Answer:
30,000,000
Step-by-step explanation:
You are given the number 8,531,980,112.45.
Look at attached place value chart.
If we go to the left from decimal point, then we have
- ones;
- tens;
- hundreds;
- thousands;
- ten thousands;
- hundred thousands;
- millions;
- ten millions;
- and so on.
Digit 3 in the number 8,5 <u>3</u> 1,980,112.45 is on ten millions position, so the value of this digit is 30,000,000
Answer:
y = 16x/65
Step-by-step explanation:
Given:
Triangle ABE is similar to triangle ACD. AED and ABC are straight lines
EB and DC are parallel
The area of quadrilateral BCDE = xcm²
The area of triangle ABE = ycm²
Find attached the diagram from the above information.
In similar triangles, the ratio of their corresponding angles are equal.
Also, the ratio of the area of the two triangles = square of ratio of the corresponding sides of the two triangles.
Area ∆ACD/area of ∆ABE = (DC/EB)²
Area ∆ACD/area of ∆ABE = [(area of quadrilateral BCDE +
area of ∆ABE)]/(area of ∆ABE)
(x+y)/y = (DC/EB)²
(x+y)/y = (9/4)²
x+y = (81/16)y
x = (81/16)y - y
x = (81y - 16y)/16
x = 65y/16
Making y subject of formula
16x = 65y
y = 16x/65
An expression for y in terms of x:
y = 16x/65
Answer:
make those cheeks go in circuler motion
Step-by-step explanation:
(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.
The answer is alternate interior angles theorem and substitution property of equality.
