Mrs. Robertson is taking a picture of three students-Zach, John, and

9.9 divided by 80.19 Google is usually wrong ._.

lions [1.4K]

0.123456789 not even joking thats the answer I did it on a calculator ‍♀️

5 0

3 years ago

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How do you solve this ???

galina1969 [7]

First you have to figure out how much a video game costs and how much a used one costs
Then you plug in the costs of the video games into Janets equation 120=3x+y
(x= video games and y=used video games)
Then you subtract the cost of video games from 120 and then divide that answer by the cost of used video games and that should give you how many used video games she can get

3 0

3 years ago

Ryan goes the mall with his friends after school. At the mall he purchases four pairs of basketball shorts and a pair of sneaker

Korvikt [17]

Answer:

$25.00

Step-by-step explanation:

200-100=100

100/4=25

4 0

3 years ago

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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.

(c) The probability that exactly 1 of the next three vehicles passes is 0.189.

(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

3 years ago

HELP PLEASE!!!! WILL GIVE BRAINLIEST ANY LINKS OR WRONG ANSWERS WILL BE REPORTED!!!!

jarptica [38.1K]

Answer:

The numerator is 70.

Step-by-step explanation:

70% converts to the fraction of 70/100. Therefore, the numerator is 70.

Hope you have a great day,

Ginny

3 0

3 years ago

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