All categories

2 years ago

Please help me with this thanks

Mathematics

1 answer:

geniusboy [140]2 years ago

5 0

<span> When area is equal 1120. We can write an equation
(x+4)(-x+64)=1120
-x²-4x+64x+256=1120
-x²+60x+256-1120=0
-x²+60x-864=0
D=b² - 4ac= 3600-4*864=144, √D=12
x= (-b+/-√D)/2a
x=(-60+/-12)/(-2)
x=24, x=36
For x=24
(x+4)=24+4=28
(-x+64)=(-24+64)=40
For x=36
(x+4)=36+4=40
(x-64) =(-36+64)=28
So sides should be 28 and 40 in.
We did not get any extraneous solutions. They could be if we get negative length side, for example. They can come because a quadratic equation can
give positive and negative numbers because a^2 and (-a)^2 give the same positive number.
We chose to solve this equation using formula for quadratic equations, because this equation has too big numbers to solve it using other methods.

</span>

You might be interested in

Rosa earns $120 per week tutoring math. Each week, she puts $36 from her paycheck in her bank account to save for college. Rosa

natita [175]

Answer:

36/120 x 100

= 30%

..............

3 0

2 years ago

In order to tell if a graph is a function it must

nikitadnepr [17]

You can use the vertical line test (the blue lines on the drawing) and if the other end or if any part of the line touches the blue line, then it isn’t a function.

5 0

2 years ago

Find the distance between the two points in simplest radical form. (-5,8) (-8,6)

Tanzania [10]

Answer:

$\sqrt{445}$ cannot be reduced any further, so that's the answer.

Step-by-step explanation:

Use the distance formula:

$\sqrt{(x2-x1)^{2} +(y2-y1)^{2}} \\\\\sqrt{(-8-10)^{2}+(6--5)^{2}}\\\\\sqrt{(-18)^{2}+(11)^{2}}\\\\\sqrt{(324)+(121)}\\\\= \sqrt{445}\\$

5 0

2 years ago

Read 2 more answers

Estimate the minimum and maximum ages for typical textbooks currently used in college courses, then use the range rule of thumb

Maksim231197 [3]

Answer:

$n=(\frac{1.64(4)}{0.25})^2 =688.537 \approx 689$

So the answer for this case would be n=689 rounded up to the nearest integer

Step-by-step explanation:

Assuming this complete question: "Suppose that the minimum and maximum ages for typical textbooks currently used in college courses are 0 and 8 years. Use the range rule of thumb to estimate the standard deviation.

Estimate the minimum and maximum ages for typical textbooks currently used in college courses, then use the range rule of thumb to estimate the standard deviation. Next, find the size of the sample required to estimate the mean age (in years) of textbooks currently used in college courses. Use a 90% confidence level and assume that the sample mean will be in error by no more than 0.25 year."

Solution for the problem

First we need ti find the estimation for the standard deviation using the Rule of thumb, with the following formula:

$s \approx \frac{R}{4}$

Where R is the range defined as :

$R = Max - Min = 8-0 = 8$

So then the deviation would be approximately:

$s \approx 4$

Important 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 (ME) 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".

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

And on this case we have that $ME =\pm 0.25$ and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (2)

We can assume that the estimator for the population deviation from the rule of thumb is $\hat \sigma = s= 4$

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got $z_{\alpha/2}=1.64$ , replacing into formula (2) we got:

$n=(\frac{1.64(4)}{0.25})^2 =688.537 \approx 689$

So the answer for this case would be n=689 rounded up to the nearest integer

6 0

2 years ago

Malcolm took 18 minutes to do 15 math problems. Diedra took 17 minutes to do 16 math problems. Which student did more problems p

boyakko [2]

The student that did more maths problems per minute is Diedra.

In order to determine which student did more maths problems per minute, the rate of each student has to be determined. Rate is the total questions solved divided by the time.

Rate = total questions / total time

Malcolm = 15 / 18 = 0.833 problems per minute

Diedra = 16/ 17 = 0.941 problems per minute

To learn more about rate, please check: brainly.com/question/9834403

3 0

1 year ago

Read 2 more answers