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nignag [31]
3 years ago
9

A builder is installing a Norman window made up of a rectangle topped by a semicircle. If the radius of the semicircle is r, and

the rectangle is 3r high, find the area of the window as a function of r.

Mathematics
1 answer:
Nookie1986 [14]3 years ago
3 0
Notice the picture

the semicircle, has a radius of "r",  you can use that to get the area of a circle, then half it, that's the area of that semicircle in "r" terms,

now, look at the rectangle the semicircle is sitting on top of, to make up
the normal window, it has a height of 3r, and notice its width

so, get the area of the semicircle, and the area of the rectangle,
both in r-terms
add them up, and that's the area of the window

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A bag contains x counters. 7 of the counters are blue. Sam takes at random a counter from the bag and does not replace it. Jill
defon

Answer:

0.2x² - 0.2x - 42 = 0

Step-by-step explanation:

Number of counters = x

Number of blue counters =7

Probability, p = required outcome / Total possible outcomes

P(Sam picks blue ) : = 7/x

P(jill picks blue) = 6/(x - 1)

P(Sam picks blue) * P(Jill picks blue) = 0.2

7/x * 6/(x-1) = 0.2

42/x(x - 1) = 0.2

42 / x² - x= 0.2

42 = 0.2(x² - x)

42 = 0.2x² - 0.2x

0.2x² - 0.2x - 42

6 0
3 years ago
18. El equipo de docentes del Área de Formación Humanística decide repartir 195 ejemplares de libros de lectura a 4 estudiantes
irina [24]

Answer:

Each of the students absent for 1, 2, 4 and 5 days get 57, 53, 45 and 40 copies of the the reading books respectively.

Step-by-step explanation:

English Translation

The team of teachers from the Humanistic Training Area decides to distribute 195 copies of reading books to 4 students previously classified on their own merits. But the distribution will be made taking into account absences from classes during the first cycle. How many books does each student receive if their absences have been 1; two; 4 and 5 days, respectively?

Solution

Because the more the number of days absent, the less the number of books each of the 4 students deserve, we have to employ an elaborate scheme.

If 195 copies are distributed equally between 4 People, each person gets 48.75 copies of the books.

Now, if we add another 48.75 copies of books to the total number of books, each person gets 60.9375 books.

But, we can then divide 48.75 extra books on top according to the number of days absent and subtract each person's number from the assumed 60.9375 books, with an extra 48.75 books.

Total number of days absent = 1 + 2 + 4 +5 = 12

(48.75/12) = -4.0625 copies for each absent day.

The student with only 1 absent day gets 60.9375 - 4.0625 = 56.875 copies = 57 copies

The student with 2 absent days, gets = 60.9375 - (2×4.0625) = 52.8125 copies = 53 copies

The student with 4 absent days, gets = 60.9375 - (4×4.0625) = 44.6875 copies = 45 copies

The student with 5 absent days, gets = 60.9375 - (5×4.0625) = 40.425 copies = 40 copies

Hence, each of the students absent for 1, 2, 4 and 5 days get 57, 53, 45 and 40 copies of the the reading books respectively.

Hope this Helps!!!

6 0
3 years ago
Which are the solutions of the quadratic equation x^2=-5x-3
frutty [35]
The correct answer would be \frac{-5 +/-  \sqrt{13} }{2}

In order to solve for this you must first get the equation equal to 0. 

<span>x^2=-5x-3 ----> add 5x to both sides. 
x^2 + 5x = -3 ----> add 3 to both sides.
x^2 + 5x + 3 = 0

Now knowing this we can use the coefficients of each one in descending order of power as a, b and c. 

a = 1 (because it is the coefficient to x^2)
b = 5</span> (because it is the coefficient to x)
c = 3 (because it is the end number)

Now we can plug these values into the quadratic equation. 

\frac{-b +/-  \sqrt{b^{2} - 4ac } }{2a}

\frac{-5 +/- \sqrt{5^{2} - 4(3)(1) } }{2(1)}

\frac{-5 +/- \sqrt{13} }{2}

And those would be your two answers.
5 0
4 years ago
Suppose the weights of tight ends in a football league are normally distributed such that σ2=1,369. A sample of 49 tight ends wa
Darya [45]

Answer:

Step-by-step explanation:Answer Explanation

Correct answers:

$\left(241.42,\ 262.14\right)$

A 95% confidence interval for μ is (x¯−zα/2σn‾√,x¯+zα/2σn‾√). Here, α=0.05, σ=37, and n=49. Use Excel to calculate the 95% confidence interval.

1. Open Excel, enter the given data in column A, and find the sample mean, x¯, using the AVERAGE function. Thus, the sample mean, rounded to two decimal places, is x¯=251.78.

2. Click on any empty cell, enter =CONFIDENCE.NORM(0.05,37,49), and press ENTER.

3. The margin of error, rounded to two decimal places, is zα/2σn‾√≈10.36. The confidence interval for the population mean has a lower limit of 251.78−10.36=241.42 and an upper limit of 251.78+10.36=262.14.

Thus, the 95% confidence interval for μ is (241.42, 262.14).

6 0
3 years ago
In a clinical study of an allergy drug, 108 of the 200 subjects reported experiencing significant relief from their symptoms. Te
DochEvi [55]

Answer:

The decision is to not reject the null hypothesis.

At a significance level of 0.01, there is not enough evidence to support the claim that the proportion of all those using the drug that experience relief is significantly higher than 50% (P-value = 0.1443).

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that the proportion of all those using the drug that experience relief is significantly higher than 50%.

Then, the null and alternative hypothesis are:

H_0: \pi=0.5\\\\H_a:\pi>0.5

The significance level is 0.01.

The sample has a size n=200.

The sample proportion is p=0.54.

p=X/n=108/200=0.54

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.5*0.5}{200}}\\\\\\ \sigma_p=\sqrt{0.00125}=0.035

Then, we can calculate the z-statistic as:

z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.54-0.5-0.5/200}{0.035}=\dfrac{0.038}{0.035}=1.061

This test is a right-tailed test, so the P-value for this test is calculated as:

\text{P-value}=P(z>1.061)=0.1443

As the P-value (0.1443) is greater than the significance level (0.01), the effect is  not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.01, there is not enough evidence to support the claim that the proportion of all those using the drug that experience relief is significantly higher than 50%.

6 0
4 years ago
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