0. 6 0 0
5 3. 0 0 0
− 0
3 0
− 3 0
0 0
− 0
0 0
− 0
0
The 3 lbs of candy is irrelevant, so we can discard that. All you need to do is divide 147 by 26, which is 5.65 pieces of candy. Each student will get 5 pieces, with a few left over.
On #29 x = 12 and <apb = 90° and <dcb = 40° and on #30 all I dont even understand, sorry
Answer:
A tree with a height of 6.2 ft is 3 standard deviations above the mean
Step-by-step explanation:
⇒
statement: A tree with a height of 5.4 ft is 1 standard deviation below the mean(FALSE)
an X value is found Z standard deviations from the mean mu if:

In this case we have: 

We have four different values of X and we must calculate the Z-score for each
For X =5.4\ ft

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.
⇒
statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean.
(FALSE)
For X =4.6 ft

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean
.
⇒
statement:A tree with a height of 5.8 ft is 2.5 standard deviations above the mean
(FALSE)
For X =5.8 ft

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.
⇒
statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean.
(TRUE)
For X =6.2\ ft

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.