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NeTakaya
2 years ago
8

39 is what percent of 260

Mathematics
1 answer:
Lemur [1.5K]2 years ago
7 0
N/100*260=39

260n=3,900

n=3,900/260

=(390*10)/(26*10)

=390/26

=15

39 is 15% of 260.
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What is the minimum value of C = 7x + 8y, given the constraints: 2x + y ≥ 8, x + y ≥ 6, x ≥ 0, y ≥ 0. A. 32 B. 42 C. 46 D. 64
marshall27 [118]

Answer: The minimum value of C is 46.

Step-by-step explanation:

Since, Here, We have to find out Min C = 7x+8y

Given the constraints are 2x+y\geq 8 -------(1)

x+y \geq 6   ------------- (2)

x \geq 0, y \geq 0  -------- (3)

Since, For equation 1) x-intercept, (4, 0) and y-intercept (0,8)

And, 2\times 0+0\geq 8⇒0\geq 8 ( false)

Therefore the area of line 1) does not contain the origin.

For equation 2) x-intercept, (6, 0) and y-intercept (0,6)

And, 0+0\geq 6⇒0\geq 6 ( false)

Therefore the area of line 2) does not contain the origin.

Thus after plotting the constraints 1) 2) and 3) we get Open Shaded feasible region AEB ( Shown in below graph)

At A≡(0,8) , C= 64

At E≡(2,4),  C= 46

At B≡(6,0),  C= 42

Thus at B, C is minimum, And its minimum value = 42


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PLS HELP MEEE !!!!!
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Answer:

5 terms

Step-by-step explanation:

x is one term

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Rearrange the formula C = 5/9 (F − 32) for F.
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The answer is (9c/5)+32=F
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a tap can fill a bucket in 2 minutes, another in 3 minutes. If both are turned on how long does it take to fill the bucket?
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Answer:

It takes 1 minute 12 seconds to fill the bucket if both taps are turned on.

Step-by-step explanation:

  1. One tap fills the bucket in 2 minutes, thus fills 1/2 of the bucket in one minute.
  2. Other tap fills the bucket in 3 minutes, thus fills 1/3 of the bucket in one minute.
  3. Both together fill 1/2+1/3=5/6 of the bucket in one minute.
  4. If they can fill 5/6 of the bucket in 1 minute, they fill 1/6 of the bucket in 1/5 minute.
  5. They can fill the bucket (6/6) in 1+1/5 minute
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3 years ago
Suppose that a manager is interested in estimating the average amount of money customers spend in her store. After sampling 36 t
musickatia [10]

Answer:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Step-by-step explanation:

Assuming this complete question: Which statement gives a valid interpretation of the interval?

The store manager is 90% confident that the average amount spent by the 36 sampled customers is between S38.01 and $44.29.

There is a 90% chance that the mean amount spent by all customers is between S38.01 and $44.29.

There is a 90% chance that a randomly selected customer will spend between S38.01 and $44.29.

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

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3 years ago
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