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.
Answer:
whats the shape?
Step-by-step explanation:
Answer:
13/6
Step-by-step explanation:
1 Simplify \sqrt{8}
8
to 2\sqrt{2}2
2
.
\frac{2}{6\times 2\sqrt{2}}\sqrt{2}-(-\frac{18}{\sqrt{81}})
6×2
2
2
2
−(−
81
18
)
2 Simplify 6\times 2\sqrt{2}6×2
2
to 12\sqrt{2}12
2
.
\frac{2}{12\sqrt{2}}\sqrt{2}-(-\frac{18}{\sqrt{81}})
12
2
2
2
−(−
81
18
)
3 Since 9\times 9=819×9=81, the square root of 8181 is 99.
\frac{2}{12\sqrt{2}}\sqrt{2}-(-\frac{18}{9})
12
2
2
2
−(−
9
18
)
4 Simplify \frac{18}{9}
9
18
to 22.
\frac{2}{12\sqrt{2}}\sqrt{2}-(-2)
12
2
2
2
−(−2)
5 Rationalize the denominator: \frac{2}{12\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{2\sqrt{2}}{12\times 2}
12
2
2
⋅
2
2
=
12×2
2
2
.
\frac{2\sqrt{2}}{12\times 2}\sqrt{2}-(-2)
12×2
2
2
2
−(−2)
6 Simplify 12\times 212×2 to 2424.
\frac{2\sqrt{2}}{24}\sqrt{2}-(-2)
24
2
2
2
−(−2)
7 Simplify \frac{2\sqrt{2}}{24}
24
2
2
to \frac{\sqrt{2}}{12}
12
2
.
\frac{\sqrt{2}}{12}\sqrt{2}-(-2)
12
2
2
−(−2)
8 Use this rule: \frac{a}{b} \times c=\frac{ac}{b}
b
a
×c=
b
ac
.
\frac{\sqrt{2}\sqrt{2}}{12}-(-2)
12
2
2
−(−2)
9 Simplify \sqrt{2}\sqrt{2}
2
2
to \sqrt{4}
4
.
\frac{\sqrt{4}}{12}-(-2)
12
4
−(−2)
10 Since 2\times 2=42×2=4, the square root of 44 is 22.
\frac{2}{12}-(-2)
12
2
−(−2)
11 Simplify \frac{2}{12}
12
2
to \frac{1}{6}
6
1
.
\frac{1}{6}-(-2)
6
1
−(−2)
12 Remove parentheses.
\frac{1}{6}+2
6
1
+2
13 Simplify.
\frac{13}{6}
6
13
Done
The volume of a rectangular prism is represented by the following equation:
Where the variables are for volume, width, height, and length, respectively.
We are given that the area of one end is 16 cm² (units have to be correct when solving these problems, so it's 16 cm², not 16 cm as described in the problem). We know that 
Using this knowledge, we can change the volume equation to our needs.
Note: We know that A is 16 since it's given
The volume is 208 cm³ (once again, incorrect units given). Insert this into the equation.
Divide both sides of the equation by 16.
The length is 13 cm.
Let me know if you need any clarifications, thanks!
The sample space is 22 words and out of it, 14 words are known to have meanings for the student. In the problem, we use 10 C8 + 10 C9 + 10 C10 to associate at least 8 words. The complete formula is (10 C8 * (14/22)^8 * (8/22)^2 + 10 C9 (14/22)^9 * (8/22)^1 + 10 C 10 (14/22)^10 * (8/22)^)0) that is equal to 0.233
