Answer:
12
Step-by-step explanation:
There are 4 different digits to choose from. Since we want 2-digit number, we will see what digits can we put in place of 1st digit and what digit can we place in place of 2nd digit.
Hence, <u><em>how many digits (from 1,5,6 and 8) can be placed in the 1st digit of the number we want?</em></u>
Any one of the four digits (1,5,6, or 8).
Now, <em><u>How many digits (from 1,5,6 and 8) can be placed as the 2nd digit of the number we want?</u></em>
Any one of the THREE digits (since repetition is not allowed, we disregard the initial digit).
Thus the number of two-digit positive integers is 4 * 3 = 12
Answer: 1st is if it’s 2 legs second is if it’s a leg and hypotenuse
Step-by-step explanation:
Answer:
x ≤ 4
Step-by-step explanation:
6x + 3 ≤ 27 subtract 3 both sides
6x ≤ 24 divide 6 both sides
x ≤ 4 There's your answer
Answer:
Step-by-step explanation:
I don't think this can be done without the diagram. You do not know what HD is opposite. I will take a guess that it is opposite TU which makes TU = 220 because both H and D are midpoints and that makes TU twice as large as HD.
If this is incorrect, post the diagram.
Answer:
0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.
Step-by-step explanation:
Problems of normally distributed samples are solved using 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 the diameter of a selected bearing is greater than 85 millimeters.
This is 1 subtracted by the pvalue of Z when X = 85. Then



has a pvalue of 0.7486.
1 - 0.7486 = 0.2514
0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.