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Softa [21]
3 years ago
11

The ratio of the area to the circumference is A. 2r B.r/2 C. 2/r

Mathematics
2 answers:
weqwewe [10]3 years ago
4 0
It's nothing because e=mc^2
alexgriva [62]3 years ago
3 0
E=mc^2 ...............................................
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PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!!
Alina [70]

Yo sup??

We can solve this problem by applying trigonometric ratios.

sin36=XY/WY

WY=5/sin36

=8.5

Hope this helps.

6 0
3 years ago
Sketching a Hyperbola In Exercises 19-32, find the center, vertices, foci, and the equations of the asymptotes of the hyperbola.
zlopas [31]

Step-by-step explanation:

her this is the answer for the question

7 0
3 years ago
What is the value of the expression (9+15)*3+2
dimulka [17.4K]
First do parenthesis the multiplication then add
9+15=24*3=72+2=74 :)hope this helps
3 0
3 years ago
Using a number cube and this hat with 5 different-colored marbles, which simulation would help you answer this question? During
Arlecino [84]

Answer:

  see the attachment

Step-by-step explanation:

We assume that the question is interested in the probability that a randomly chosen class is a Friday class with a lab experiment (2/15). That is somewhat different from the probability that a lab experiment is conducted on a Friday (2/3).

Based on our assumption, we want to create a simulation that includes a 1/5 chance of the day being a Friday, along with a 2/3 chance that the class has a lab experiment on whatever day it is.

That simulation can consist of choosing 1 of 5 differently-colored marbles, and rolling a 6-sided die with 2/3 of the numbers being designated as representing a lab-experiment day. (The marble must be replaced and the marbles stirred for the next trial.) For our purpose, we can designate the yellow marble as "Friday", and numbers greater than 2 as "lab-experiment".

The simulation of 70 different choices of a random class is shown in the attachment.

_____

<em>Comment on the question</em>

IMO, the use of <em>70 trials</em> is coincidentally the same number as the first <em>70 days</em> of school. The calendar is deterministic, so there will be exactly 14 Fridays in that period. If, in 70 draws, you get 16 yellow marbles, you cannot say, "the probability of a Friday is 16/70." You need to be very careful to properly state the question you're trying to answer.

5 0
3 years ago
The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of days an
solniwko [45]

Answer:

(a) 283 days

(b) 248 days

Step-by-step explanation:

The complete question is:

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 268 days and a standard deviation of 12 days. ​(a) What is the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths? ​(b) What is the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths?

Solution:

The random variable <em>X</em> can be defined as the pregnancy length in days.

Then, from the provided information X\sim N(\mu=268, \sigma^{2}=12^{2}).

(a)

The minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths implies that:

P (X > x) = 0.11

⇒ P (Z > z) = 0.11

⇒ <em>z</em> = 1.23

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\1.23=\frac{x-268}{12}\\\\x=268+(12\times 1.23)\\\\x=282.76\\\\x\approx 283

Thus, the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths is 283 days.

(b)

The maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths implies that:

P (X < x) = 0.05

⇒ P (Z < z) = 0.05

⇒ <em>z</em> = -1.645

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\-1.645=\frac{x-268}{12}\\\\x=268-(12\times 1.645)\\\\x=248.26\\\\x\approx 248

Thus, the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths is 248 days.

8 0
3 years ago
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