Problem 3: Let x = price of bag of pretzels Let y = price of box of granola bars
We have Lesley's purchase: 4x+2y=13.50
And Landon's: 1x+5y=17.55
We can use the elimination method. Let's negate Landon's purchase by multiplying by -1. -1x-5y=-17.55
We add this four times to Lesley's purchase to eliminate the x variable.
2y-20y=13.50-70.2
-18y=-56.7
y = $3.15 = Price of box of granola bars
Plug back into Landon's purchase to solve for pretzels.
x+5*3.15=17.55
x+15.75=17.55
x = $1.80 = price of bag of pretzels
Problem 4.
Let w = number of wood bats sold
Let m = number of metal bats sold
From sales information we have: w + m = 23
24w+30m=606
Substitution works well here. Solve for w in the first equation, w = 23 - m, and plug this into the second.
24*(23-m)+30m=606
552-24m+30m=606
6m=54
m=9 = number of metal bats sold
Therefore since w = 23-m, w = 23-9 = 14. 14 wooden bats were sold.
Given:
425 school children
Teacher : 80
Doctor : 105
Lawyer : 70
Police Officer : 70
Firefighter : 100
Total = 425
After applying Pearson's chi-squared test, it is concluded that this data reflects a real difference in the population.
<span>5x-6y=-15
5x+6y=12
---------------add
10x = -3
x = -3/10
x = -0.3
</span>5x+6y=12
5(-0.3) + 6y = 12
6y = 12 + 1.5
6y = 13.5
y = 13.5 / 6
y = 2.25
(1/36) = 0.0277777777778
(1/108)^3 = 7.9383224102<span> x 10^-7 </span>
(1/9)^4 = 0.000152415790276
(1/6)^2 = 0.0277777777778
(1/2)^5 = <span>0.03125
The only one that matches with the value of 1/36 is (1/6)^2. Therefore, your answer is C. (1/6)^2
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