The simplest way to find percent is to take the partial number, divide by the whole number, and multiply the answer by 100%.
= 0.4
0.4 x 100% = 40%.
<u>Answer:</u>
P(3 blue) = 1/5
<u>Steps:</u>
P(3 blue) = 2/3 × 3/5 × 1/2
P(3 blue) = 6/30
P(3 blue) = 1/5
<em>that's 20%</em>
Answer:
144
Step-by-step explanation:
<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u><u> </u>
- Least Common denominator .
<u>Solution</u><u> </u><u>:</u><u>-</u><u> </u>
We have ,
> 1/8 , 2/9 , 3/12 .
The denominator of the fractions are ,
> 8 , 9 , 12
The LCM of 8,9,12 will be ,
2 |<u> </u><u>8</u><u> </u><u>,</u><u> </u><u>9</u><u> </u><u>,</u><u> </u><u>1</u><u>2</u><u> </u>
2 <u>| 4 , 9 , 6</u>
2 |<u> </u><u>2</u><u> </u><u>,</u><u> </u><u>9</u><u> </u><u>,</u><u> </u><u>3 </u>
3 |<u> </u><u>2</u><u> </u><u>,</u><u> </u><u>3 </u><u>,</u><u> </u><u>1</u><u> </u>
Therefore , LCM will be ,
> 2⁴ × 3² = 16 × 9 = 144
Answer:
The probability that the total weight of the passengers exceeds 4222 pounds is 0.0018
Step-by-step explanation:
The Central limit Theorem stays that for a large value of n (21 should be enough), the average distribution X has distribution approximately normal with mean equal to 182 and standard deviation equal to 30/√21 = 6.5465. Lets call W the standarization of X. W has distribution approximately N(0,1) and it is given by the formula

In order for the total weight to exceed 4222 pounds, the average distribution should exceed 4222/21 = 201.0476.
The cummulative distribution function of W will be denoted by
. The values of
can be found in the attached file.

Therefore, the probability that the total weight of the passengers exceeds 4222 pounds is 0.0018.
Equivalent fractions for 3/4 are 6/8 and 9/12
Equivalent fractions for 4/5 is 8/10 and 12/15