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2 years ago

Ike plans to save the same amount in the 12th month as in each of the other months.

1 answer:

6 0

People who never develop symptoms or are not yet showing symptoms (pre-symptomatic) might not know that they are infected but can still spread rona to others. Masks also offer protection to the wearer.

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What is the average of 4 1/2, 3 1/3, 2 , 2 1/6

Zinaida [17]

Answer:

Step-by-step explanation:

To find the average add all these numbers together, then divide by how many numbers you have.

$4\frac{1}{2} +3\frac{1}{3} +2+2\frac{1}{6} =12$

$\frac{12}{4} =3$

$Answer: 3$

4 0

3 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let X = number of vehicles that pass the inspection.

The probability of the random variable X is P (X) = 0.70.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

$=(0.70)^{3}\\=0.343$

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

$=1-0.343\\=0.657$

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

+ P (Only 3rd vehicle passes)

$=(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189$

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

+ P (0 vehicles passes)

$=0.189+(0.30\times0.30\times0.30)\\=0.216$

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let X = all 3 vehicle passes and Y = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

$P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525$

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0

2 years ago

What is the value of log^625^5

Tems11 [23]

Answer:

170.84

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

5/x+1 ÷ 6/3x+3 divide

Shalnov [3]

Here is how to work that problem out. (check the picture)

7 0

2 years ago

(-12)/(-3+16/8)^3+2 whats the anwer

Naddika [18.5K]

Answer: I think it's 110.

Step-by-step explanation: I think it's 110, but I can't exactly tell the problem, so do please let me know if this isn't it.

4 0

2 years ago