The perimeter is 49 miles, which means that all the sides added up must equal 49. Let's write and equation with the information given, where x is the magnitude of the two sides:
Simplify this a little bit to get x on one side:
Divide by 2 on both sides, and your answer is:
Therefore, the length of the two sides of the trapezoid are both 7 miles.
The answer is 252 pi units.
The surface area of a right cylinder is
So, given height is = 15 units
Given radius is = 6 units.
So, putting these values in the formula, we get:
Hence, the answer is option D.
0.5 < t < 2
The function reaches its maximum height at ...
t = -b/(2a) = -16/(2(-16)) = 1/2 . . . . . . where a=-16, b=16, c=32 are the coefficients of f(t)
The function can be factored to find the zeros.
f(t) = -16(t^2 -1 -2) = -16(t -2)(t +1)
The factors are zero for ...
x = -1 and x = +2
The ball is falling from its maximum height during the period (0.5, 2), so that is a reasonable domain if you're only interested in the period when the ball is falling.
y = x + 6
The y-intercept is (0, 6). This is where the line crosses the y-axis.
The slope is 4/4, or 1: m = 1
The equation of the line is therefore y = x + 6
At a carnival, the probability that a person will win a prize at the ring-toss game has been found to be (1/20). What is the probability that a prize will be won by exactly
a. 1 of the next 5 players?
b. 3 of the next 5 players?
c. 0 of the next 5 players?
d. 2 of the next 20 players?
This is a binomial distribution problem
A binomial experiment is one in which the probability of success doesn't change with every run or number of trials.
It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure.
The outcome of each trial/run of a binomial experiment is independent of one another.
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = 5, 20
x = Number of successes required = 1, 3, 0, 2
p = probability of success = probability of winning a prize in the ring-toss = (1/20) = 0.05
q = probability of failure = probability of NOT winning a prize in the ring-toss = 1 - (1/20) = (19/20) = 0.95
n = total number of sample spaces = 5
x = Number of successes required = 1
p = probability of success = 0.05
q = probability of failure = 0.95
P(X = 1) = ⁵C₁ (0.05)¹ (0.95)⁵⁻¹ = 0.2036265625 = 0.20363
b) 3 of the next 5 players?
x = Number of successes required = 3
P(X = 3) = ⁵C₃ (0.05)³ (0.95)⁵⁻³ = 0.001128125 = 0.00113
x = Number of successes required = 0
P(X = 0) = ⁵C₀ (0.05)⁰ (0.95)⁵⁻⁰ = 0.7737809375 = 0.77378
n = total number of sample spaces = 20
x = Number of successes required = 2
P(X = 2) = ²⁰C₂ (0.05)² (0.95)²⁰⁻² = 0.18867680127 = 0.18868
Hope this Helps!!!