Let A, B and C be the number of pounds of each brand of coffee.
A+B+C=51 (weight) which is 3A+B=51 because C=2A, B=51-3A
15.99A+13.21B+19.90A=51×12.82=653.82 (cost) I’ve used C=2A
35.89A+13.21B=653.82
Substitute for B:
35.89A+13.21(51-3A)=653.82
35.89A+673.71-39.63A=653.82
3.74A=19.89, A=117/22 pounds, B=771/22 pounds, C=117/11 pounds.
Approximate weights are A: 5.32lb, B: 35.05lb, C: 10.64lb.
Answer:
18° and 72°
Step-by-step explanation:
The angle measures are unknown, so they will be represented with the variable, x. Since, one angle is 4 times the other, this will be represented with 4x.
To find complementary angles, they are added together and equal 90. So let's create an equation.
x + 4x = 90
Now solve for x.
5x = 90
5x/5 = 90/5
x = 18
One angle is 18°.
Now find the second angle.
4x
plugin 18 for x and multiply
4(18) = 72
The answer is -3. Adding a negative onto a positive is just like subtracting from the positive. So, first off, 5+(-3) is 2. Then, subtracting 6 from the 2 will give you a negative number, -4. Subtracting a negative changes the sign to be a plus, so you plus one onto the -4. This leaves you with the answer of -3.
I don't know if you have to do this on a graph, or just on a blank space. But, noncollinear means, "not on the same line". So, if you were to draw four points in a blank spot on a piece of paper, just draw four points in random places, making sure they aren't all directly in line with eachother. For a graph, use points like: (-6,2), (5,7), (8,3), and (1,-3). Notice that none of them have the same x or y coordinates, so they can't possibly be collinear.
Answer:
a. 0.76
b. 0.23
c. 0.5
d. p(B/A) is the probability that given that a student has a visa card, they also have a master card
p(A/B) is the probability that given a student has a master card, they also have a visa card
e. 0.35
f. 0.31
Step-by-step explanation:
a. p(AUBUC)= P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBnC)
=0.6+0.4+0.2-0.3-0.11-0.1+0.07= 0.76
b. P(AnBnC')= P(AnB)-P(AnBnC)
=0.3-0.07= 0.23
c. P(B/A)= P(AnB)/P(A)
=0.3/O.6= 0.5
e. P((AnB)/C))= P((AnB)nC)/P(C)
=P(AnBnC)/P(C)
=0.07/0.2= 0.35
f. P((AUB)/C)= P((AUB)nC)/P(C)
=(P(AnC) U P(BnC))/P(C)
=(0.11+0.1)/0.2
=0.21/0.2 = 0.31
