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tankabanditka [31]
3 years ago
13

A circle has a circumference of 6. It has an arc of length 17/3. What is the central angle of the arc, in degrees?

Mathematics
1 answer:
zepelin [54]3 years ago
7 0

Answer:

340 degrees

Step-by-step explanation:

So the key thing here is to notice that we are given the circumference which will allow us to find a value for the radius of the circle and hence the angle subtended by the arc (the central angle).

So the circumference of a circle = 2pi(r)

This means:

6 = 2pi(r)

Which means that

r = 6/2pi or r = 3/pi

Now we can use this value of r to find our angle in conjunction with the value of the arc length. So:

Arc length is defined by: length = θr

Where θ is our angle value.

So lets plug in:

\frac{17}{3} = (angle)\frac{3}{\pi }

Multiply by pi to get:

\frac{17\pi }{3} = 3(angle)

Divide by 3 to get that:

θ = 17pi/9

So if we convert that from radians to degrees we get 340 degrees.

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Write a quadratic equation in standard form that has two solutions, -5 and 7
nadya68 [22]

Answer:

{b}^{2}  -  {2b}  - 35

Step-by-step explanation:

1) find two factors of c that sum up to -2.

2) 5 and -7 are factors of -35 that sum up to -2.

3) (x+5)(x-7)

4) x+5= 0, x= -5

5) x-7= 0, x= 7

4) the two solutions are -5 and 7.

4 0
4 years ago
Do not send links or files!! i really need help lol
AlekseyPX

Answer:

x²+11x+30

Step-by-step explanation:

for this question, all we need to do is multiply (x+5) by (x+6)

we can use the distributive property.

(x+5)(x+6)

x² + 5x + 6x + 30

x² + 11x + 30

4 0
3 years ago
A scale drawing of Joshua’s rectangular room is 4 inches by 5 inches. The scale is 1 inch = 4 feet. Enter the Area, in square fe
gulaghasi [49]

Answer:

320 sq.feet

Step-by-step explanation:

1 inch* 4= 4 feet

4 inches*4= 16 feet

5 inches*4= 20 feet

16*20= 320 sq.feet

5 0
3 years ago
Read 2 more answers
Find the value of the variable if P is between J &amp; K. <br><br> JP = 2x ; PK = 7x ; JK = 27
Nana76 [90]
J ---- P --- K

JP = 2x
PK = 7x
JK = 27

2x + 7x = 27
9x = 27
x = 27/9
x = 3

The value of P is 3.

JP = 2x = 2(3) = 6
PK = 7x = 7(3) = 21

7 0
3 years ago
The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean 1263 and a stan
DENIUS [597]

Answer:

(a) The 29th percentile for the number of chocolate chips in a bag is 1198.65.

(b) The number of chocolate chips in a bag that make up the middle 95​% of bags are [1146, 1380].

(c) The inter-quartile range of the number of chocolate chips in a bag of chocolate chip​ cookies is 157.83.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of chocolate chips in a bag of chocolate chip cookies.

The random variable <em>X</em> is normally distributed with mean, <em>μ </em>= 1263 and a standard deviation, <em>σ </em>= 117.

(a)

Compute the 29th percentile for the number of chocolate chips in a bag as follows:

P (X < x) = 0.29

⇒ P (Z < z) = 0.29

The value of <em>z</em> for the above probability is, <em>z</em> = -0.55.

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sogma}\\-0.55=\frac{x-1263}{117}\\x=1263-(117\times 0.55)\\x=1198.65

Thus, the 29th percentile for the number of chocolate chips in a bag is 1198.65.

(b)

According to the Empirical rule 95% of the normally distributed data lies within 2 standard deviations of the mean.

P (μ - σ < X < μ + σ) = 0.95

P (1263 - 117 < X < 1263 + 117) = 0.95

P (1146 < X < 1380) = 0.95

Thus, the number of chocolate chips in a bag that make up the middle 95​% of bags are [1146, 1380].

(c)

The inter-quartile range of the normal distribution is:

IQR = 1.349 <em>σ</em>

Compute the inter-quartile range of the number of chocolate chips in a bag of chocolate chip​ cookies as follows:

IQR = 1.349 <em>σ</em>

      = 1.349 × 117

      = 157.833

Thus, the inter-quartile range of the number of chocolate chips in a bag of chocolate chip​ cookies is 157.83.

5 0
3 years ago
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