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charle [14.2K]
4 years ago
14

A candy distributor needs to mix 10% a fat content chocolate with 40% fat content chocolate to create 200 kg of a 16% fat conten

t chocolate how many kilograms of each kind of chocolate must they use

Mathematics
1 answer:
Luden [163]4 years ago
6 0
I like to use an X diagram to solve problems like this. The constituent percentages are on the left, the desired mix is in the middle and the numbers on the diagonals are the difference between the mix and the constituent.

The numbers 6 and 24 here represent the ratio of 40% to 10% chocolate. That 6:24 ratio can be reduced to 1:4. The sum of those ratio units is 1+4 = 5, so each must represent (200 kg)/5 = 40 kg of chocolate.

The mix must consist of
   40 kg of 40% fat chocolate
   160 kg of 10% fat chocolate

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Frank needs to fill juicecups for his little sister's birthday party there are 35 juice cups in each juice cup holds 12 fluid ou
ICE Princess25 [194]

Answer:

He would need 3.3 gallons

Step-by-step explanation:

1 gallon = 128 ounces

35 * 5 = 420

420 / 128 = 3.28

5 0
3 years ago
When simplified, the expression (x Superscript one-eighth Baseline) (x Superscript three-eighths Basleline) is 12. Which is a po
timofeeve [1]
<h2>The required 'option C) 144' is correct.</h2>

Step-by-step explanation:

We have,

(x^{\dfrac{1}{8}})(x^{\dfrac{3}{8}})=12

To find, the possible value of x = ?

∴ (x^{\dfrac{1}{8}})(x^{\dfrac{3}{8}})=12

⇒ x^{\dfrac{1}{8}+\dfrac{3}{8}}=12

Using the exponential identity,

a^{m} a^{n} =a^{m+n}

⇒ x^{\dfrac{1+3}{8}}=12

⇒ x^{\dfrac{4}{8}}=12

⇒ x^{\dfrac{1}{2}}=12

Squaring both sides, we get

(x^{\dfrac{1}{2}})^2=(12)^2

⇒ x^{{\dfrac{1}{2}}\times2}=144

⇒ x = 144

∴ The possible value of x = 144

Thus, the required 'option C) 144' is correct.

5 0
3 years ago
Read 2 more answers
Each week the manager of a coffee shop orders three more cases of soy milk than almond milk. Let x represent the cases of almond
noname [10]

Answer:

1 4

2 5

4 7

7 10

9 12

Step-by-step explanation:

7 0
3 years ago
Suppose that the national average for the math portion of the College Board's SAT is 510. The College Board periodically rescale
Marta_Voda [28]

Answer:

a) 0.1587 b) 0.023 c) 0.1587 d) 1.15 e)-0.95      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 510

Standard Deviation, σ = 100

We are given that the distribution of SAT score is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(score greater than 610)

P(x > 610)

P( x > 610) = P( z > \displaystyle\frac{610 - 510}{100}) = P(z > 1)

= 1 - P(z \leq 1)

Calculation the value from standard normal z table, we have,

P(x > 610) = 1 - 0.8413 = 0.1587 = 15.87\%

b) P(score greater than 710)

P(x > 710) = P(z > \displaystyle\frac{710-510}{100}) = P(z > 2)\\\\P( z > 2) = 1 - P(z \leq 2)

Calculating the value from the standard normal table we have,

1 - 0.977 = 0.023 = 2.3\%\\P( x > 710) = 2.3\%

c)P(score between 410 and 510)

P(410 \leq x \leq 510) = P(\displaystyle\frac{410 - 510}{100} \leq z \leq \displaystyle\frac{510-510}{100}) = P(-1 \leq z \leq 0)\\\\= P(z \leq 0) - P(z < -1)\\= 0.500 - 0.159 = 0.341 = 34.1\%

P(410 \leq x \leq 510) = 34.1\%

d) x = 625

z_{score} = \displaystyle\frac{625 - 510}{100} = \displaystyle\frac{115}{100} = 1.15

e) x = 415

z_{score} = \displaystyle\frac{415 - 510}{100} = \displaystyle\frac{-95}{100} = -0.95

8 0
3 years ago
Write an equation for the line in slope- intercept form <br>points are:<br>-2,-2<br>-1,1<br>0,4<br>​
egoroff_w [7]

Answer:

y=3x+4

Step-by-step explanation:

Slope-intercept form is written as y=mx+b

m = the slope

slope is rise over run or (y2-y1)/(x2-x1). We need to choose which y value is y2 and the corresponding x value is the x2. If -2 is y2, then:

x2 = -2

y1 = 1

x1 = -1

then plug in the values.

(-2-1)/(-2-(-1))

-3/-1 = 3

m =3

then plug in values for the first equation. you can use any point on the line.

y = 3x+b

4 = 3*0 +b

4=0+b

b = 4

then write it with out x or y

y = 3x+4

3 0
3 years ago
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