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3 years ago

A merry-go-round has a radius of 18 feet. If a passenger gets on a

Mathematics

1 answer:

bearhunter [10]3 years ago

6 0

Answer:

The angle of rotation is 2.11 °

Step-by-step explanation:

Given as :

The radius of the round path = r = 18 feet

The length of the distance cover by wheel = l = 38 feet

Let The angle of rotation = Ф

Now, According to question

∵ length of the arc of a circle subtending an angle at center = Ф

So,

length of the distance cover by wheel = π × radius × $\frac{\Theta }{180^{\circ}}$

Since 180° = π radian

And π = 3.14

So, length of the distance cover by wheel = 180 °× radius × $\frac{\Theta }{180^{\circ}}$

i.e l = r × Ф

Or, Ф = $\frac{l}{r}$

Or, Ф = $\frac{38 feet}{18 feet}$

Or, Ф = 2.11 °

So, The angle of rotation = Ф = 2.11 °

Hence, The angle of rotation is 2.11 ° Answer

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A pharmacist has an 18% alcohol solution and a 40% alcohol solution. How much of each should he mix together to make 10 liters o

TiliK225 [7]

Lets call x the amount of 18% solution and y the amount of 40% solution, and write as equations the info of the problem:
18x + 40y = 10(20)
x + y = 10
lets multiply the second equation by -18 and add to the first:
18x + 40y = 200
-18x - 18y = -180
----------------------
0 + 22y = 20
y = 20/22 = 10/11
and substitute in the original equation:
x + y = 10
x = 10 - y
x = 10 - 10/11
x = 110/11 - 10/11
x = 100/11
so they have to use 100/11 liters of 18% solution and 10/11 liters of 40% solution

3 0

2 years ago

Combine these unlike fractions, then simplify. 4 2/3 - 2 2/5 need answer asap and plz explain

Zielflug [23.3K]

Answer: 2 4/5

Step-by-step explanation:

14/3 - 2 2/5

14/3 - 12/5

= 34/15

= 2 4/15

3 0

3 years ago

I NEED HELP WITH A PROBLEM

postnew [5]

Answer:

the first one I guess

Step-by-step explanation:

6 0

3 years ago

A fruit company delivers its fruit in two types of boxes: large and small. a delivery of 6 large boxes and 5 small boxes has a t

Afina-wow [57]

6 large + 5 small = 127 kg ------------ (1)
2 large + 3 small = 51kg ------------ (2)

(2) x 3 :
6 large + 9 small = 153 ---------------- (2a)

(2a) - (1) :
4 small = 26
1 small = 6.5kg ------------ (sub into equation 1)

6 large + 5 small = 127 kg
6 large + 5(6.5) = 127
6 large + 32.5 = 127
6 large = 127 - 32.5
6 large = 94.5
1 large = 15.75 kg

Answer: The small box weighs 6.5kg and the big box weighs 15.75 kg

5 0

3 years ago

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose

Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

In which

x is the number of sucesses

$
e = 2.71828$ is the Euler number

$\mu$ is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Poisson mean:

$\mu = 0.2$

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

$P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819$

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017$

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so $p = 0.181$

We want to find P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671$

0.671 = 67.1% probability that neither contains a missing pulse

8 0

2 years ago