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IrinaVladis
2 months ago
7

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

Mathematics
1 answer:
Zina [12.3K]2 months ago
6 0

Answer:

The rotation angle measures 2.11 °

Step-by-step explanation:

Stated as follows:

The radius of the circular path = r = 18 feet

The distance rolled by the wheel = l = 38 feet

Let us denote the angle of rotation as Ф

Now, according to the problem:

∵ the length of an arc at the center corresponds to an angle Ф

Thus,

distance rolled by the wheel = π × radius × \frac{\Theta }{180^{\circ}}

As 180° represents π radians

And π approximates to 3.14

Thus, distance rolled by the wheel = 180 °× radius × \frac{\Theta }{180^{\circ}}

That is l = r × Ф

So, Ф = \frac{l}{r}

Consequently, Ф = \frac{38 feet}{18 feet}

Therefore, Ф = 2.11 °

Thus, the rotation angle is Ф = 2.11 °

Hence, the rotation angle is 2.11 ° Answer

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Answer:

Refer to the explanation

Step-by-step explanation:

As approximately 10% of users fail to close Windows properly, we determine:

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A. On average,

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C^{10}_8\cdot q^{8}p^{10-8}=\dfrac{10!}{8!(10-8)!}\cdot (0.9)^8\cdot (0.1)^2=45\cdot (0.9)^8\cdot (0.1)^2\approx 0.1937

6 0
1 month ago
Assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. Based on this assumption,
Svet_ta [12734]

Answer:

Here’s the response provided

Step-by-step explanation:

Referring to the flask diagram, the diameter of the cylinder measures 1 inch and its height (h) is 3 inches. Thus, the radius of the cylinder (r) = diameter / 2 = 1/2 = 0.5 inch

The volume of the cylinder can be calculated as πr²h = π(0.5)² × 3 = 2.36 in³

As for the sphere, its diameter is 4.5 in. Hence, the radius of the sphere R = diameter / 2 = 4.5/2 = 2.25 in

The volume of the sphere is calculated as 4/3 (πR³) = 4/3 × π × 2.25³ = 47.71 in³

The total volume of the flask = Volume of the cylinder  + Volume of the sphere = 2.36 + 47.71 = 50.07 in³

<pWhen the cylinder and the sphere are expanded by a scale factor of 2, the height (h') of the cylinder becomes 3/2 = 1.5 inches and the radius (r') becomes 0.5/2 = 0.025 inches.

The new volume for the cylinder = πr'²h' = π(0.25)² × 1.5 = 0.29 in³

For the sphere, the new radius is R' = 2.25 / 2 = 1.125 in.

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6 0
1 month ago
Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribu
zzz [12365]

Response:

Detailed explanation:

Greetings!

You have the variable

X: Area eligible for painting with a can of spray paint (feet²)

This variable is normally distributed with a mean of μ= 25 feet² and a standard deviation of δ= 3 feet²

As this variable has a normal distribution, it needs to be converted into the standard normal form to utilize tabulated cumulative probabilities.

a.

P(X>27)

The first step involves standardizing the X value using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Having determined the Z value, you can find it in the table, but since the table includes probabilities for P(Z, the following conversion must be applied:

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b.

A sample of 20 cans was taken, and you need to ascertain the probability of averaging a coverage area of 540 feet².

The sample mean maintains the same distribution as its source variable, but its variance is influenced by sample size, thus it is normally distributed with parameters:

X[bar]~N(μ;δ²/n)

To cover 540 feet² with 20 cans, the average coverage must be approximately 540/20= 27 feet² per can.

c.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

d.

No, if the distribution is not normal and skewed, the normal distribution should not be applied for calculating probabilities. While the central limit theorem might approximate the sampling distribution to normal when the sample size is 30 or larger, that isn’t applicable here.

I trust this information is helpful!

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Answer:

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type 2

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Step-by-step explanation:

4 0
1 month ago
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