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3 years ago

If I you have to mix a gallon of liquid solution 3 parts water and 1 part alcohol, how much alcohol do you mix in?

Mathematics

2 answers:

Ivahew [28]3 years ago

5 0

1 quart alcohol and 3 quarts water

Veseljchak [2.6K]3 years ago

5 0

If the alcohol is absolute alcohol to make 1 gallon (4 quarts), then mix in 3 quarts water and 1 quart alcohol.

However, most of the time, alcohol comes in 96.5% alcohol, then we will define x=volume of alcohol in gal, then 1-x=volume of water, and solve following equation for x:
0.965x+(1-x)*0=0.25
x=0.25*0.965=0.259 gal
will need 0.259 gal of 96.5% alcohol and 0.741 gal. of water.

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There are 12 students in a graduate class. The students are to be divided into three groups of 3, 4, and 5 members for a class p

exis [7]

Answer:

27720

Step-by-step explanation:

Given that there are 12 students in a graduate class. The students are to be divided into three groups of 3, 4, and 5 members for a class project.

From 12 students 3 students for group I can be selected in 12C3 ways.

Now from remaining 9, 4 students can be selected for II group in 9C4 ways

The remaining 5 have to be placed in III group.

Hence possible divisions for grouping the 12 students in the class into three groups

= 12C3 *9C4

= $=220*126=27720$

8 0

3 years ago

A cylinder has a base diameter of 8ft and a height of 20ft. What is its volume in cubic ft, to the nearest tenths place?

jenyasd209 [6]

Answer:

$4022.9$

Step-by-step explanation:

Volume of a cylinder is given by cross sectional area multiplied by height hence

V=Ah

Since the cross sectional area of

$A=\pi r^{2}$

Substituting the formula for A into formula for volume then

$V=Ah=\pi r^{2}h$

Substituting 8ft for r and 20 ft for h then

$V=Ah=\pi 8^{2}\times 20=4022.8571428571\ ft^{3}\approx 4022.9\ ft^{3}$

Rounded off to the nearest tenths, the volume is equivalent to

$4022.9$

5 0

3 years ago

If we replace y with x2cos x and solve for y', we can conclude that the derivative is

marusya05 [52]

Y = x^2 cos x

y' = x^2 * -sin x + cos x * 2x

= 2x cos x - x^2 sin x

= x ( 2 cos x - x sin x)

5 0

3 years ago

Susie spent $78 in 3 hours. How much did Susie spend per hour?

mart [117]

Answer:

$26

Step-by-step explanation:

78 divided by 3

3 0

3 years ago

Read 2 more answers

Which expressions are equivalent to R16? Check all that apply.

Ilya [14]

Option C: $(6^{-4})^{-4}$

Option D: $(6^{-2})^{-8}$

Option F: $(6^8)^2$

Solution:

Given expression: $6^{16}$

To find which expression is equivalent to the given expression.

Let us solve this using exponent rule: $\left(a^{b}\right)^{c}=a^{b c}$

Option A: $(6^{0})^{16}$

$(6^{0})^{16}=6^{0 \times 16}= 6^0$

It is not equivalent expression.

Option B: $(6^8)^8$

$(6^{8})^{8}=6^{8 \times 8}= 6^{64}$

It is not equivalent expression.

Option C: $(6^{-4})^{-4}$

$(6^{-4})^{-4}=6^{(-4 )\times (-4)}= 6^{16}$

It is equivalent expression for the given expression.

Option D: $(6^{-2})^{-8}$

$(6^{-2})^{-8}=6^{(-2) \times (-8)}= 6^{16}$

It is equivalent expression for the given expression.

Option E: $(6^{-1})^{16}$

$(6^{-1})^{16}=6^{(-1) \times 16}= 6^{-16}$

It is not equivalent expression.

Option F: $(6^8)^2$

$(6^{8})^{2}=6^{8 \times 2}= 6^{16}$

It is equivalent expression for the given expression.

Hence $(6^{-4})^{-4}, \ (6^{-2})^{-8}, \ (6^8)^2$ are the equivalent expressions.

Option C, Option D and Option F are correct answers.

6 0

2 years ago