Answer:
The answer to your question is the third histogram
Step-by-step explanation:
What we must check in a histogram is that the x-axis is represented the intervals and in the y-axis is represented the frequency.
The first histogram is incorrect just by observing the first bar, we notice that the correct frequency from 0 to 4 is 3, not 14. This histogram is incorrect.
Also, the second histogram is incorrect, the frequency of the first category is 3, not 12. This histogram is wrong.
The third histogram is correct because all the bars are in agreement with their frequencies.
The last histogram is incorrect, for example, the last frequency is 12, not 3.
Answer:
Problem 15)
$40.91
Problem 16)
12.446 cm^2
Step-by-step explanation:
Problem 15)
Gala.
12.8 * 1.35 = 17.28
Honeycrisp.
9 * 1.68 = 15.12
Fuji
7.4 * 1.15 = 8.51
17.28 + 15.12 + 8.51 = 40.91
Problem 16)
(area1) + (area2) + (area3) = (total area)
(1.25*4.38) + (1.36*2.15) + (1.14*3.55) = x
5.475 + 2.924 + 4.047 = x
12.446 = x
Answer:
400
Step-by-step explanation:
Answer:
a = 1
Step-by-step explanation:
1. distribute the -6 on the left into the parentheses. you should get -36a + 30.
2. on the side right of the =, distribute the -2 to get -16 + 10a.
now it should look like -36a + 30 = -16 + 10a.
3. move the terms over to isolate the a value. subtract 30 to the right side to get -46 (-36a=-46+10a)
4. move the 10a on the right to the left by subtracting it to the other side.
(-46a=-46)
5.divide by -46 to get a=1
Answer:
140
Step-by-step explanation:
To construct a subset of S with said property, we have two choices, include 3 in the subset or include four in the subset. These events are mutually exclusive because 3 and 4 can not both be elements of the subset.
First, let's count the number of subsets that contain the element 3.
Any of such subsets has five elements, but since 3 is already an element, we only have to select four elements to complete it. The four elements must be different from 3 and 4 (3 cannot be selected twice and the condition does not allow to select 4), so there are eight elements to select from. The number of ways of doing this is 
.
Now, let's count the number of subsets that contain the element 4.
4 is already an element thus we have to select other four elements . The four elements must be different from 3 and 4 (4 cannot be selected twice and the condition does not allow to select 3), so there are eight elements to select from, so this can be done in ways.
We conclude that there are 70+70=140 required subsets of S.
