Answer:
How much water is in an empty glass that is 10cm high and has a diameter of 5 cm ?
= 196,25 cm
Step-by-step explanation:
diameter = 5 cm
radius = 1/2 × diameter
radius = 1/2 × 5 = 2,5
How much water is in an empty glass that is 10cm high and has a diameter of 5 cm?
= π × r × r × high
= 3,14 × 2,5 × 2,5 × 10
= 196,25 cm
Answer:
0.5%/year
24.2%
Step-by-step explanation:
Estimate the average yearly increase in the percentage of first-year college females claiming no religious affiliation
Percentage of females by year:
1980 = 6.2%
1990 = 10.8%
2000 = 13.6%
2012 = 21.7%
Average yearly increase :
Percentage increase between 1980 - 2012 :
2012% - 1980% = ( 21.7% - 6.2%) = 15.5% increase over [(2012 - 1980)] = 32 years
15.5 % / 32 years = 0.484375% / year = 0.5%/year
b. Estimate the percentage of first-year college females who will claim no religious affiliation in 2030,
Given an average increase of 0.484375% / year
(2030 - 1980) = 50 years
Hence by 2030 ; ( 50 years × 0.484375%/year) = 24.218% will claim no religious affiliation.
=24.2% (nearest tenth)
<u>Answer:</u>
If PQ=RS then PQ and RS have the same length. Hence option D is correct
<u>Solution:</u>
Given that, pq = rs
And, we have to find which of the given options are true.
<u><em>a) pq and rs form a straight angle
</em></u>
We can’t decide the angle in between pq and rs just by the statement pq = rs.
So this statement is false.
<u><em>b) pq and rs form a zero angle.
</em></u>
We can’t decide the angle in between pq and rs just by the statement pq = rs.
So this statement is false.
<u><em>c) pq and rs are same segment.
</em></u>
If two things equal then there is no condition that both represents a single item.
So this statement is false.
<u><em>d) pq and rs have the same length
</em></u>
As given that pq = rs, we can say that they will have the same length
Hence, option d is true.
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
