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andrew-mc [135]

3 years ago

5

What is the integral of

frac{x}{1 + 3x} " alt=" \frac{x}{1 + 3x} " align="absmiddle" class="latex-formula">

1 answer:

kvv77 [185]3 years ago

3 0

Answer:

√76. im sorry i dont know the step by step explanation but this is the best i could do

Step-by-step explanation:

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The line y=15x Intersex the quadratic function y=15x^2 at two points. what are the coordinates of the two points of intersection

REY [17]

Answer:

D

Step-by-step explanation:

we can set up the equation as 15x = 15x^2

factoring we get

15x(x-1) = 0

So the solutions are when x are 0 and 1

this gives the coordinates

(0,0) and (1,15)

8 0

2 years ago

Sean and sarah spend $70 a month eating out how much money would they have after 15 years if they invested the money instead at

SpyIntel [72]

Answer:

Future Value= $17,170.12

Step-by-step explanation:

Giving the following information:

Monthly deposit (A)= $70

Interest rate (i)= 0.04/12= 0.0033

Number of periods (n)= 15*12= 180 months

<u>To calculate the future value (amount saved), we need to use the following formula:</u>

FV= {A*[(1+i)^n-1]}/i

A= monthly deposit

FV= {70*[(1.0033^180) - 1]} / 0.0033

FV= $17,170.12

5 0

2 years ago

3 consecutive odd integers such that the sum of twice the first and three times the second is 55 more than twice the third

den301095 [7]

Answer:

Consecutive odd integers are 19 , 21 & 23

Step-by-step explanation:

Let the first 3 consecutive odd intergers be x , (x + 2) and (x + 4).

According to the question,

$2x + 3(x + 2) = 2(x + 4) + 55$

$= > 2x + 3x + 6 = 2x + 8 + 55$

Eliminating 2x from both the sides,

$= > 3x + 6 = 63$

$= > 3x = 63 - 6 = 57$

$= > x = \frac{57}{3} = 19$

So, the consecutive odd integers are = 19 , 21 & 23.

3 0

1 year ago

First answer gets brainiest!! I need answers for number eight and nine

nalin [4]

Answer:

8) a, d, e

9) yes, he will have $5.75 (or 5 /34)

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

8 0

3 years ago