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Goryan [66]
2 years ago
13

Wich statement is true about the graphs of the two lines y=-6 and ×=1/6?No links ​

Mathematics
1 answer:
bagirrra123 [75]2 years ago
7 0
Is it multiple choice? If so please say the the possible answers in the comments
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While watching a parade I saw some clowns and horses. I counted 30 legs and 10 heads. How many horses did I see in the parade?
emmasim [6.3K]
The answer is 5 horses
7 0
3 years ago
PLEASE HELP ME <br>...This is my homework and I don't understand how to do it.​
musickatia [10]

The probability that a point chosen at random lies in the shaded region is 0.28

<h3>Calculating the area of a shaded region and probability</h3>

From the question, we are to find the probability that a point chosen at random lies in the shaded region.

The shaded region is a triangle.

The probability that a point chosen at random lies in the shaded region = Area of the triangle / Area of the circle

First, we will calculate the unknown side of the triangle

Let the unknown side be x.

Then, from the <em>Pythagorean theorem</em>, we can write that

12² = 6² + x²

144 = 36 + x²

x² = 144 - 36

x² = 108

x = √108

x = 6√3

From formula,

Area of a triangle = 1/2 × base × height

Area of the triangle = 1/2 × 6 × 6√3

Area of the triangle = 18√3 square units

Now, we will determine the area of the circle

Area of a circle = πr²

Where r is the radius

From the given information,

Diameter of the circle = 12

But,

Radius = Diameter / 2

Therefore,

r = 12/2

r = 6

Thus,

Area of the circle = 3.14 × 6²

Area of the circle = 113.04 square units

Now,

The probability that a point chosen at random lies in the shaded region = 18√3 / 113.04

The probability that a point chosen at random lies in the shaded region = 0.2758

The probability that a point chosen at random lies in the shaded region ≈ 0.28

Hence, the probability is 0.28

Learn more on Calculating area of a shaded region here: brainly.com/question/23629261

#SPJ1

4 0
1 year ago
Select the statements that describe a normal distribution.
Snezhnost [94]

Answer: The statements that describe a normal distribution are;

a. The density curve is symmetric and bell-shaped.

b. The normal distribution is a continuous distribution.

Step-by-step explanation: The normal distribution is the most commonly used and important statistic tool. It is referred to as the "Bell Curve" because of its bell-shape and the the fact that it is symmetric density curve. A continuous distribution defines the possibilities of a continuous random variable and a prime example of a continuous distribution is the Normal distribution.

The normal distribution is not a discrete distribution because it does not have discrete variables. The normal distribution is not a flat line that extends from a minimum to a maximum but it is a continuous distribution that extends in a bell shape from one minimum value going up to a maximum value before descending back to another minimum value.

68% of a normal distribution curve falls with one standard deviation from the mean not 32%.

The two parameters that define a normal distribution is the mean and the standard deviation.

3 0
3 years ago
Geoffrey draws a triangle. One of the angles measures 75° and one of the angles measures 40°. What is the measure of the third a
valina [46]

Answer:

D) 65°

Step-by-step explanation:

Remember, all the angles of a triangle are 180 degrees and 75 + 40 + 65 = 180 degrees. Please if you can, support my channel. In the images you can find the information ;)

4 0
2 years ago
A college basketball player makes 80% of his freethrows. Over the course of the season he will attempt 100 freethrows. Assuming
BARSIC [14]

Answer:

The probability that the number of free throws he makes exceeds 80 is approximately 0.50

Step-by-step explanation:

According to the given data we have the following:

P(Make a Throw) = 0.80%

n=100

Binomial distribution:

mean:   np = 0.80*100= 80

hence, standard deviation=√np(1-p)=√80*0.20=4

Therefore, to calculate the probability that the number of free throws he makes exceeds 80 we would have to make the following calculation:

P(X>80)= 1- P(X<80)

You could calculate this value via a normal distributionapproximation:

P(Z<(80-80)/4)=1-P(Z<0)=1-50=0.50

The probability that the number of free throws he makes exceeds 80 is approximately 0.50

3 0
3 years ago
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