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2 years ago

44+88= __ (2+4) 11 6 2 22

Mathematics

2 answers:

DENIUS [597]2 years ago

8 0

Dang I was just about to say 22

Naddik [55]2 years ago

5 0

Answer:

ist 22

Step-by-step explanation:

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In a box of 25 external hard disks, there are 2 defectives. An inspector examines 5 of these hard disks. Find the probability th

Salsk061 [2.6K]

Answer:

0.3667 or 36.67%

Step-by-step explanation:

The probability of getting at least 1 defective hard disk among the 5 (P(X>0)) is equal to 100% minus the probability of getting none defectives (1-P(X=0)).

If 23 out of 25 hard disks are non-defective, the probability is:

$P(X>0) = 1 -P(X=0)\\P(X>0) = 1 -\frac{23}{25}*\frac{22}{24} *\frac{21}{23} *\frac{20}{22} *\frac{19}{21}\\ P(X>0) = 0.3667=36.67\%$

The probability that there is at least 1 defective hard disk is 0.3667 or 36.67%.

8 0

3 years ago

Given: O is the midpoint of MN OM = OW Prove: OW = ON

mafiozo [28]

Answer is in the attachment below.

3 0

3 years ago

Read 2 more answers

Use substitution to solve the system of equations :

mote1985 [20]

Answer:

x=1

y=-4

Step-by-step explanation:

4x+y=0 -> y= -4x

2(-4x)+x=-7

-8x+ x=-7

-7x=-7

x=-7/-7=1

y=-4 . 1 = -4

3 0

2 years ago

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection.

This means that,

Probability that at least one of them needs replacement at the first inspection =

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

5 0

3 years ago

Solve for x. What is x/15 = 21/5?

melisa1 [442]

Answer: x = 63

Step-by-step explanation:

15/3 = 5

3 * 21 = 63

63/15 = 21/5

8 0

3 years ago

Read 2 more answers