = 0.1633

P(X = 3) = C_{6,3}.(0.1416)^{3}.(8584)^{3} = 0.0359

P(X = 4) = C_{6,4}.(0.1416)^{4}.(8584)^{2} = 0.0044

P(X = 5) = C_{6,5}.(0.1416)^{5}.(8584)^{1} = 0.0003

P(X = 6) = C_{6,6}.(0.1416)^{6}.(8584)^{0} = 0.00001

(b) Compute the probability that an even number of eggs is broken.

0, 2, 4 or 6

P = P(X = 0) + P(X = 2) + P(X = 4) + P(X = 6) = 0.4 + 0.1633 + 0.0044 + 0.00001 = 0.5677

56.77% probability that an even number of eggs is broken.

(c) Compute the expectation and variance of X.

Expectation:

E(X) = np = 6*0.1416 = 0.8496

Variance:

V(X) = np(1-p) = 6*0.1416*0.8584 = 0.7293

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Have a nice day!

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