All categories

3 years ago

Make D the subject of S=D/T You may use the flowchart to help you if you wish.

1 answer:

8 0

Given the equation, S = D/T:

To make the D the subject of the equation, we can start by multiplying both sides by T to isolate D:

(T) S = D/T (T)

TS = D

Therefore, the final answer is D = TS

You might be interested in

There are 115 students in a music club 73 of them are girls what is the ratio of the number of boys and girls

REY [17]

Answer:

73:42

Step-by-step explanation:

1115-73=42

G : B

73 : 42

8 0

3 years ago

What is the area of the triangle

NikAS [45]

First find the length of the longest leg

x^2 = 13^2 - 5^2 ( by the Pythagoras theorem)
x^2 = 144
x = 12

Area of the traingle = 1/2 * base * height
= 1/2 * 5 * 12
= 30 units^2

3 0

3 years ago

15 points ! Find m < RQP

Phoenix [80]

Answer:

<RQP = 70°

Step-by-step explanation:

<RQP = 1/2 (140)

<RQP = 70°

4 0

3 years ago

Which shows 50% of the area shaded?

Murljashka [212]

Answer:

B shows 50% shaded

Step-by-step explanation:

Count how many triangles are in it (4) then see how many are shaded (2/4).

Simplify it to 1/2=50% then you got your answer.

4 0

4 years ago

Read 2 more answers

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

3 years ago