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2 years ago

How much alfalfa does Helmut need to add to 30L of ostrich pellets to make a feed mix that is 40% alfalfa by volume?

Mathematics

1 answer:

Nutka1998 [239]2 years ago

8 0

Answer:

a = 20 . . . . liters

Step-by-step explanation:

To solve this, multiply the equation by 30+a:

a = 0.4(30 +a)

a = 12 + 0.4a . . . . . eliminate parentheses

0.6a = 12 . . . . . . . . subtract 0.4a

a = 12/0.6 = 20 . . . divide by the coefficient of a

20 L of alfalfa must be added

_____

Check

20/(30+20) = 20/50 = 40/100 = 0.4 . . . . . . answer checks OK

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Twenty-one boys and eight girls go on a camping trip. In how many ways can a group of six be selected to gather firewood, given

lorasvet [3.4K]

(a) there are 8C2 = 28 ways of picking 2 girls from 8
And there are 21C4 = 5985 ways of picking 4 boys

Required number of ways for 2g / 4b = 28 * 5985 = 167,580

(b) at least 2 girls means combinations of 2g/4b , 3g,3b , 4g/2b , 5g 1b or
6 girls.

2g/4b = 167,580 ways
3g/3b = 8C3 * 21C3 = 56 * 1330 = 74,480
4g/2b = 8C4* 21C2 = 70 * 210 = 14,700
5g 1b = 8C5* 21 = 56*21 = 1176
6 girls = 8C6 = 28

adding these up we get the answer to (b) which is 257,964

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2 years ago

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ahrayia [7]

The answer is 1/125.

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Possibly 124%.........

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2 years ago

WXYZ is a trapezoid with WX ≅ YZ. Isosceles trapezoid W X Y Z is shown. Sides X Y and W Z are parallel. The length of X Y is 3 a

mash [69]

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Step-by-step explanation:

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Elanso [62]

Answer:

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We can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.

Step-by-step explanation:

The marginal productivy is the instant rate of change in the result for an increase in one unit of a factor.

In this case, the productivity is the time he last in the 100-yard. The factor is the amount of yards he train per week.

The marginal productivity can be expressed as:

$\frac{dt}{dL}$

where dt is the variation in time and dL is the variation in training yards.

We can not derive the function because it is not defined, but we can approximate with the last two points given:

$\frac{dt}{dL}\approx\frac{\Delta t}{\Delta L} =\frac{t_2-t_1}{L_2-L_1}=\frac{44.6-46.4}{70,000-60,000}=\frac{-2.0}{10,000}=-0.0002$

Then we can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.

This is an approximation that is valid in the interval of 60,000 to 70,000 yards of training.

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