For the first digit you are choosing from 2 digits
for the second digit you are choosing from 2 digits
for the third digit you are choosing from 2 digits
2*2*2=8
111
110
011
101
000
001
100
010
However......
A number doesn't usually start with a 0 or 0s.
Therefore, if you want 3-digit numbers and not just permutations using 0 and 1, then you must eliminate
011, 000, 001, and 010
Seeing that the first digit can't be 0, you choose from 1, then 2, and then 2 digits again; 1*2*2=4 numbers
You choose which answer best suits your problem.
Answer:
Step-by-step explanation:
A = 2(½(20)(15)) + 16(15 + 20 + 25) = 1,260 in²
Answer:
13) x = 30
14) x = 22.5
15) x = 36
Step-by-step explanation:
13) 4x+2x = 180
6x = 180
x = 30
14) 3x+5x = 180
8x = 180
x = 22.5
15) 2x+3x+2x+3x = 360
10x = 360
x = 36
Answer:
57.49% probability that a randomly selected individual has an IQ between 81 and 109
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean
and standard deviation
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:

Find the probability that a randomly selected individual has an IQ between 81 and 109
This is the pvalue of Z when X = 109 subtracted by the pvalue of Z when X = 81. So
X = 109



has a pvalue of 0.67
X = 81



has a pvalue of 0.0951
0.67 - 0.0951 = 0.5749
57.49% probability that a randomly selected individual has an IQ between 81 and 109