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Radda [10]
3 years ago
12

Find the arc length for a circle with diameter 10 inches and central angle 60 degrees. Round to the nearest tenth.

Mathematics
1 answer:
UkoKoshka [18]3 years ago
6 0

Answer:

C. 5.2 cm

Step-by-step explanation:

Arc length formula:

s = \dfrac{n}{360^\circ} 2 \pi r

where s = arc length

n = central angle of arc

r = radius of circle

diamter = d = 10 cm

r = d/2 = 10 cm/2 = 5 cm

n = 60 deg

s = \dfrac{n}{360^\circ} 2 \pi r

s = \dfrac{60^\circ}{360^\circ} 2 \times 3.14159 \times 5 ~cm

s = 5.2~cm

Answer: 5.2 cm

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well, let's take a peek when x = 0 and x = 1.

\bf g(x) = ab^x \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x = 0\\ y = 1 \end{cases}\implies 1=ab^0\implies 1=a(1)\implies \boxed{1=a} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x = 1\\ y = 4 \end{cases}\implies 4 = ab^1\implies 4=1b^1\implies \boxed{4=b} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill g(x) = 4^x\qquad \qquad \qquad \begin{array}{|c|c|ll} \cline{1-2} x&y\\ \cline{1-2} -2&\frac{1}{4^2}\to \frac{1}{16}\\ -1&\frac{1}{4}\\ 0&1\\ 1&4\\ 2&16\\ \cline{1-2} \end{array}~\hfill

8 0
3 years ago
X+7+3x+5=180 bro help
KATRIN_1 [288]

Answer:

x = 42

Step-by-step explanation:

Collect like terms (all x together and constants together)

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4 0
3 years ago
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You always need some time to get up after the alarm has rung. You get up from 10 to 20 minutes later, with any time in that inte
Mama L [17]

Answer:

a) P(x<5)=0.

b) E(X)=15.

c) P(8<x<13)=0.3.

d) P=0.216.

e) P=1.

Step-by-step explanation:

We have the function:

f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \,  otherwise }} \right.

a)  We calculate  the probability that you need less than 5 minutes to get up:

P(x

Therefore, the probability is P(x<5)=0.

b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.

E(X)=15.

c) We calculate  the probability that you will need between 8 and 13 minutes:

P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3

Therefore, the probability is P(8<x<13)=0.3.

d)  We calculate the probability that you will be late to each of the 9:30am classes next week:

P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6

You have 9:30am classes three times a week.  So, we get:

P=0.6^3=0.216

Therefore, the probability is P=0.216.

e)  We calculate the probability that you are late to at least one 9am class next week:

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Therefore, the probability is P=1.

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