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chubhunter [2.5K]
3 years ago
10

Of new calculators tested,8 were defective and 42 passed inspection. What ratio compares the number of defective calculators to

the number of new calculators?
Mathematics
2 answers:
sesenic [268]3 years ago
4 0
The number of new calculators tested was 50 . . . 8 were bad and 42 were good.

The ratio of defective ones to new ones tested is 8:50 or 4:25 or 16% .
tatuchka [14]3 years ago
3 0
You simply put the 2 numbers together to find the answer, which is 8:42.  

You want to find the ratio in its simplest form.  Therefore, you want to find the lowest common multiple of 8 and 42.  
2 goes into both numbers.  
8 / 2 = 4.  
42 / 2 = 21

The answer is 4:21. 
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3 years ago
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8 over 15? is it closer to 0, 1 over 2 or 1
ruslelena [56]
8/15 is about 0.53
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4 0
3 years ago
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By what number should-15/56 be divided to get -5/6<br><br>pls pls pls pls I need your help ​
mr_godi [17]

Answer:

The desired divisor is n = 21

Step-by-step explanation:

Let that number be n.  Then:

-15/56        -5

---------- = ---------

     n            6

Through cross-multiplication we get:

(-15/56)*6 = -5n, which reduces to:

(-15)(7) = -5n, or 3(7) = n

The desired divisor is n = 21

6 0
3 years ago
Donna the trainer has two solo workout plans that she offers her clients: Plan A and Plan B. Each client does either one or the
kykrilka [37]
X=hours of plan A
y=hours of plan B

wed: 5x+6y=12
thu: 3x+2y=6

so
5x+6y=12 and
3x+2y=6
we can eliminate y's by multiplying 2nd equation by -3 and adding to first equation

-9x-6y=-18
<u>5x+6y=12 +</u>
-4x+0y=-6

-4x=-6
divid both sides by -4
x=-6/-4
x=3/2
x=1.5

sub back

3x+2y=6
3(1.5)+2y=6
4.5+2y=6
2y=1.5
y=0.75



Plan A=1.5 hours
Plan B=0.75 hours
3 0
3 years ago
If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w
MatroZZZ [7]

Let x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11} be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.

2. If the mean weight of the 7 other players is 202 lb, then

\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.

3. From the previous statements you have that

x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.

Add these two equalities and then divide by 11:

x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.

Answer: the mean weight of the 11-person team is 208\dfrac{10}{11}\ lb.

4 0
3 years ago
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