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crimeas [40]
4 years ago
14

Carbon-14 is a radioactive isotope of carbon that is used to date fossils. There are about 1.5

Mathematics
1 answer:
Rom4ik [11]4 years ago
6 0
I’m so sorry I would try my best to answer but I can’t :(
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Find the length of the darkened arc. Leave your answer in terms of pi.​
jonny [76]

Answer:

\frac{15 }{4}  \pi \:

Step-by-step explanation:

Radius of circle (r) = 18/2 = 9 ft

Central angle = 180° - 30° = 150°

length \: of \: arc  \\ =  \frac{150 \degree}{360 \degree}  \times 2\pi \: r \\  \\  = \frac{150 \degree}{360 \degree}  \times 2 \times \pi \: \times  9\\  \\  = \frac{5 }{12 }  \times 18 \times \pi \: \\  \\  = \frac{5 }{4}  \times 3\times \pi \:\\  \\  = \frac{15 }{4}  \pi \:

6 0
3 years ago
An adult house cat could be about ____ high. A. 1 inch B. 1 foot C. 4 inches D. 4 feet
sladkih [1.3K]
The answer is B. 1 foot
7 0
3 years ago
Read 2 more answers
Suppose you have a random variable that is uniformly distributed between 1 and 29. what is the expected value for this random va
Kryger [21]
Assuming the distribution is continuous, you have

\mathbb E(X)=\displaystyle\int_1^{29}\frac x{29-1}\,\mathrm dx=\dfrac1{28}\int_1^{29}x\,\mathrm dx=15

If instead the distribution is discrete, the value will depend on how the interval of number between 1 and 29 are chosen - are they integers? evenly spaced rationals? etc
3 0
3 years ago
PLEASE HELP I cannot get this wrong WILL MARK BRAINLIEST
Nana76 [90]

The answer is graph number 2

8 0
3 years ago
Suppose that the value of a stock varies each day from $13 to $24 with a uniform distribution. (a) Find the probability that the
bogdanovich [222]

Answer:

a. P= 0.6364

b. P = 0.3636

c. Q = $21.25

d. P = 0.5

Step-by-step explanation:

given data

value of a stock varies = $13 to $24

solution

P (stock value is more than $17)

P = \frac{(24-17)}{(24-13)}

P =  \frac{7}{11}

P = 0.6364

and

P (value of the stock is between $17 and $21)

P = \frac{(21-17)}{(24-13)}

P = \frac{4}{11}

P = 0.3636  

and  

Let the upper quartile be Q

\frac{(24 - Q)}{(24 - 13)} = 0.25

\frac{(24 - Q)}{11} = 0.25

(24 - Q) = 2.75

so

Q = $21.25

and

P(X > 20 | X > 16)

P = \frac{(24-20)}{(24-16)}

P = \frac{4}{8}

P = 0.5

3 0
4 years ago
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