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Sonbull [250]
2 years ago
6

The sector shows the area of a lawn that will be watered by a sprinkler. What is the area, rounded to the nearest tenth? Use 3.1

4 for

Mathematics
1 answer:
a_sh-v [17]2 years ago
3 0

Answer:

95.5\ ft^2

Step-by-step explanation:

step 1

Find the radius of the circle

we know that

A circumference of the circle subtends a central angle of 360 degrees

so

using proportion

\frac{2\pi r }{360^o}=\frac{10}{30^o} \\\\r=\frac{360(10)}{2(3.14)(30)}\\\\r= 19.1\ ft

step 2

Find the area of sector

we know that

The area of the circle subtends a central angle of 360 degrees

so using proportion

Let

x ----> the area of the sector

\frac{\pi r^{2}}{360^o}=\frac{x}{30^o}\\\\x=\frac{3.14(19.1^2)(30)}{360}\\\\x=95.5\ ft^2

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The senior class purchased pizza to sell after school for $320. If they sell pizza by the slice they make $2 per slice. If they
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Answer:30

Step-by-step explanation:

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2 years ago
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Every day, a lecture may be canceled due to inclement weather with probability 0.05. Class cancelations on different days are in
andrezito [222]
<h3>(a)Probability of cancelling at least 4 classes  is 0.0055.</h3><h3>(b)Probability of 10th class is third class that gets cancelled  = 0.0031</h3>

Step-by-step explanation:

Here, the question is<u> incomplete</u>.

Every day, a lecture may be canceled due to inclement weather with probability 0.05. Class cancellations on different days are independent.

(a) There are 15 classes left this semester. Compute the probability that at least 4 of them get canceled.

(b) Compute the probability that the tenth class this semester is the third class that gets canceled.

Now, here:

The probability of cancelling each class  = 0.05

Now, probability of cancelling at least 4 classes

= 1 - P(Cancelling  at max 3 classes)

=  1 - P(0 ≤ x  ≤  3)   = 1 - Binomial (15,0.05,3)

= 0.0055

Hence, probability of cancelling at least 4 classes  is 0.0055.

(b) As given, 10th class is third class that gets cancelled.

So, the first and second classes that get cancelled in between 1 - 9.

P(2 class cancelled in 1st 9) = Binomial (15,0.05,3) = 0.0629

Also, as given  P(10th class cancelled) = 0.05

⇒ P(10th class is third class that gets cancelled ) = 0.0629 x 0.05 = 0.0031

5 0
1 year ago
The line plot shows the number of hours students studied for a final exam. A number line with one x above .5, one x above 1.5, o
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To read the line plot in order to answer this question, you will need to understand that each X above the numbers represents the number of hours 1 student studied. If you count the total number of Xs, that would represent the total number of students surveyed.

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1 year ago
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Ava had $28.50 to spend at the farmer's market. After buying 3 pumpkins Ava, has $12 left. Which equation could you use to find
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Answer:

Step-by-step explanation:

Ava had $28.50 to spend at the farmer's market. She bought three pumpkins from the $28.50 that she had to spend.

After buying the three pumpkins, she had $12 left. This means that the amount that she spent on the pumpkin would be the total amount she had to spend - the amount she had left. This becomes

28.50 -12 = $16.5

This means that the cost pet pumpkin would be total amount that she spent on pumpkin / the number of pumpkin that she bought. It becomes

The equation that will be used to find the price of pumpkin,P would be

P = cost of n pumpkins /number of pumpkins

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In this case

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8 0
2 years ago
A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe
xxMikexx [17]

Answer:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

Step-by-step explanation:

For the random variable N_1 we define the possible values for this variable on this case [1,2,3,4,5] . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable N_1 like this:

P(N_1 = a) = \frac{5-a C 1}{5C2}

For the distribution of N_2 we need to take in count that we are finding a conditional distribution. N_2 given N_1 =a, for this case we see that N_2 \in [1,2,...,5-a], so then exist 5-a C 1 ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}

And if we want to find the joint probability we just need to do this:

P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)

And if we multiply the probabilities founded we got:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

8 0
1 year ago
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