Okay so I am going to summarize the work out process because its a lot to
Here we go
1/3 (t) + 3/4 - 2/4 - t = ?
1/2 (simplify )
(1/3 (T)+3/4 - 1/2 - (t) = ?
t (2) / 2
1 - 2(t) / 2 = ?
3/4 (simplify this )
1/3(t)+ 3/4 - [1 - 2(t) / 2 = ?
1/3 (this is re last one you have to simplify)
L (Denominator): 3
R (Denominator): 4
L: [L.C.M] : 4
R: [L.C.M] : 3
Basically , we just switched the dominators around
So, Therefore The of t is -3/16
T = -3/16
Answer:
25/64
Step-by-step explanation:
ratio of perimeters = 15/24 = 5/8
ratio of areas = square of ratio of perimeters
ratio of areas = (5/8)^2 = 25/64
Answer: 8√3
<u>Step-by-step explanation:</u>
ΔXVW is a 30°-60°-90° triangle. This special triangle has corresponding sides of lengths: b - b√3 - 2b. So,
VW = b
XW = b√3
XV = 2b
Since, YV = 24, then XW = 24 and we stated above that XW = b√3. So,
24 = b√3

= b
= b
= b
8√3 = b
We are looking for side YX which is congruent (equal) to VW and we stated above that VW = b, So, VW = 8√3
Answer:
2.28% probability that a person selected at random will have an IQ of 110 or higher
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 problem, we have that:

What is the probability that a person selected at random will have an IQ of 110 or higher?
This is 1 subtracted by the pvalue of Z when X = 110. So



has a pvalue of 0.0228
2.28% probability that a person selected at random will have an IQ of 110 or higher