The system:
y = 3 x - 7
15 x - 5 y = 14
---------------------
Using the substitution method:
15 x - 5 · ( 3 x - 7 ) = 14
15 x - 15 x + 35 = 14
0 · x = 14 - 35
0 · x = - 21
x , y ∈ ∅
Answer:
D ) There is no solution.
So,
We'll just use A to represent both Jan and Mya's miles, since they ran the same number.
We have the equations:
1. Jan (J) = Mya (M)
2. Sara (S) = M - 8
3. 2A + S = 64
J = M
S = M - 8
We'll just use A to represent both J and M.
S = M - 8
We'll use Elimination by Substitution.
2A + A - 8 = 64
Collect Like Terms
3A - 8 = 64
Add 8 to both sides
3A = 72
Divide both sides by 3
A = 24
Since
A = J
and
A = M
and
J = M
then
J = 24
M = 24
Substitute
S = 24 - 8
S = 16
Check
24 + 24 + 16 = 64
64 = 64 This checks.
So,
J = 24
M = 24
S = 16
Answer:
(0.102, -0.062)
Step-by-step explanation:
sample size in 2018 = n1 = 216
sample size in 2017 = n2 = 200
number of people who went for another degree in 2018 = x1 = 54
number of people who went for another degree in 2017 = x2 = 46
p1 = x1/n1 = 0.25
p2 = x2/n2 = 0.23
At 95% confidence level, z critical = 1.96
now we have to solve for the confidence interval =
<h2>
</h2>
= 0.02 ± 1.96 * 0.042
= 0.02 + 0.082 = <u>0.102</u>
= 0.02 - 0.082 = <u>-0.062</u>
<u>There is 95% confidence that there is a difference that lies between - 0.062 and 0.102 on the proportion of students who continued their education in the years, 2017 and 2018.</u>
<u></u>
<u>There is no significant difference between the two.</u>
Answer:
Bag 1: 20
Bag 2: 40
Step-by-step explanation:
Let x be the amount taken out of bag 2
Then the amount left in each bag can be written as:
Bag 1: 50-3x
Bag 2: 50-x
Since we know that half of bag 2 is bag 1, that gives us:
50-3x = 1/2(50-x)
-> 50-3x = 25-x/2
Now lets isolate x and solve:
25 = 5x/2
-> 50 = 5x
-> x = 10
So plug x bag in for the original equations:
Bag 1: 50-3x = 50-3(10) = 20
Bag 2: 50-x = 50-10 = 40
Then that would be 3 and 9.
