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Vaselesa [24]
3 years ago
13

Last winter Armand had StartFraction 5 Over 6 EndFraction of a row of stacked logs. At the end of the winter, he had StartFracti

on 8 Over 15 EndFraction of the same row left. How much wood did he burn over the winter? 1 and StartFraction 9 Over 16 EndFraction rows 1 and StartFraction 11 Over 30 EndFraction rows StartFraction 4 Over 9 EndFraction row StartFraction 3 Over 10 EndFraction row
Mathematics
2 answers:
Nataly_w [17]3 years ago
6 0

Answer:

3/10

Step-by-step explanation:

According to the Question,

  • Given, Last winter Armand had 5/6 of a row of stacked logs & At the end of the winter he had 8/15 of the same row left.

Now, First We have to make the Fraction even so their denominators must have to be equal.

So. 15×2 = 30(denominator), Multiply with numerator too 8×2=16.

And, 6× 5 = 30(denominator), Multiply with numerator too 5×5=25.

Now. We get New Ratio's now with equal denominator, We can say that

⇒Armand had 25/30 of a row of stacked logs & At the end of the winter he had 16/30 of the same row left.

So, on Simply we get (25-16)/30 ⇒ 9/30 ⇒ 3/10 Wood he burns Over the winter.

hjlf3 years ago
5 0

Answer:

The answer is 3/10

Step-by-step explanation:

took the quiz recently and aced it with a 100 .

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3. What is the radius of the figure if the Area of the Sector is 288*pi? *
Alla [95]

Answer:

The radius of the circle is 36

x = 36

Step-by-step explanation:

Area of sector formula is:

A=\frac{\theta}{360}*\pi r^2

Given  \theta=80, and area (A) is 288\pi, we substitute and solve for r:

A=\frac{\theta}{360}*\pi r^2\\288\pi=\frac{80}{360}*\pi r^2\\288\pi=\frac{2}{9}*\pi r^2\\288\pi=\frac{2\pi r^2}{9}\\2\pi r^2= 2592 \pi \\r^2= 1296\\r=\sqrt{1296}\\r=36

Thus,

The radius of the circle is 36

x = 36

7 0
3 years ago
Which range of time values describes the entire interval over which she would be interpolating
horsena [70]
Answer: first option 0 to 30 minutes.

Jusitification:

Interpolation is the prediction of an intermediate value using a linear model.

In this case, she has the data for 0, 5, 10, 15, and 30 minutes.

So, she can interpolate between any pair of consecutive data to find an intermediate value.

For example: i) if she wants the value for 20 minutes, she can interpolate between 15 and 30 minutes, ii) if she wants the value for 7 minutes, she can interpolate between 5 and 10 minutes, iii) if she wants the value for 12 mintues, she will interpolate between 10 and 15 minutes.

7 0
4 years ago
You spend a $20 per turn on a fair game to win $50 for each win. you lose the first round but win the next two rounds.
Brums [2.3K]

Answer:

70

Step-by-step explanation:

ok, so since you want to find out what the profit is, first you take the times you won and multiply the amount you got for winning so 50*2=100. now that you have 100 you take the money you spent and multiply it by the number of times you played the game so 20*3=30. now that you have those you have everything you need to finish the question. you take the amount you spent per play (30) and subtract it from what you won (100). 100-30=70

3 0
3 years ago
Read 2 more answers
A sheet metal manufacturer is making 10-gauge sheet metal, which is supposed to be 3.416 mm thick. One of the pieces of manufact
satela [25.4K]

Answer:

Step-by-step explanation:

We need to evaluate if thickness of a 10 gauge metal sheet is a requiered in the process that means   3.416 mm thick. So that should be our null hypothesis;  μ₀ = 3.416    and based on sample data, we will formulate an alternative hypothesis taking into account the mean, and standard deviation obtained. We surely will have better  information to take a decision

4 0
3 years ago
The frequency distribution shows the number of registered motor vehicles for a random sample of 200 California households.
erastovalidia [21]

Answer:

1). Mean = 2.275

2). Median = 2 vehicles

Step-by-step explanation:

From the given table,

Number of vehicles (x)         Frequency (f)      (f)×(x)             Cumulative freq.

                  0                                    11                  0                           11

                  1                                     52                52                        63

                  2                                    66                132                      129

                  3                                    35                105                      164

                  4                                    19                  76                       183

                  5                                    12                  60                       195

                  6                                     5                  30                       200

                                                                          \sum (f\times x) = 455

Number of households = 200

Mean = \sum \frac{f.x}{n}

          = \frac{455}{200}

         = 2.275

Median = value of \frac{(n+1)}{2}th observation

             = value of \frac{201}{2}th observation

             = Value of 100.5th observation

             = Since 100.5th observation lies in the row of cumulative freq. = 129

             = 2

Therefore, median number of registered vehicles per California household

= 2 vehicles

6 0
3 years ago
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