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Oliga [24]
3 years ago
5

The diameter of a semicircle is 8 miles. What is the semicircle's perimeter? Use 3.14 for ​.

Mathematics
2 answers:
deff fn [24]3 years ago
8 0

Answer:

Circumference of the semicircle=12.56 miles

Step-by-step explanation:

Diameter of the semicircle= 8 miles

Radius of the semicircle=8/2=4 miles

Perimeter of a circle or semicircle is called its Circumference:

Circumference of a circle= 2*\pi *r

   Circumference of a semicircle= \frac{2*\pi *r}{2}=\pi  *r

As,   \pi =\frac{22}{7}=3.14

Putting the values in the formula of the circumference of the semicircle:

Circumference of the semicircle= 3.14*4

                                               =12.56 miles

The Circumference of the semicircle with the radius of 4 miles =12.56 miles

                                           

wlad13 [49]3 years ago
7 0

Answer:

41.13 cm

Step-by-step explanation:

ITSSSS RIGHTTTTTTTTT

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Answer:

92 degrees

Step-by-step explanation:

The sum of the exterior angles will be 360.  

67 x 4 = 268.  That means that the last angles must be 92 (360 - 268)

5 0
2 years ago
You deposit $2000 in an account that earns 5% annual interest compounded quarterly Find the balance after five years. Show work.
evablogger [386]
500 , 2000x.05 is 100 10x5 = 500
8 0
3 years ago
What is the sale price of a car that has an original price of 15,500 with a 14% off sale
Mars2501 [29]
14% of $15,500 is $2,170. $15,500 minus $2,170 is $13,330.

When the car is on sale, the price is reduced from $15,500 to $13,330.
5 0
3 years ago
The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the f
Verizon [17]

Answer:

CI(95%): [205,5;400.8]dolars

Step-by-step explanation:

Hello!

So you need to construct a confidence interval for the average family dental expenses (μ) using the sample given in the problem. To estimate it you need to first choose a statistic. For this small sample, considering that the variable has a normal distribution (I made a quick Shapiro Wilks test, with p-value 0.3234, you can assume normality) the best statistic to use is the Student's t-test.

t= [x(bar)-μ]/S/√n ≈ t₍ₙ₋₁₎

The formula for the confidence interval to estimate the mean is

x(bar)±t_{n-1; 1-\alpha/2}* (S/√n)

<u>The critical value is from a t-distribution with 11 degrees of freedom </u>

±t_{n-1; 1-\alpha/2} = t_{11; 0.975} = 2.201

<em> >remember since it's two-tailed, to get the right critical value you have to divide α by 2. So in the text, you received a confidence level of 1-α=0.95 so α=0.05 then α/2=0.025 and 1-α/2=0.975</em>

To construct the interval, you need to first calculate the sample mean and the standard derivation.

<u>Sample</u>

115; 370; 250; 593; 540; 225; 117; 425; 318; 182; 275; 228

n= 12

∑xi = 3638 ∑xi² = 1362670

<u>Sample mean</u>

x(bar): (∑xi)/n = 3638/12 = 303.17 dolars

<u>Standard derivation</u>

S²= 1/n-1*[∑xi²- (∑xi)²/n] = 1/11 * [1362670-((3638)²/12)] = 23614.47 dolars²

S= 153.67 dolars

<u>Confidence interval (95%)</u>

303.17± 2.201* (153.67/√12)

[205,5;400.8]dolars

I hope you have a SUPER day!

3 0
4 years ago
Find a matrix representation of the transformation L(x, y) = (3x + 4y, x − 2y).
mario62 [17]

Answer:

\left[\begin{array}{cc}x&y\end{array}\right] * \left[\begin{array}{cc}3&1\\4&-2\end{array}\right] = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]

Step-by-step explanation:

The general matrix representation for this transformation would be:

\left[\begin{array}{cc}x&y\end{array}\right] * A = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]

As the matrix A should have the same amount of rows as columns in the firs matrix and the same amount of columns as the result matrix it should be a 2x2 matrix.

\left[\begin{array}{cc}x&y\end{array}\right] * \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]

Solving the matrix product you have that the members of the result matrix are:

3x+4y = a*x + c*y

x - 2y = b*x + d*y

So the matrix A should be:

\left[\begin{array}{cc}3&1\\4&-2\end{array}\right]

8 0
3 years ago
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