John has two different sugar syrups. Syrup A is 2 parts of sugar and 8 parts of water, and Syrup B is 3 parts of sugar and 5 parts of water. How much of each should john mix together to get 280 quarts of syrup that is 3 parts sugar and 7 parts water.

Answers

Answer 1

Answer:

Volume of syrup A = 120 quarts

Volume of syrup B = 280 - 120 = 160 quarts .

Step-by-step explanation:

Syrup A : 2 parts sugar , 8 parts of water .

Total = 10 parts

Syrup B : 3 parts sugar , 5 parts of water .

Total = 8 parts .

Mixture required 280 quarts , 3 parts sugar , 7 parts water

In 280 quarts , sugar = 280 x 3  / ( 3 + 7 )

sugar = 84 parts

Let volume of syrup A be x , volume of syrup B = 280 - x

sugar A =  2 x / 10 = 0.2 x

sugar B = 3 ( 280 - x ) / 8 = .375 ( 280 - x )

Total sugar =  .2 x + .375 ( 280 - x ) = 84 .

.2 x + 105 - .375 x = 84

105 - 84 = .375 x - .2 x = .175 x

.175 x = 21

x = 21 / .175 = 120

Volume of syrup A = 120 quarts

Volume of syrup B = 280 - 120 = 160 quarts .

Answer 2

John mixes 120 quarts of syrup A and 160 quarts of syrup of B together to get 280 quarts of syrup that is 3 parts sugar and 7 parts water.

Given that,

John has two different sugar syrups.

Syrup A is 2 parts of sugar and 8 parts of water, and Syrup B is 3 parts of sugar and 5 parts of water.

We have to determine,

How much of each should john mix together to get 280 quarts of syrup that is 3 parts sugar and 7 parts water.

According to the question,

Syrup A: 2 parts sugar, 8 parts of water.

Total number of parts = 10 parts

Syrup B : 3 parts sugar, 5 parts of water.

The total number of parts  = 8 parts.

The mixture required 280 quarts, 3 parts sugar, 7 parts water,

In 280 quarts, parts of sugar are,

[tex]= \dfrac{280 \times 3 }{ ( 3 + 7 )}\\\\= \dfrac{280 \times 3}{10}\\\\= 84[/tex]

 

Let the volume of syrup A be x,

And the volume of syrup B be 280 - x

[tex]Sugar\ A = \dfrac{2x}{10} \\\\Sugar \ A = 0.2 x[/tex]

[tex]Sugar \ B = \dfrac{3 ( 280 - x ) }{8 }\\\\Sugar \ B= .375 ( 280 - x )[/tex]

The total amount of sugar is,

[tex]0.2 x + 0.375 ( 280 - x ) = 84 \\\\0.2 x + 105 -0 .375 x = 84\\\\105 - 84 = 0.375 x - 0.2 x = 0.175 x\\\\ 0.175 x = 21\\\\x = \dfrac{21}{0.175}\\\\x = 120[/tex]

Therefore,

The volume of syrup A = 120 quarts

Volume of syrup B = 280 - 120 = 160 quarts

Hence, John mixes 120 quarts of syrup A and 160 quarts of syrup of B together to get 280 quarts of syrup that is 3 parts sugar and 7 parts water.

For more details, refer to the link.

https://brainly.com/question/11921451


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A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift. A quality control consultant is to select 6 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 slips without replacement from among 45). (Enter your answers to four decimal places.)

Required:
a. How many selections result in all 7 workers coming from the day shift?
b. What is the probability that all 7 selected workers will be from the day shift?
c. What is the probability that all 7 selected workers will be from the same shift?
d. What is the probability that at least two different shifts will be represented among the selected workers?
e. What is the probability that at least one of the shifts will be unrepresented in that sample of workers?

