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Nady [450]
3 years ago
10

Clyde has a cone-shaped party hat. The height of the hat is 10 inches, and the radius of the base of the hat is 4 inches. What i

s the area of a vertical cross section through the center of the base of the party hat?
Mathematics
1 answer:
SSSSS [86.1K]3 years ago
7 0

Check the picture below.

\textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b=base\\ h=height\\[-0.5em] \hrulefill\\ b=8\\ h=10 \end{cases}\implies A=\cfrac{1}{2}(8)(10)\implies A=40

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Find the surface area of each figure. Round to the nearest tenth if necessary.​
Lady bird [3.3K]

Answer:

Step-by-step explanation:

Base of triangle = b = 8 cm

Height of triangle = h = 6 cm

Length of rectangle = l = 6 cm

Sides of the triangle :  S1 = 6 cm ;  S2 = 8 cm and S3 = 10 cm

Surface area of triangular prism= bh + l *(s1+ S2 + S3)

                                                    = 8 * 6 + 6*(6+8+10)

                                                    = 48 + 6*24

                                                     = 48 + 144

                                                     = 192 square cm

3 0
3 years ago
Suppose the diameter at breast height (in.) of a certain type of tree is distributed normally with a mean of 8.8 and a standard
SpyIntel [72]

Answer:14

Step-by-step explanation:

beacsue i just kno i have been working on it ofro days 7+7 1+7  

and plase read the pargraphh if you don't i will (; , cryyyyyyyyyyyyyyyyyyyingggggggggggggg

aragraphs are the building blocks of papers. Many students define paragraphs in terms of length: a paragraph is a group of at least five sentences, a paragraph is half a page long, etc. In reality, though, the unity and coherence of ideas among sentences is what constitutes a paragraph. A paragraph is defined as “a group of sentences or a single sentence that forms a unit” (Lunsford and Connors 116). Length and appearance do not determine whether a section in a paper is a paragraph. For instance, in some styles of writing, particularly journalistic styles, a paragraph can be just one sentence long. Ultimately, a paragraph is a sentence or group of sentences that support one main idea. In this handout, we will refer to this as the “controlling idea,” because it controls what happens in the rest of the paragraph.

8 0
3 years ago
Suppose n people, n ≥ 3, play "odd person out" to decide who will buy the next round of refreshments. The n people each flip a f
blondinia [14]

Answer:

Assume that all the coins involved here are fair coins.

a) Probability of finding the "odd" person in one round: \displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}.

b) Probability of finding the "odd" person in the kth round: \displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left( 1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}.

c) Expected number of rounds: \displaystyle \frac{2^{n - 1}}{n}.

Step-by-step explanation:

<h3>a)</h3>

To decide the "odd" person, either of the following must happen:

  • There are (n - 1) heads and 1 tail, or
  • There are 1 head and (n - 1) tails.

Assume that the coins here all are all fair. In other words, each has a 50\,\% chance of landing on the head and a

The binomial distribution can model the outcome of n coin-tosses. The chance of getting x heads out of

  • The chance of getting (n - 1) heads (and consequently, 1 tail) would be \displaystyle {n \choose n - 1}\cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left(\frac{1}{2}\right)^{n - (n - 1)} = n\cdot \left(\frac{1}{2}\right)^n.
  • The chance of getting 1 heads (and consequently, (n - 1) tails) would be \displaystyle {n \choose 1}\cdot \left(\frac{1}{2}\right)^{1} \cdot \left(\frac{1}{2}\right)^{n - 1} = n\cdot \left(\frac{1}{2}\right)^n.

These two events are mutually-exclusive. \displaystyle n\cdot \left(\frac{1}{2}\right)^n + n\cdot \left(\frac{1}{2}\right)^n  = 2\,n \cdot \left(\frac{1}{2}\right)^n = n \cdot \left(\frac{1}{2}\right)^{n - 1} would be the chance that either of them will occur. That's the same as the chance of determining the "odd" person in one round.

<h3>b)</h3>

Since the coins here are all fair, the chance of determining the "odd" person would be \displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1} in all rounds.

When the chance p of getting a success in each round is the same, the geometric distribution would give the probability of getting the first success (that is, to find the "odd" person) in the kth round: (1 - p)^{k - 1} \cdot p. That's the same as the probability of getting one success after (k - 1) unsuccessful attempts.

In this case, \displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}. Therefore, the probability of succeeding on round k round would be

\displaystyle \underbrace{\left(1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}}_{(1 - p)^{k - 1}} \cdot \underbrace{n \cdot \left(\frac{1}{2}\right)^{n - 1}}_{p}.

<h3>c)</h3>

Let p is the chance of success on each round in a geometric distribution. The expected value of that distribution would be \displaystyle \frac{1}{p}.

In this case, since \displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}, the expected value would be \displaystyle \frac{1}{p} = \frac{1}{\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}}= \frac{2^{n - 1}}{n}.

7 0
3 years ago
What is the answer to 2.25×0.5×0.25?
slega [8]
The answer is 0.28125
8 0
3 years ago
If g(x) = -2x5 - 4, find g(2).
miskamm [114]

g(x)+-2x^5-4


Substitute the value of x = 2 to the equation:


g(2)=-2\cdot2^5-4=-2\cdot32-4=-64-4=-68

7 0
3 years ago
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