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3 years ago

How many zeroes are in the product of (6 x 5) x10 to the power of 3

Mathematics

1 answer:

Karo-lina-s [1.5K]3 years ago

4 0

4 zeros as:
(6x5) x 10^3
(30) x 10^3
30 x 1000
30,000

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Read 2 more answers

What is the value of x?

Tom [10]

Answer:

x=21

Step-by-step explanation:

100+3x+x-4=180

100+4x-4=180

96+4x=180

4x=180-96

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3 years ago

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

How do i do this? and what’s the answer?

Trava [24]

This is a parallelogram, and the answer is 8.04

8 0

3 years ago