Answers

Answer:

Step-by-step explanation:

Total number of workers = 20+ 15 + 10 = 45

Suppose 6 workers are selected from 45 workers, the number of ways they can be selected is [tex](^{45}_6)[/tex] ways

If 7 workers are selected from 20 workers on a day shift;

Then, the number of ways to select 7 members from a group of 20 is [tex](^{20}_7)[/tex] ways;

P( all 7 selected workers will be from the day shift) is:

[tex]= \dfrac{(^{20}_7)}{(^{45}_6)}[/tex]

[tex]=\dfrac{ \dfrac{20!}{7!(20-7)!} } { \dfrac{45!}{6!(45-6)!}}[/tex]

= 0.0095

The possible number of ways to select all day shift workers is:

[tex]=(^{20}_{7})[/tex]

The  possible number of ways to select all swing shift workers is:

[tex]=(^{15}_{7})[/tex]

The possible number of ways to select all graveyard shift workers is:

[tex]= ( ^{10}_{7})[/tex]

The probability that all selected workers are from the same shift is:

[tex]= \Bigg [ \dfrac{ (^{20}_7) }{(^{45}_6) } + \dfrac{ (^{15}_7) }{(^{45}_6) } + \dfrac{ (^{10}_7) }{(^{45}_6) } \Bigg][/tex]

[tex]= \begin {bmatrix} \dfrac{ \dfrac{20!} {7!(20-7)!}}{ \dfrac{ 45! }{6!(45-6)!} } + \dfrac{ \dfrac{15!} {7!(15-7)!}}{ \dfrac{ 45! }{6!(45-6)!} } + \dfrac{ \dfrac{10!} {7!(10-7)!}}{ \dfrac{ 45! }{6!(45-6)!} } \end {bmatrix}[/tex]

= 0.0095 + 0.00079 + 0.000015

= 0.010305

P(selected workers are from at least two different shifts)

= 1 - P(selected workers are from 1 shift from any of the 3 shifts)

= 1 - 0.010305

= 0.989695

≅ 0.9897

We can determine the probability that at least one shift is unrepresented through the following ways:

Let assume that:

[tex]X_1 =[/tex] event that day shift is unrepresented

[tex]X_2 =[/tex] event that swing shift is unrepresented

[tex]X_3 =[/tex] event that graveyard shift is unrepresented

Thus; [tex](X_1 \cap X_2)[/tex] = event that all of them are drawn from graveyard shift.

[tex](X_1 \cap X_3) =[/tex] event that all are drawn from swing shift

[tex](X_2 \cap X_3) =[/tex] event that they are all drawn from day shift.

So;

[tex]P(X_1) = \dfrac{(^{25}_{7})} { (^{45}_{6}) } = 0.0590[/tex]

[tex]P(X_2) = \dfrac{(^{30}_{7})} { (^{45}_{6}) } = 0.2499[/tex]

[tex]P(X_3) = \dfrac{(^{35}_{7})} { (^{45}_{6}) } = 0.8256[/tex]

[tex]P(X_1 \cap X_2) = \dfrac{(^{10}_{7})} { (^{45}_{6}) } = 0.000015[/tex]

[tex]P(X_1 \cap X_3) = \dfrac{(^{15}_{7})} { (^{45}_{6}) } = 0.00079[/tex]

[tex]P(X_2 \cap X_3) = \dfrac{(^{20}_{7})} { (^{45}_{6}) } = 0.0095[/tex]

[tex]P(X_1 \cap X_2 \cap X_3) =0[/tex]

Probability (that at least one shift is unrepresented) is:

[tex]= P(X_1) + P(X_2) +P(X_3) - P(X_1 \cap X_2) - P(X_1 \cap X_3) - P(X_2 \cap X_3) + P(X_1 \cap X_2 \cap X_3)[/tex]

= 0.0590 + 0.2499 + 0.8256 - 0.000015 - 0.00079 - 0.0095 + 0

≅ 1.124

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Answers

Answer:

x = ¹¹⁄₃₆ Or 0.30555555555

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Simplify

x = ¹¹⁄₃₆ Or 0.30555555555

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Answer:

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720 ways can the team choose one captain and then two assistants.

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