How much times does 18 go into 109

Tommy will spend start fraction, 1/5 of his time this weekend studying for his 4 final exams. What fraction of the weekend will

maxonik [38]

Answer:

1/20

Step-by-step explanation:

(1/5) ÷ 4

*remember, you don't divide fractions, you multiply the first fraction by the inverse (flip it upside down) of the second fraction.*

(1/5) × (1/4) = (1/20)

8 0

3 years ago

Read 2 more answers

Please help‍♂️ I forgot

vladimir2022 [97]

12 because the 23 3/8 is rounded down and the 11 2/9 is rounded down as well

3 0

3 years ago

On Martin's first stroke, his golf ball traveled \dfrac45 5 4 start fraction, 4, divided by, 5, end fraction of the distance t

makvit [3.9K]

Answer: 0.395 km

Step-by-step explanation:

Let Martin distance from the hole be X

On first stroke, his golf ball traveled 4/5 of the distance to the hole. That is 4/5 X

On his second stroke, the ball traveled 79 meters and went into the hole

Total distance covered will be

X = 79 + 4/5X

X - 4/5X = 79

X - 0.8X = 79

0.2X = 79

X = 79/0.2 = 395 meters

How many kilometers from the hole was Martin when he started

X = 395/1000 = 0.395 km

6 0

3 years ago

Asia new boat has a cruising speed of 24 miles per hour at that rate how long will it take her to travel the 6 miles across old

Igoryamba

Answer:

1/4 of an hour, or 60/4=15, 15 minutes.

Step-by-step explanation:

24/1=6/x

cross product

1*6=24*x

6=24x

x=6/24

x=1/4

3 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.

(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