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3 years ago

Magic Video Games, Inc., sells an expensive video games

Mathematics

1 answer:

iVinArrow [24]3 years ago

4 0

Answer:

Magic Video Games, Inc., sells an expensive video games

package. Because the package is so expensive, the company wants to advertise an impressive guarantee

for the life expectancy of its computer control system.

The guarantee policy will refund the full purchase price if the computer fails during the guarantee period.

The research department has done tests that show that the mean life for the computer is 35 months, with standard deviation of 5 months. The computer

life is normally distributed. How long can the guarantee period be if management does not want to refund the purchase price on more than 6.596 of the

Magic Video packages?

Step-by-step explanation:

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A certain virus infects one in every 200 people. a test used to detect the virus in a person is positive 70% of the time when t

V125BC [204]

We're told that

$P(A)=\dfrac1{200}=0.005\implies P(A^C)=0.995$

$P(B\mid A)=0.7$

$P(B\mid A^C)=0.05$

a. We want to find $P(A\mid B)$ . By definition of conditional probability,

$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$

By the law of total probability,

$P(B)=P(B\cap A)+P(B\cap A^C)=P(B\mid A)P(A)+P(B\mid A^C)P(A^C)$

Then

$P(A\mid B)=\dfrac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid A^C)P(A^C)}\approx0.0657$

(the first equality is Bayes' theorem)

b. We want to find $P(A^C\mid B^C)$ .

$P(A^C\mid B^C)=\dfrac{P(A^C\cap B^C)}{P(B^C)}=\dfrac{P(B^C\mid A^C)P(A^C)}{1-P(B)}\approx0.9984$

since $P(B^C\mid A^C)=1-P(B\mid A^C)$ .

4 0

3 years ago

A company that produces fine crystal knows from experience that 17% of its goblets have cosmetic flaws and must be classified as

Alex73 [517]

Answer:

0.3891 = 38.91% probability that only one is a second

Step-by-step explanation:

For each globet, there are only two possible outcoes. Either they have cosmetic flaws, or they do not. The probability of a goblet having a cosmetic flaw is independent of other globets. 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.

17% of its goblets have cosmetic flaws and must be classified as "seconds."

This means that $p = 0.17$

Among seven randomly selected goblets, how likely is it that only one is a second

This is P(X = 1) when n = 7. So

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

$P(X = 1) = C_{7,1}.(0.17)^{1}.(0.83)^{6} = 0.3891$

0.3891 = 38.91% probability that only one is a second

7 0

3 years ago

Simplify 6^7/6^5 in index form

Oksanka [162]

Answer:

6²

Step-by-step explanation:

Using the rule of exponents

$\frac{a^{m} }{a^{n} }$ ⇔ $a^{(m-n)}$ , then

$\frac{6^{7} }{6^{5} }$

= $6^{(7-5)}$

= 6²

6 0

3 years ago

Read 2 more answers

Given that f(x) = x2 – 14x + 45 and g(x) = 2 – 9, find f(x) – g(x) and

seropon [69]

Answer:

x2-14x+52 is the answer in my calculation.

8 0

3 years ago

The sum of the tens digit and the hundreds digit of a number is three times the units digit. 1/5 of the sum of all three digits

Crazy boy [7]

Answer: The numbers are:

965

875

785

695

Step-by-step explanation:

We have a 3 digit number that can be written as:

a*100 + b*10 + c

Where a, b, and c are single-digit numbers.

a is the hundreds

b is the tens

c is the unit.

We know that:

"The sum of the tens digit and the hundreds digit of a number is three times the units digit."

a + b = 3*c

" 1/5 of the sum of all three digits is 1 less than the units digit."

(a + b + c)/5 = c - 1

Then we have the two conditions:

a + b = 3*c

(a + b + c)/5 = c - 1

From the first one, we can write:

(a + b)/3 = c

Replacing that in the other equation we get:

(a + b + a/3 + b/3)/5 = a/3 + b/3 - 1

(4/3)*(a + b)/5 = (a/3 + b/3) - 1

(4/15)*(a + b) = (a + b)/3 - 1

(4/15)*(a + b) - (a + b)/3 = - 1

-(1/15)*(a + b) = -1

a = 15 - b

Then we can give different values for b, and find the values of a and c, where a and c must be positive.

b = 9

a = 15 - 9 = 6

Then:

c = (9 + 6)/3 = 5

(notice that a + b = 15, then c will be always equal to 5, reggardless of the values of b and a).

This number will be:

695.

if b = 8, then:

a = 15 - 8 = 7

and c = 5, same as before.

the number is 785.

If b = 7, then:

a = 15 - 7 = 8

The number is 875

if b = 6, then:

a = 15 - 6 = 9

The number is:

965

4 0

3 years ago