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3 years ago

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.

_____

Comment on the question

IMO, the use of 70 trials is coincidentally the same number as the first 70 days 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 X 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

⇒ z = 1.23

Compute the value of x 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

⇒ z = -1.645

Compute the value of x 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