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2 years ago

CAN SOMEONE HELP ME WITH THIS EQUATION?

Mathematics

1 answer:

Scilla [17]2 years ago

5 0

Answer:

82.96 people/km^2

Step-by-step explanation:

people / square km

3.70 x 10^9

----------------- =

4.46 x 10^7

3.70 x 10^2

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

4.46

0.8295... x 100 = 82.959641 or rounded 82.96

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Can you answer this please?

Nat2105 [25]

Answer: 112.5 square units

Step-by-step explanation:

I think:

For the triangle:

1/2base(height) = 1/2(9)(7)= 31.5 sq. Units

For the small rectangle:

Length x width= 3 x 9 = 27 sq. Units

For the larger rectangle:

The length must be 16 - 7 - 3 = 6

So length x width = 6 x 9 = 54 sq units

Now add them together:

31.5 + 27 + 54 = 112.5 square units

7 0

3 years ago

Does anyone know this?

tresset_1 [31]

Answer:

f(x) = -x^2 + x + 13 find f(9) = -59

Hope this helps :)

4 0

3 years ago

Read 2 more answers

I have nine white t-shirts and 11 colored t-shirts what is the probability of me picking a color T-shirt

ozzi

55 percent, I believe. Hope this helps!

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3 years ago

The population of a particular type of bacteria is known to triple in 22 days. What is the daily growth rate

Paul [167]

Answer: ∛(3)

Step-by-step explanation:

Suppose that at day 0 the population was A.

at day 3, the population will be: 3*A

at day 6, the population will be 3*(3*A) = A*3^2

then, at day N = 3*n (where n = 0, 1, 2.....) the population will be:

P(N) = A*3^n

Particularly, if we take n = 1/3 we will have:

N = 3*1/3 = 1

This means that this is the first day after the day 0.

P(1) = A*3^(1/3)

(then in day one, the population grow by a factor of 3^(1/3))

N = 2 is when n = 2/3, then:

P(2) = A*3^(2/3)

The quotient between P(2) and P(1) is equal to the growth between day one and day two, this should be the same as the growth between day zero and day one.

A*3^(2/3)/(A*3^(1/3)) = 3^( 2/3 - 1/3) = 3^(1/3)

So we found that the daily growth rate is 3^(1/3) or ∛(3)

3 0

3 years ago

A survey on British Social Attitudes asked respondents if they had ever boycotted goods for ethical reasons (Statesman, January

Blababa [14]

Answer:

a) 27.89% probability that two have ever boycotted goods for ethical reasons

b) 41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) 41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) The expected number is 2.3 and the standard deviation is 1.33.

Step-by-step explanation:

We use the binomial probability distribution to solve this question.

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.

23% of the respondents have boycotted goods for ethical reasons.

This means that $p = 0.23$

a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons?

This is P(X = 2) when n = 6. So

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

$P(X = 2) = C_{6,2}.(0.23)^{2}.(0.77)^{4} = 0.2789$

27.89% probability that two have ever boycotted goods for ethical reasons

b) In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?

Either less than two have, or at least two. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = 0) = C_{6,0}.(0.23)^{0}.(0.77)^{6} = 0.2084$

$P(X = 1) = C_{6,1}.(0.23)^{1}.(0.77)^{5} = 0.3735$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2084 + 0.3735 = 0.5819$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5819 = 0.4181$

41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) In a sample of ten British citizens, what is the probability that between 3 and 6 have boycotted goods for ethical reasons?

Now n = 10.

$P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

$P(X = 3) = C_{10,3}.(0.23)^{3}.(0.77)^{7} = 0.2343$

$P(X = 4) = C_{10,4}.(0.23)^{4}.(0.77)^{6} = 0.1225$

$P(X = 5) = C_{10,5}.(0.23)^{5}.(0.77)^{5} = 0.0439$

$P(X = 6) = C_{10,6}.(0.23)^{6}.(0.77)^{4} = 0.0109$

$P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.2343 + 0.1225 + 0.0439 + 0.0109 = 0.4116$

41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) In a sample of ten British citizens, what is the expected number of people that have boycotted goods for ethical reasons? Also find the standard deviation.

$E(X) = np = 10*0.23 = 2.3$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.23*0.77} = 1.33$

The expected number is 2.3 and the standard deviation is 1.33.

5 0

3 years ago