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2 years ago

If you can buy 15 bags of potatoes for $20, then how many can you buy with $4?

Mathematics

2 answers:

Mekhanik [1.2K]2 years ago

8 0

Answer:

Step-by-step explanation:

15 bags for $20 can be represented with 15/20
so x bags for $4 can be represented with x/4

now if you proportion it out (ricochet), 4x15 is 60, 60 divided by 20 is 3, so you can buy 3 bags of potatoes with $4.

you could do 15/20 is 0.75

0.75 times 4 is 3, so $4 can buy you 3 bags.

timofeeve [1]2 years ago

6 0

Answer:

Step-by-step explanation:

Remark

This is a proportion problem. It assumes that if you know 3 of the four parts of a 2 ratio proportion, you can solve for the 4th one.

Givens

number of bags = 15

Cost = 20

Different cost = 4

number of bags = ?

Formula

Number of bags / Cost = different number of bags/ New Cost

Solution

15/20 = x / 4 Multiply both sides by 4

15*4/20 = x / 4 * 4

60/20 = x

Answer x = 3

That is 4 dollars gets you 3 bags of potatoes

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6a-2(4a-5)=40<br>How do I solve it

Elza [17]

Answer:

Distribute, then combine like terms, subtract 10 from both sides of the equation, simplify, divide both sides of the equation by the same term, and simplify to get your solution.

Step-by-step explanation:

1.) Distribute

6−2(4−5)=40

6−8+10=40

2.) Combine like terms

6a-8a+10=40

-2a+10=40

3.) Subtract 10 from both sides of the equation

-2a+10=40

-2a+10-10=40-10

4.) Simplify

subtract the numbers

-2a+10-10=40-10

-2a=40-10

subtract the numbers

-2a=40-10

-2a=30

5.) Divide both sides of the equation by the same term

-2a=30

$\frac{-2a}{-2} =\frac{30}{-2}$

6.) Simplify

cancel terms that are in both the numerator or denominator

$\frac{-2a}{-2} =\frac{30}{-2}$

a= $\frac{30}{-2}$

Divide the numbers

a= $\frac{30}{-2}$

a=-15

a=-15

7.) Solution

a=-15

8 0

3 years ago

How do I write 3/4 as a percent

Pani-rosa [81]

Answer:

3/4 written as a percent would be 75%

7 0

3 years ago

Read 2 more answers

Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. (a) Descri

OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

$\mu = 74$

and

$\sigma = 6$

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

$\mu = 74$

The standard error of the means becomes the standard deviation of the sampling distribution.

$\sigma_ { \bar X } = \frac{ \sigma}{ \sqrt{n} } \\ \sigma_ { \bar X } = \frac{ 6}{ \sqrt{36} } = 1$

Part b) We want to find

$P(\bar X \:>\:75.9)$

We need to convert to z-score.

$P(\bar X \:>\:75.9) = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: 1.9) \\ = 0.0287$

Part c)

We want to find

$P(\bar X \: < \:71.95)$

We convert to z-score and use the normal distribution table to find the corresponding area.

$P(\bar X \: < \:71.95) = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: - 2.05) \\ = 0.0202$

Part d)

We want to find :

$P(73\:$

We convert to z-scores and again use the standard normal distribution table.

$P( \frac{73 - 74}{1} \:< \: z$

5 0

3 years ago

I NEED HELP ON NUMBER 2 pleaseeee helpppppp

exis [7]

Answer:

Step-by-step explanation:

The slope is -2/4. Don't forget the x!

4 0

4 years ago

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. The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to nd

Blababa [14]

The question is:

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation

x²y'' - 7xy' + 16y = 0; y1 = x^4

Answer:

The second solution y2 is

A(x^4)lnx

Step-by-step explanation:

Given the homogeneous differential equation

x²y'' - 7xy' + 16y = 0

And a solution: y1 = x^4

We need to find a second solution y2 using the method of reduction of order.

Let y2 = uy1

=> y2 = ux^4

Since y2 is also a solution to the differential equation, it also satisfies it.

Differentiate y2 twice in succession with respect to x, to obtain y2' and y2'' and substitute the resulting values into the original differential equation.

y2' = u'. x^4 + u. 4x³

y2'' = u''. x^4 + u'. 4x³ + u'. 4x³ + u. 12x²

= u''. x^4 + u'. 8x³ + u. 12x²

Now, using these values in the original equation,

x²(u''. x^4 + u'. 8x³ + u. 12x²) - 7x(u'. x^4 + u. 4x³)+ 16(ux^4) = 0

x^6u'' + 8x^5u' + 12x^4u - 7x^5u' - 28x^4u + 16x^4u = 0

x^6u'' + x^5u' = 0

xu'' = -u'

Let w = u'

Then w' = u''

xw' = -w

w'/w = -1/x

Integrating both sides

lnw = -lnx + C

w = Ae^(-lnx) (where A = e^C)

w = A/x

But w = u'

So,

u' = A/x

Integrating this

u = Alnx

Since

y2 = uy1

We have

y2 = (Alnx)x^4 = (Ax^4)lnx

Therefore, the second solution y2 is

A(x^4)lnx

6 0

3 years ago