The perimeter of a rectangle is 34 feet. if the length is 2 more than 4 times the width, find the dimensions of this rectangle

A pyramid of lcm x 1cm x 1cm cubes is formed. The bottom layer has 1 anrrangement of these cubes. The second layer has a 9 angan

loris [4]

Answer:

1444 uf FC fun with that one is a little bit ago and now I have to do it again in a bit of the day I will be there in a few days

6 0

2 years ago

HELP PLEASE ASAP 55 POINTS I BEG YOU!!!!!!

Alja [10]

Answer:1 b+1T+1 U=5617b+1 T =9062b+7T+5U = 1758

Step-by-step explanation:

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

2 years ago

Three cards are chosen at random from a standard 52-card deck. What is the probability that they are not all the same color?

rewona [7]

Three cards are selected from a standard deck of 52 cards. Disregarding the order in which they are drawn, the possible outcomes are (52/3). Out of these, how many include all cards of the same color (say red)? There are (13/3) ways in which you can get all 13 red cards.

8 0

3 years ago

At a store, notebooks are on sale for 78% off the regular price. The cost of a notebook

NARA [144]

Answer: C. n-0.78n

n−0.78n

=n+−0.78n

=(n+−0.78n)

=0.22n

4 0

3 years ago