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2 years ago

Some help with the circle problem would be much aprecated.

Mathematics

1 answer:

Oksana_A [137]2 years ago

8 0

Try this solution:
for the circle A: circumference=6π, area=9π
for the circle B: circumference=12π, area=36π

PS. formula for circumference is 'L=2πr', for area is 'S=πr²'.

You might be interested in

Simplify surds √50+√√72-√√128

34kurt

Answer:

3 $\sqrt{2}$

Step-by-step explanation:

Using the rule of radicals

$\sqrt{a}$ × $\sqrt{b}$ ⇔ $\sqrt{ab}$

Simplify the given radicals

$\sqrt{50}$

= $\sqrt{25(2)}$

= $\sqrt{25}$ × $\sqrt{2}$

= 5 $\sqrt{2}$

------------------------

$\sqrt{72}$

= $\sqrt{36(2)}$

= $\sqrt{36}$ × $\sqrt{2}$

= 6 $\sqrt{2}$

-----------------------

$\sqrt{128}$

= $\sqrt{64(2)}$

= $\sqrt{64}$ × $\sqrt{2}$

= 8 $\sqrt{2}$

Then

$\sqrt{50}$ + $\sqrt{72}$ - $\sqrt{128}$

= 5 $\sqrt{2}$ + 6 $\sqrt{2}$ - 8 $\sqrt{2}$

= 11 $\sqrt{2}$ - 8 $\sqrt{2}$

= 3 $\sqrt{2}$

3 0

2 years ago

4 times what equals 204? Explanation please.

omeli [17]

To find the answer to your question you just have to divide 204 by 4:

204 / 4 = 51

It means that 4 times 51 equals 204:

4 * 51 = 204

I hope it will help you :)

5 0

3 years ago

Read 2 more answers

4.37) A set of data whose histogram is extremely skewed yields a mean and standard deviation of 70 and 12, respectively. What is

Brums [2.3K]

Answer:

75%

88.89%

Step-by-step explanation:

Given :

Mean = 70

Standard deviation = 12

Since the data is said to be extremely skewed, we apply Chebyshev's theorem rather than the empirical rule :

The minimum proportion of observation between 46 and 94

Chebyshev's theorem :

1 - 1 / k²

k = number of standard deviations from the mean

k = (94 - 70) / 12 = 24 / 12 = 2

Hence, we have ;

1 - 1/2²

1 - 1/4

1 - 0.25 = 0.75

Hence, The minimum proportion of observation between 46 and 94 is 75%

Between 36 and 106 :

k = (106 - 70) / 12 ;

k = 36/12 = 3

Hence,

1 - 1/3² = 1 - 1/9 = 8/9 = 0.8888 = 88.89%

The minimum proportion of observation between 34 and 106 is 88.89%

7 0

2 years ago

it looks as though you made 72 of your goal so far for the month you only have 5 days left to reach 100% of your goal let me rem

dezoksy [38]

72% can also be written as 0.72, so take 0.72 x 45 = 32.4

Subtract 32.4 from 45 45 - 32.4 = 12.6 You need 12.6 renewals in 5 days.

8 0

3 years ago

Read 2 more answers

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

3 years ago