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3 years ago

Is 6 a divisor of 18 why

1 answer:

7 0

Six is a divisor of 18 because 6 x 3 = 18. A number that fits perfectly into 18 is called a divisor or factor. 1, 2, 3, 6, 9 and 18 are all factors of 18

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4. The figure shows a rectangle with a width of 6k inches. The length of the rectangle is 4 inches shorter than the width.

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Answer:

Step-by-step explanation:

sorry if its not correct but i hope it helps

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3 years ago

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Evaluate 13⋅13....<br> We are working on exponents X3 <br> Please answer ^^<br>

andrey2020 [161]

Answer:

13*13= 169

Step-by-step explanation:

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4 years ago

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A television production company charges $22001 for 3 hours of work and $40574 for 7 hours of work. Find the hourly rate.

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A box of Georgia peaches has 3 bad and 12 good peaches. (a) If you make a peach cobbler of 12 peaches randomly selected from the

Eddi Din [679]

Answer:

a) 0.21% probability that there are no bad peaches in the peach cobbler.

b) 99.79% probability of having at least 1 bad peach in the peach cobbler

c) 7.91% probability of having exactly 2 bad peaches in the peach cobbler.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the peaches are chosen is not important. So the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

(a) If you make a peach cobbler of 12 peaches randomly selected from the box, what is the probability that there are no bad peaches in the peach cobbler?

Desired outcomes:

12 good peaches, from a set of 12. So

$D = C_{12,12} = \frac{12!}{12!(12 - 12)!} = 1$

Total outcomes:

12 peaches, from a set of 15. So

$T = C_{15,12} = \frac{15!}{12!(15 - 12)!} = 455$

Probability:

$p = \frac{D}{T} = \frac{1}{455} = 0.0021$

0.21% probability that there are no bad peaches in the peach cobbler.

(b) What is the probability of having at least 1 bad peach in the peach cobbler?

Either there are no bad peaches, or these is at least 1. The sum of the probabilities of these events is 100%. So

p + 0.21 = 100

p = 99.79

99.79% probability of having at least 1 bad peach in the peach cobbler

(c) What is the probability of having exactly 2 bad peaches in the peach cob- bler?

Desired outcomes:

2 bad peaches, from a set of 3.

One good peach, from a set of 12.

$D = C_{3,2}*C_{12,1} = \frac{3!}{2!(3-2)!}*\frac{12!}{1!(12 - 1)!} = 36$

Total outcomes:

12 peaches, from a set of 15. So

$T = C_{15,12} = \frac{15!}{12!(15 - 12)!} = 455$

Probability:

$p = \frac{D}{T} = \frac{36}{455} = 0.0791$

7.91% probability of having exactly 2 bad peaches in the peach cobbler.

3 0

3 years ago

Two fair dice are rolled.

Natalka [10]

Answer:

1/12

Step-by-step explanation:

There are two dices being rolled. A dice has six faces numbered from one to six. A fair dice means that the probability of each number to appear is equal. Thus the probability of any number showing up is:

P(a number appears on the dice) = how much times the number has been displayed on the dice/number of faces of the dice.

Since all numbers appear once, and there are six sides of a dice. therefore:

P(a number appears on the dice) = 1/6. Thus, P(6 appears on a dice) = 1/6.

As far as the odd numbers are concerned, there are three even numbers and three odd numbers on the dice. So P(odd number appears on the dice) = 3/6 = 1/2.

Assuming that the probabilities of both the dices are independent, we can safely multiply both the probabilities. Thus:

P(first dice lands on a 6 and second dice lands on an odd number) = 1/6 * 1/2 = 1/12.

Thus, the final probability is 1/12!!!

3 0

3 years ago