Plllllease help guuuys!!??

A box contains fifteen billiard balls, numbered 1 through 15. Anna draws a ball, records the number, and then returns it to the

Drupady [299]

The correct end part of the question says;

Find the probability of rolling an even number "then" an odd number

Answer:

Probability of rolling even, then odd = 56/225

Step-by-step explanation:

Now, the numbers are from 1 to 15.

Since we want find probability of even, then odd number, let's list all the even numbers first;

Even numbers from 1 to 15 are;

2, 4, 6, 8, 10, 12, 14

Odd numbers from 1 to 15 are;

1, 3, 5, 7, 9, 11, 13, 15

So, we have 7 even numbers and 8 odd numbers.

Probability of even is = 7/15

Probability of odd is = 8/15

Thus,probability of rolling even, then odd is;

P(even, then odd) = 7/15 x 8/15 = 56/225

6 0

4 years ago

A cell phone plan charges a monthly fee of $35 plus $0.05 per minute. Another company does not charge a monthly fee but charges

ehidna [41]

Answer:

it would be 375 minutes

Step-by-step explanation:

7 0

3 years ago

HELP!!

MrRa [10]

Answer:

the correct option is B

Step-by-step explanation:

AC and AB are both radii of circle A

an equilateral triangle is one with all three sides the same length and angle.

from point A of circle A, a compass is use to bisect.

it works because the compasss width is not changed between drawing each side, guaranteeing they are all equal with the same length

using an object and a circular base

1 . Use any cylindrical object with a circular base, each side of the equilateral triangle will be as long as the radius (half the diameter) of the circular object.

2. Draw the first side. It should be exactly as long as the radius of the circular object – the distance halfway across. Make sure that it is perfectly straight!

If you have a ruler: simply measure the diameter of the object and draw a line that is half as long.

If you don't have a ruler: place the circular object onto a sheet of paper, then carefully trace the circumference with your pencil. Remove the object, and you should have a perfect circle. Use a straight edge to draw a line across the exact center of the circle: the point that is completely equidistant from any point around the circumference of the circle.

3. Trace another arc. Now, shift the circular object so that the edge touches the other end of the line segment. Make sure that the line segment runs through the exact center of the circle. Draw another quarter-arc that crosses the first arc at a point directly above the line segment. This point is the apex of your triangle.

4. Complete the triangle. Draw the remaining sides of the triangle: two more straight lines will connect the apex with the two open ends of the line segment.

reference: https://www.wikihow.com/Draw-an-Equilateral-Triangle

3 0

3 years ago

"A supplier of digital memory cards claims that no more than 1% of the cards are defective. In a random sample of 600 memory car

Nady [450]

Answer:

Both the values fall in the critical region so we reject H0: p ≤ 0.01 and conclude that the results are significantly different from chance identification and it is not a sample fluctuation.

Step-by-step explanation:

1) Formulate the hypothesis

H0: p ≤ 0.01 against the claim Ha: p > 0.01

2) Choose the significance level ∝= 0.01

3) The test statistics under H0 is

z= x -np /√npq ( without continuity correction)

z= x ± 1/2 -np /√npq ( with continuity correction)

4) the critical region is z> ±2.33 because the alternate hypothesis is stated on greater than basis.

5)Computations:

Z= x -np /√npq ( without continuity correction)

Z= 6- 600( 0.03)/√ 600( 0.03)(0.97)

z= - 2.82 ( without continuity correction)

z= x ± 1/2 -np /√npq ( with continuity correction)

Z= (6-0.5)- 600( 0.03)/√ 600( 0.03)(0.97)

z= -12.5/4.1785

z= -0.29915 ( with continuity correction)

6) Both the values fall in the critical region so we reject H0: p ≤ 0.01 and conclude that the results are significantly different from chance identification and it is not a sample fluctuation.

8 0

3 years ago

1.Suppose that scores on a knowledge test are normally distributed with a mean of 71 and a standard deviation of 6

Degger [83]

Answer:

a) For this case we can see the distribution in the figure attached is a bell shaped graph and symmetrical around 71

b) $z = \frac{76-71}{6}= 0.833$