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2 years ago

The population of a town increases every year by 2% of the population

Mathematics

1 answer:

OLga [1]2 years ago

5 0

Answer:

Step-by-step explanation:

A = Pe^rt

A = 1.4P

r = 0.02

t = ?

e^0.02t = 1.4

ln both sides

0.02t = ln(1.4)

t = ln(1.4)/0.02

t ~ 16.8 years

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Express 42 base 7 in base 10

xxTIMURxx [149]

Answer:

correct 42 in bsae 10 is 60

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2 years ago

Which function is graphed below

Svetlanka [38]

I think the answer is a. f(x)= cos(x).

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3 years ago

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Please solve with explanation I’ve been asking all day (this is not a multiple choice question)

Anastaziya [24]

Answer:

a) $SA = 522.9~cm^2$

b) $V_{cone} = 670.2~cm^3$

c) $V_{empty} = 1340.4~cm^3$

Step-by-step explanation:

For a cone,

$SA = \pi r (L + r)$

where L = slant height

$L = \sqrt{r^2 + h^2}$

We have r = 8 cm; h = 10 cm

$L = \sqrt{(8~cm)^2 + (10~cm)^2}$

$L = \sqrt{164~cm^2}$

$SA = (\pi)(8~cm)(\sqrt{164~cm^2} + 8~cm)$

$SA = 522.9~cm^2$

$V_{cone} = \dfrac{1}{3}\pi r^2 h$

$V_{cone} = \dfrac{1}{3}(\pi)(8~cm)^2(10~cm)$

$V_{cone} = 670.2~cm^3$

$V_{cylinder} = \pi r^2 h$

empty space = volume of cylinder - volume of cone

$V_{empty} = V_{cylinder} - V_{cone}$

$V_{empty} = \pi r^2 h - \dfrac{1}{3}\pi r^2 h$

$V_{empty} = (\pi)(8~cm)^2(10~cm) - \dfrac{1}{3}(\pi)(8~cm)^2(10~cm)$

$V_{empty} = 1340.4~cm^3$

3 0

2 years ago

Please help

dedylja [7]

Answer:

C ( 8,-6)

Because you move the points according to the rule.

6 0

2 years ago

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A telemarketer is successful at getting people to donate money for her organization in 55% of all calls she makes. She must get

Tems11 [23]

An interesting twist to a binomial distribution problem.

Given:
p=55%=0.55 for probability of success in solicitation
x=4=number of successful solicitations
n=number of calls to be made
P(x,n,p)>=89.9%=0.899 (from context, it is >= and not =, which is almost impossible)

From context of question, all calls are assumed independent, with constant probability of success, so binomial distribution is applicable.

The number of successes, x, is then given by
$P(x)=C(n,x)p^x(1-p)^{n-x}$ where
p=probability of success
n=number of trials
x=number of successes $C(n,x)=\frac{n!}{x!(n-x)!}$

Here we need n such that
P(x,n,p)>=0.899
given
x>=4, p=0.55, which means we need to find

Method 1: if a cumulative binomial distribution table is available, we can look up n=9,10,11 and find
P(x>=4,9,0.55)=0.834
P(x>=4,10,0.55)=0.898
P(x>=4,11,0.55)=0.939
So she must make (at least) 11 calls to make sure the probability of meeting her quota is 89.9% or more.

Method 2: using technology.
Similar to method 1, we can look up the probabilities directly, for n=9,10,11
P(x>=4,9,0.55)=0.834178
P(x>=4,10,0.55)=0.8980051
P(x>=4,11,0.55)=0.9390368

Method 3: using simple calculator
Here we need to calculate the probabilities for each value of n=10,11 and sum the probabilities of FAILURE S=P(0,n,0.55)+P(1,n,0.55)+P(2,n,0.55)+P(3,n,0.55)
so that the probability of success is 1-S.
For n=10,
P(0,10,0.55)=0.000341
P(1,10,0.55)=0.004162
P(2,10,0.55)=0.022890
P(3,10,0.55)=0.074603
So that
S=0.000341+0.004162+0.022890+0.074603
=0.101995
and Probability of getting 4 successes (or more)
=1-S
=0.898005, missing target by 0.1%

So she will have to make 11 phone calls, bring up the probability to 93.9%. The work is similar to that of n=10.

8 0

2 years ago