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atroni [7]
3 years ago
15

At George Washington High School the ratio of graduating seniors that go on to college, as compared to those who do not, is 7 :

4. Last year 28 students went on to college. How large was the graduating class?
Mathematics
1 answer:
wariber [46]3 years ago
7 0
The solution to the problem is as follows:

<span>7(4)/4(4) = 28/16
 
28+16= 44
</span>
Therefore, the size of the graduating class would be 44 students in all.

I hope my answer has come to your help. God bless and have a nice day ahead!
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I think I know this much so far for A (but I could be wrong):
Eduardwww [97]
Part A. You have the correct first and second derivative.

---------------------------------------------------------------------

Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.

-------------------------------------------------------------

Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out

To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h  ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0. 
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well. 
6 0
3 years ago
A number decreased by three square
Leto [7]
A number decreased by three square is: (where x is the number)

x - 3^2
8 0
3 years ago
How do you find DE, Round to nearest 10th
Sergeu [11.5K]

Answer:  DE\approx6.0

Step-by-step explanation:

Observe in the figure given in the exercise that four right triangles are formed.

In this case you can use the following Trigonometric Identity to solve this exercise:

cos\alpha=\frac{adjacent}{hypotenuse}

From the figure you can identify that:

\alpha =53\°\\\\hypotenuse=10\\\\adjacent=BE=DE

Then, you can substitute values:

cos(53\°)=\frac{DE}{10}

The next step is to solve for DE in order to find its value. This is:

10*cos(53\°)=DE\\\\DE=6.01

Finally, rounding the result to the nearest tenth, you get that this is:

DE\approx6.0

7 0
3 years ago
The realtor and her clients do not know the average home sale price for all of Guelph (500 was actually just a guess). However,
strojnjashka [21]

Answer:

a) The 99% confidence interval would be given by (346.708;453.292)

b) The 99% confidence interval would be given by (338.445;461.555)

Step-by-step explanation:

1) 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".

\bar X=400 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=80 represent the population standard deviation

n=15 represent the sample size  

2) Part a

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

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that z_{\alpha/2}=2.58

Now we have everything in order to replace into formula (1):

400-2.58\frac{80}{\sqrt{15}}=346.708    

400+2.58\frac{80}{\sqrt{15}}=453.292

So on this case the 99% confidence interval would be given by (346.708;453.292)    

3) Part b

For this case we don't know the population deviation so we need to use the t distribution instead the normal standard distribution.

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

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

We need to find the degrees of freedom first df=n-1=15-1=14

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that t_{\alpha/2}=2.98

Now we have everything in order to replace into formula (1):

400-2.98\frac{80}{\sqrt{15}}=338.445    

400+2.98\frac{80}{\sqrt{15}}=461.555

So on this case the 99% confidence interval would be given by (338.445;461.555)    

4 0
2 years ago
Two poles are connected by a wire that is also connected to the ground. The first pole is 18 ft tall and the second pole is 24 f
9966 [12]

Answer:

  72 feet from the shorter pole

Step-by-step explanation:

The anchor point that minimizes the total wire length is one that divides the distance between the poles in the same proportion as the pole heights. That is, the two created triangles will be similar.

The shorter pole height as a fraction of the total pole height is ...

  18/(18+24) = 3/7

so the anchor distance from the shorter pole as a fraction of the total distance between poles will be the same:

  d/168 = 3/7

  d = 168·(3/7) = 72

The wire should be anchored 72 feet from the 18 ft pole.

_____

<em>Comment on the problem</em>

This is equivalent to asking, "where do I place a mirror on the ground so I can see the top of the other pole by looking in the mirror from the top of one pole?" Such a question is answered by reflecting one pole across the plane of the ground and drawing a straight line from its image location to the top of the other pole. Where the line intersects the plane of the ground is where the mirror (or anchor point) should be placed. The "similar triangle" description above is essentially the same approach.

__

Alternatively, you can write an equation for the length (L) of the wire as a function of the location of the anchor point:

  L = √(18²+x²) + √(24² +(168-x)²)

and then differentiate with respect to x and find the value that makes the derivative zero. That seems much more complicated and error-prone, but it gives the same answer.

7 0
2 years ago
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