Step-by-step explanation:
If half of the tuna sandwiches were on white bread then 21 of them would be white, meaning the other half would be on brown bread.
Then if 25% of ham sandwiches were on brown bread then 25% of 32 would be 8, so the final 75% would be on white bread, which would be 24.
Finally, with 20 more white bread sandwiches sold than brown bread it couldn’t be 80 and 20, it would be 40 brown bread in total and 60 white bread in total adding up to 100, as it says white sells 20 more than brown not only 20.
To finish this question you would add up the tuna and ham for white (21+24=45) then minus that from 60 (60-45=15) so there were 15 cheese sandwiches on white bread. And then for brown (21+8=29) then minus that from 40 (40-29=11) so therefore there are 11 cheese sandwiches on brown bread.
Answer:
Brown:
Tuna-21
Cheese-11
Ham-8
Total-40
White:
Tuna-21
Cheese-15
Ham-24
Total-60
Hope that helps!
Answer:
m∠E= 37°
m∠F=37°
Step-by-step explanation:
Total angle value: 180°
180-106=74
The lengths are the same, so we can determine that the angles E and F are the same. So we divide 74 by 2
74/2= 37
m∠E= 37°
m∠F=37°
Hope this helps! Have a nice day :)
Answer:
Step-by-step explanation:
Given
The attached triangle
Required
Find cos V
In trigonometry:
In this case:
Where
and
-- Pythagoras
Take square roots
So:
Rationalize:
Answer:
1716 ;
700 ;
1715 ;
658 ;
1254 ;
792
Step-by-step explanation:
Given that :
Number of members (n) = 13
a. How many ways can a group of seven be chosen to work on a project?
13C7:
Recall :
nCr = n! ÷ (n-r)! r!
13C7 = 13! ÷ (13 - 7)!7!
= 13! ÷ 6! 7!
(13*12*11*10*9*8*7!) ÷ 7! (6*5*4*3*2*1)
1235520 / 720
= 1716
b. Suppose seven team members are women and six are men.
Men = 6 ; women = 7
(i) How many groups of seven can be chosen that contain four women and three men?
(7C4) * (6C3)
Using calculator :
7C4 = 35
6C3 = 20
(35 * 20) = 700
(ii) How many groups of seven can be chosen that contain at least one man?
13C7 - 7C7
7C7 = only women
13C7 = 1716
7C7 = 1
1716 - 1 = 1715
(iii) How many groups of seven can be chosen that contain at most three women?
(6C4 * 7C3) + (6C5 * 7C2) + (6C6 * 7C1)
Using calculator :
(15 * 35) + (6 * 21) + (1 * 7)
525 + 126 + 7
= 658
c. Suppose two team members refuse to work together on projects. How many groups of seven can be chosen to work on a project?
(First in second out) + (second in first out) + (both out)
13 - 2 = 11
11C6 + 11C6 + 11C7
Using calculator :
462 + 462 + 330
= 1254
d. Suppose two team members insist on either working together or not at all on projects. How many groups of seven can be chosen to work on a project?
Number of ways with both in the group = 11C5
Number of ways with both out of the group = 11C7
11C5 + 11C7
462 + 330
= 792
