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Sauron [17]
2 years ago
10

Roberto’s scooter traveled 746 miles before needing fuel. If he started with 4 gallons of fuel in the tank, describe the scooter

’s gas mileage
Mathematics
2 answers:
arlik [135]2 years ago
4 0
I divided 746 by 4 which equals 186.5 I think that’s the answer
sammy [17]2 years ago
3 0

Answer:

186.5mi/gal

Step-by-step explanation:

You might be interested in
which ordered pairs in the form (x, y) are solutions to the equation 3x−4y=21 ? Select each correct answer. (−1, −6) (−3, 3) (11
Valentin [98]

Answer:

see explanation

Step-by-step explanation:

To determine which ordered pairs are solutions to the equation

Substitute the x and y values into the left side of the equation and if equal to the right side then they are a solution.

(- 1, - 6)

3(- 1) - 4(- 6) = - 3 + 24 = 21 = right side ← thus a solution

(- 3, 3)

3(- 3) - 4(3) = - 9 - 12 = - 21 ≠ 21 ← not a solution

(11, 3)

3(11) - 4(3) = 33 - 12 = 21 = right side ← thus a solution

(7, 0)

3(7) - 4(0) = 21 - 0 = 21 = right side ← thus a solution

The ordered pairs (- 1, - 6), (11, 3), (7, 0) are solutions to the equation

3 0
2 years ago
Read 2 more answers
Which row in the chart correctly identifies the functions of structures A, B, and C? *
3241004551 [841]
Structure A is used for (4) cell communication. The cell membrance has receptros for cell communcation.
Structure B is used to (1) extract energy from nutrients. The mitochondria is the power house of the cell.
Structure C is used for (1) protein synthesis. The golgi apparatus is where protein is created.
5 0
2 years ago
In constructing a 95 percent confidence interval, if you increase n to 4n, the width of your confidence interval will (assuming
Damm [24]

Answer:

about 50 percent of its former width.

Step-by-step explanation:

Let's assume that our parameter of interest is given by \theta and in order to construct a confidence interval we can use the following formula:

\hat \theta \pm ME(\hat \theta)

Where \hat \theta is an estimator for the parameter of interest and the margin of error is defined usually if the distribution for the parameter is normal as:

ME = z_{\alpha} SE

Where z_{\alpha/2} is a quantile from the normal standard distribution that accumulates \alpha/2 of the area on each tail of the distribution. And SE represent the standard error for the parameter.

If our parameter of interest is the population proportion the standard of error is given by:

SE= \frac{\hat p (1-\hat p)}{n}

And if our parameter of interest is the sample mean the standard error is given by:

SE = \frac{s}{\sqrt{n}}

As we can see the standard error for both cases assuming that the other things remain the same are function of n the sample size and we can write this as:

SE = f(n)

And since the margin of error is a multiple of the standard error we have that ME = f(n)

Now if we find the width for a confidence interval we got this:

Width = \hat \theta + ME(\hat \theta) -[\hat \theta -ME(\hat \theta)]

Width = 2 ME (\hat \theta)

And we can express this as:

Width =2 f(n)

And we can define the function f(n) = \frac{1}{\sqrt{n}} since as we can see the margin of error and the standard error are function of the inverse square root of n. So then we have this:

Width_i= 2 \frac{1}{\sqrt{n}}

The subscript i is in order to say that is with the sample size n

If we increase the sample size from n to 4n now our width is:

Width_f = 2 \frac{1}{\sqrt{4n}} =2 \frac{1}{\sqrt{4}\sqrt{n}} =\frac{2}{2} \frac{1}{\sqrt{n}} =\frac{1}{\sqrt{n}} =\frac{1}{2} Width_i

The subscript f is in order to say that is the width for the sample size 4n.

So then as we can see the width for the sample size of 4n is the half of the wisth for the width obtained with the sample size of n. So then the best option for this case is:

about 50 percent of its former width.

7 0
2 years ago
A model that describes the population of a fishery in which harvesting takes place at a constant rate is given by dP dt = kP − h
Vadim26 [7]
It increases at T because the behavior increased its speed
8 0
2 years ago
Find the z-scores that bound the middle 74% of the area under the standard normal curve.
aniked [119]

Since 74% of the middle area is bounded, this means that there is 13% on the left side, and another 13% on the right side.

P (left) = 0.13

P (right ) = 1 - 0.13 = 0.87

 

At this P values, the z scores are approximately:

z score (left) = -2.22

<span>z score (right) = 1.13</span>

5 0
2 years ago
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