Answer:
4x+4y=$40 4 pins and 4 rings
Step-by-step explanation:
Making a table is the easiest for this equation.
6/4=10
12/8=20
18/12=30
24/16=40
Let
A = event that the student is on the honor roll
B = event that the student has a part-time job
C = event that the student is on the honor roll and has a part-time job
We are given
P(A) = 0.40
P(B) = 0.60
P(C) = 0.22
note: P(C) = P(A and B)
We want to find out P(A|B) which is "the probability of getting event A given that we know event B is true". This is a conditional probability
P(A|B) = [P(A and B)]/P(B)
P(A|B) = P(C)/P(B)
P(A|B) = 0.22/0.6
P(A|B) = 0.3667 which is approximate
Convert this to a percentage to get roughly 36.67% and this rounds to 37%
Final Answer: 37%
Answer:
$28
Step-by-step explanation:
40% of 20 is 8
20 + 8 = 28
Answer:
x-coordinate = 0
Step-by-step explanation:
(-10,1) , (5,-5)
m = (y₂ - y₁)/(x₂ - x₁) = ( -5 - 1 ) / ( 5 - [-10] )
= -6 / ( 5 + 10)
= -6/15 = -2/5
y - y₁ = m (x - x₁)
y - 1 = (-2/5) ( x - [-10] )
y - 1 = (-2/5) ( x + 10 )
5 * (y -1) = -2 (x + 10)
5y - 5 = -2x -20
2x + 20 + 5y -5 = 0
2x + 5y + 15 = 0
Another point (a, -3) in on this line
2 * a + 5 * (-3) = -15
2a -15 = -15
2a = -15 +15
2a = 0
a = 0
x-coordinate = 0
Answer:
The z-score for the trainee is of 2.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
The mean of the test scores is 72 with a standard deviation of 5.
This means that 
Find the z-score for a trainee, given a score of 82.
This is Z when X = 82. So



The z-score for the trainee is of 2.