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riadik2000 [5.3K]
3 years ago
13

Mon wants to make 5 lbs of the sugar syrup. How much water and how much sugar does he need to make 1.5% syrup?

Mathematics
1 answer:
Alona [7]3 years ago
6 0

Answer:

He needs 0.075 lbs of sugar and 4.925 lbs of water.

Step-by-step explanation:

Let x lbs be the amount of sugar in the syrup. Then 5-x lbs is the amount of water in this syrup.

Note

5 lbs - 100%

x lbs - 1.5%

Write a proportion:

\dfrac{5}{x}=\dfrac{100}{1.5}

Cross multiply:

100x=5\cdot 1.5\\ \\100x=7.5\\ \\x=\dfrac{7.5}{100}=0.075

So, he needs 0.075 lbs of sugar and 5 - 0.075 = 4.925 lbs of water.

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Basically your just swapping there positions like so:
if x=-5, then 3-2x=13
Hope this helps! :)
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What is the common denominator of 3/4 and 7/9
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On a test that has a normal distribution, a score of 19 falls one standard deviation below the mean, and a score of 40 falls two
nexus9112 [7]

Hi there!

We can use the following equation:

\large\boxed{z = \frac{x-\mu}{\sigma}}

z = amount of standard deviations away a value is from the mean (z-score)

σ = standard deviation

x = value

μ = mean

Plug in the knowns for both and rearrange to solve for the mean:

-1 = \frac{19-\mu}{\sigma}\\\\-\sigma = 19 - \mu\\\\ \sigma = -19 + \mu

Other given:

2 = \frac{40-\mu}{\sigma}\\\\2\sigma = 40- \mu\\\\ \sigma = 20 - \frac{\mu}{2}

Set both equal to each other and solve:

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2 years ago
A batch of 580 containers for frozen orange juice contains 8 that are defective. Two are selected, at random, without replacemen
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Answer:

a. 0.12109

b. 0.0001668

c .0.9726

d. 0.01038

e. 0.01211

f. 0.000001731

Step-by-step explanation:

Sample size = 580

Defective units = 8

Number of picks = 2

a) If the first container is defective, there 7 defective containers left in a population of 579. The probability of selecting a defective one is:

P=\frac{7}{579}=0.121

b) The probability that both are defective is given by:

P=\frac{8}{580}*\frac{7}{579}= 0.000167

c) The probability that both are acceptable is given by:

P=\frac{580-8}{580}*\frac{579-8}{579}= 0.9726

d) In this case, two defective units were removed from the batch, the probability that the third is also defective is:

P=\frac{6}{578}}= 0.0104

e) In this case, one acceptable and one defective unit were removed from the batch, the probability that the third is also defective is:

P=\frac{7}{578}= 0.01211

f) The probability that all three are defective is given by:

P=\frac{8}{580}*\frac{7}{579}*\frac{6}{578} = 0.000001731

7 0
3 years ago
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