Can you check my work

barxatty [35]

The given ratios 3: 5 and 15: 25 are equal. Because when you divide the ratio 15: 25 by 5 on both numerator and denominator, the first ratio 3: 5 can be obtained. Similarly, when you multiply the first ratio 3: 5 by 5, the ratio 15: 25 can be obtained.

5 0

3 years ago

PLZ HELP ME I ONLY HAVE 10 more mins left

IceJOKER [234]

thats the graph for the equation.

hope it helps

3 0

2 years ago

Function g is graphed. Find lim when x approaches 0 (2g(x)). HURRY

Tpy6a [65]

There is a good expensive lot in Minecraft

3 0

3 years ago

If Raj randomly chooses a point in the square below, what is the probability that point is not in the circle? Assume that pi = 3

hodyreva [135]

Answer:

49.7%

Step-by-step explanation:

A cdircle is located within a square.

Area of the circle

Area = $\pi$ $r^{2}$ , where r = 4 units.

Area Circle = 50.3 units^2

Area of the square

Area = l*w or l^2 for a square, since l = w

Area = (10 units)^2

Area = 100 units^2

Area in the square but outside the circle

This is the difference [Square minus Circle Areas]

Square minus Circle Areas = 100 - 50.3 or 49.7 units^2

Probability

The probability of picking a point in the square that is not in the circle is the ration of the two areas: [Outside Circle/Square]x100%

(49.7 units^2)/(100 units^2)x100% = 49.7%

6 0

2 years ago

The probabilty that a student owns a car is 0.65 the porbability that a student owns a compuer is 0.82 the probability that a st

DanielleElmas [232]

Question: The probability that s student owns a car is 0.65, and the probability that a student owns a computer is 0.82.

a. If the probability that a student owns both is 0.55, what is the probability that a randomly selected student owns a car or computer?

b. What is the probability that a randomly selected student does not own a car or computer?

Answer:

(a) 0.92

(b) 0.08

Step-by-step explanation:

(a)

Applying

Pr(A or B) = Pr(A) + Pr(B) – Pr(A and B)................. Equation 1

Where A represent Car, B represent Computer.

From the question,

Pr(A) = 0.65, Pr(B) = 0.82, Pr(A and B) = 0.55

Substitute these values into equation 1

Pr(A or B) = 0.65+0.82-0.55

Pr(A or B) = 1.47-0.55

Pr(A or B) = 0.92.

Hence the probability that a student selected randomly owns a house or a car is 0.92

(b)

Applying

Pr(A or B) = 1 – Pr(not-A and not-B)

Pr(not-A and not-B) = 1-Pr(A or B) ..................... Equation 2

Given: Pr(A or B) = 0.92

Substitute these value into equation 2

Pr(not-A and not-B) = 1-0.92

Pr(not-A and not-B) = 0.08

Hence the probability that a student selected randomly does not own a car or a computer is 0.08

8 0

2 years ago