Answer:
2 is proportional because they all equal one when I divide them
I think :)
Answer:
(a) P(X≥10) = 0.245
(b) P(X≤5) = 0.125
(c) The expected number of inmates who are serving time for drug dealing is 8.
Step-by-step explanation:
We will use the binomial distribution to solve this problem. Let X be the number of federal inmates who are serving time for drug dealing. We will use the binomial probability formula:
P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ
where n = total number of federal inmates
x = no. of federal inmates who serve time for drug dealing
p = probability that a federal inmate is serving time for drug dealing
q = probability that a federal inmate is not serving time for drug dealing
(a) P(X≥10) = 1 - P(X<10)
= 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9)]
= 1 - [²⁰C₀(0.4)⁰(0.6)²⁰⁻⁰ + ²⁰C₁(0.4)¹(0.6)²⁰⁻¹ + ²⁰C₂(0.4)²(0.6)²⁰⁻² + ²⁰C₃(0.4)³(0.6)²⁰⁻³ + ²⁰C₄(0.4)⁴(0.6)²⁰⁻⁴ + ²⁰C₅(0.4)⁵(0.6)²⁰⁻⁵ + ²⁰C₆(0.4)⁶(0.6)²⁰⁻⁶ + ²⁰C₇(0.4)⁷(0.6)²⁰⁻⁷ + ²⁰C₈(0.4)⁸(0.6)²⁰⁻⁸ + ²⁰C₉(0.4)⁹(0.6)²⁰⁻⁹
= 1 - (0.000036 + 0.00048 + 0.00308 + 0.0123 + 0.0349 + 0.0746 + 0.1244 + 0.1658 + 0.1797 + 0.1597)
= 1 - 0.7549
P(X≥10) = 0.245
(b) P(X≤5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)
since we have calculated these values in the previous part, we will simply plug them in here and then add them up.
P(X≤5) = 0.000036 + 0.00048 + 0.00308 + 0.0123 + 0.0349 + 0.0746
P(X≤5) = 0.125
(c) For a binomial distribution, the expected value can be calculated as:
μ = np
= (20)(0.4)
μ = 8
The expected number of inmates who are serving time for drug dealing is 8.
The fourth option, C=3L+50
The heights (in inches) of 13 plants are 6, 9, 10, 10, 10, 11, 11, 12, 12, 13, 14, 16, and 17. What is the interquartile range o
bonufazy [111]
The given heights:
6, 9, 10, 10, 10, 11, 11, 12, 12, 13, 14, 16, 17.
The first step is to find Median in order to divide the above series into lower and upper quartiles.
To find the median, just cross out the first and the last value until you left with the single value, as the number of elements is odd.
Like first 6 would cancel the last 17,
second number 9 would cancel the second last number 16,
and so on.
By doing above exercise, we are left with 11.
So the median = 11.
Now any number below the median 11 would be in lower quartile series, while other half would be in upper one.
To find the lower quartile, find the median of 6, 9, 10, 10, 10, 11:
Now the numbers are even; therefore, we would do the above-mentioned exercise of cancelling first and the last element until we are left with just 2 elements:
The two elements are: 10, 10.
Take average = (10 + 10) / 2 = 10
So lower quartile = 10
To find the upper quartile, find the median of 12, 12, 13, 14, 16, 17:
Now the numbers are even; therefore, we would do the above-mentioned exercise of cancelling first and the last element until we are left with just 2 elements:
The two elements are: 13, 14.
Take average = (13 + 14) / 2 = 13.5
So Upper quartile = 13.5
The interquartile range = Upper Quartile - Lower Quartile = 13.5 - 10.0 = 3.5
So the correct option is (A) 3.5
-i
Terms: 2x, 3y, x, 7
constants: 7
coefficients: 2x, 3y
like terms:2x, x
