Help me with this pls

Find the value of the variable and DF if D is between C and F if CD 4y -9, DF 2y-7, and CF 14.

Yuri [45]

CF = CD + DF (since D is simply a point on the line, and C to that point added onto F to that point is CF), so 4y-9+2y-7=14. Adding it up, we get 6y-16=14 and adding by 16 we get 30=6y. Dividing both sides by 6, we get y=5. Plugging that into DF, we get 10-7=DF=3

4 0

2 years ago

The school's computer lab goes through 5 reams of printer paper every 6 weeks. Find out how long 3 cases of printer paper is li

djyliett [7]

So to start 3*8=24 reams of paper
24 divided by 5=4.8
4.8*6=28.8 weeks.
It's up to you if you want to do 28 weeks or 29 weeks

7 0

2 years ago

3 over 5 x minus 15 = 6 over 5 x plus 12

poizon [28]

3 over 5 x minus 15 = 6 over 5 x plus 12

3/5x - 15 = 6/5x +12
-3/5x+3/5x-15=6/5x-3/5x+12 subtraction property
-15=3/5x+12 simplified
-12-15=3/5x+12-12 subtraction property
-27=3/5x simplified
(5/3)(-27)=3/5 (5/3)x multiplicative inverse
-45=x

8 0

2 years ago

Tickets for the basketball game cost 14$ each. If the sale of the tickets brought in 2,212$, how many tickets were sold?

rosijanka [135]

15.78 is the answer to the qustion that you have asked.

5 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

2 years ago