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4 years ago

What is the best approximation for the area of this circle?

Mathematics

1 answer:

Colt1911 [192]4 years ago

4 0

A=pi*r^2
A=3.14*4*4
A=50.2 m (D)

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What is the domain of the function graphed below?

DaniilM [7]

Answer:

(-2,4] and [7,α)

Step-by-step explanation:

the domain is open at -2 but closed at 4 and also closed at 7 but open till infinity..

8 0

3 years ago

What is EXACTLY the square root of 2?

Slav-nsk [51]

Answer:

1.414214

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Suppose that it rains in Spain an average of once every 10 days, and when it does, hurricanes have a 8% chance of happening in H

Law Incorporation [45]

Answer:

The probability that it rains in Spain when hurricanes happen in Hartford is 0.1127

Step-by-step explanation:

This is a question where you use must use Bayes' Theorem.

The easiest way to do Bayes' type questions is to carefully define your terms.

Let R be the event that it is raining in Spain. R' is the event it isn't.

Let H be the event that it is hurricane in Hartford. H' is the event it isn't.

We know

P(R) = 1/10,

P(H | R) = 0.08,

P(H | R') = 0.07 and we want P(R | H).

Bayes Theorem says P(R | H) = [P(H | R)×P(R)] / P(H)

where

P(H) = P(H | R)×P(R) + P(H | R')×P(R')

Therefore,

P(R | H) = [P(H | R)×P(R)] / [P(H | R)×P(R) + P(H | R')×P(R')]

P(R | H) = [0.08 × 1/10] / [(0.08 × 1/10) + (0.07 × (1 - 1/10)]

P(R | H) = 8 / 71

P(R | H) = 0.1127

Therefore, the probability that it rains in Spain when hurricanes happen in Hartford is 0.1127.

7 0

3 years ago

Which is larger 72mm or 7 cm

riadik2000 [5.3K]

72 mm. 72 mm is 7.2 cm
If this helped you, it would be appreciated if you can mark as brainliest.

7 0

3 years ago

Read 2 more answers

The numbers 0 through 9 are used in code groups of four to identify an item of clothing. Code 1083 might identify a blue blouse,

choli [55]

Answer: 5040

Step-by-step explanation:

Given : Number of choices to make a four digit code ( 0 to 9)= 10

If repetition is not allowed , then we use Permutations to find the number of different code groups can be designed.

The number of permutations of n things , taking r at a time is given by :-

$^nP_r=\dfrac{n!}{(n-r)!}$

Then, the number of permutations of 10 numbers , taking 4 at a time is given by :-

$^{10}P_{4}=\dfrac{10!}{(10-4)!}\\\\=\dfrac{10\times9\times8\times7\times6!}{6!}\\\\=\dfrac{10\times9\times8\times7}{1}\\\\=5040$

Hence, the number of different code groups can be designed= 5040

5 0

4 years ago