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wariber [46]
3 years ago
12

Does anyone know how to do this ?

Mathematics
2 answers:
lubasha [3.4K]3 years ago
8 0

Answer:

see below

Step-by-step explanation:

Find the circumference

C = 2 * pi *r

C = 2(.23) * pi

C = .46pi

Since the wheel rotated 829 times, multiply the circumference by 829

829 * .46 pi

381.34 pi

If  pi = 3.14

1197.4076 meters

If using the pi button

1198.01494 meters

Alexeev081 [22]3 years ago
3 0

Step-by-step explanation:

The circumference of a wheel is 0.46π, and we simply have to multiply that by 829 which is 381.34π or 1197.41.

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A hyperbola centered at the origin has a vertex at (0,−40) and a focus at (0, 41).
aniked [119]

\frac{(x-h)^{2} }{a^{2} } - \frac{(y - k)^{2} }{b^{2} } = 1

\frac{(x-0)^{2} }{40^{2} } - \frac{(y - 0)^{2} }{b^{2} } = 1

a² + b² = c²

40² + b² = 41²

         b² = 41² - 40²

         b² = 1681 - 1600

         b² = 81

         b = 9

slope (m) of asymptote = +/-\frac{y}{x} = +/-\frac{b}{a} = +/-\frac{9}{40}

y-intercept (b) = 0 since the center is at the origin

Answer: y = \frac{9}{40}x and y = -\frac{9}{40}x

7 0
3 years ago
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Sandra swims the 100-meter freestyle for her school’s swim team. Her state’s ranking system awards 3 points for first place, 2 p
iris [78.8K]

Complete question :

Sandra swims the 100-meter freestyle for her school’s swim team. Her state’s ranking system awards 3 points for first place, 2 points for second, 1 point for third, and 0 points if she does not place. Her coach used her statistics from last season to design a simulation using a random number generator to predict how many points she would receive in her first race this season.

What is Sandra’s expected value of points awarded for a race?

Integer Value Points Awarded Frequency

1-8 3 20

9-15 2 12

16-19 1 6

20 0 2

Answer:

expected value of points awarded for a race is 2.25

Step-by-step explanation:

Data given:

Integer Value - - Points Awarded - - Frequency

1-8 - - - - - - - - - - - - - 3 - - - - - - - - - - - - - - 20

9-15 - - - - - - - - - - - - 2 - - - - - - - - - - - - - - 12

16-19 - - - - - - - - - - - - 1 - - - - - - - - - - - - - - -6

20 - - - - - - - - - - - - - 0 - - - - - - - - - - - - - - - 2

Expected value(E) :

Score * probability of score

That is;

E = x * p(x)

From the data generated:

Probability of each score :

Probability = required outcome / Total possible outcomes

Total possible outcomes = (20+12+6+2) = 40

P(score(x) = 3) = 20/40 = 0.5

P(score(x) = 2) = 12/40 = 0.3

P(score(x) = 1) = 6/40 = 0.15

P(score(x) = 0) = 2/40 = 0.05

Expected score :

[(3*0.5) + (2*0.3) + (1*0.15) + (0*0.05)]

[1.5 + 0.6 + 0.15 + 0]

= 2.25

7 0
3 years ago
A pet store owner has a 1/2 -pound bag of dog treats treats that she divides evenly among 16 dogs.
Margarita [4]

Answer:

0.03125

Step-by-step explanation:

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2 years ago
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Let X be a random variable with probability mass function P(X = 1) = 1 2 , P(X = 2) = 1 3 , P(X = 5) = 1 6 (a) Find a function g
Goryan [66]

The question is incomplete. The complete question is :

Let X be a random variable with probability mass function

P(X =1) =1/2, P(X=2)=1/3, P(X=5)=1/6

(a) Find a function g such that E[g(X)]=1/3 ln(2) + 1/6 ln(5). You answer should give at least the values g(k) for all possible values of k of X, but you can also specify g on a larger set if possible.

(b) Let t be some real number. Find a function g such that E[g(X)] =1/2 e^t + 2/3 e^(2t) + 5/6 e^(5t)

Solution :

Given :

$P(X=1)=\frac{1}{2}, P(X=2)=\frac{1}{3}, P(X=5)=\frac{1}{6}$

a). We know :

    $E[g(x)] = \sum g(x)p(x)$

So,  $g(1).P(X=1) + g(2).P(X=2)+g(5).P(X=5) = \frac{1}{3} \ln (2) + \frac{1}{6} \ln(5)$

       $g(1).\frac{1}{2} + g(2).\frac{1}{3}+g(5).\frac{1}{6} = \frac{1}{3} \ln (2) + \frac{1}{6} \ln (5)$

Therefore comparing both the sides,

$g(2) = \ln (2), g(5) = \ln(5), g(1) = 0 = \ln(1)$

$g(X) = \ln(x)$

Also,  $g(1) =\ln(1)=0, g(2)= \ln(2) = 0.6931, g(5) = \ln(5) = 1.6094$

b).

We known that $E[g(x)] = \sum g(x)p(x)$

∴ $g(1).P(X=1) +g(2).P(X=2)+g(5).P(X=5) = \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$

   $g(1).\frac{1}{2} +g(2).\frac{1}{3}+g(5).\frac{1}{6 }= \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$$

Therefore on comparing, we get

$g(1)=e^t, g(2)=2e^{2t}, g(5)=5e^{5t}$

∴ $g(X) = xe^{tx}$

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Zinaida [17]
I believe the answer would be 2) pyramid with a rectangular base
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