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2 years ago

1) A group of five friends goes to the state fair. Each friend pays the entrance fee and buys

Mathematics

1 answer:

kari74 [83]2 years ago

8 0

Answer:

$5(d - 4)$

Step-by-step explanation:

Entrance fee per head = d

Entrance fee for the 5 friends = 5*d = 5d

Booklet of ride ticket per head = 2.5d

Booklet of ride ticket for the 5 friends = 5(2.5d)

Meal ticket per head = d - 4

Meal ticket for the 5 friends = 5(d - 4)

Total amount paid altogether by the five friends = $5d + 5(2.5d) + 5(d - 4)$

From this breakdown above, the total amount paid by the 5 friends for their meal tickets = $5(d - 4)$

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. Given f(x) = e 2x e 2x + 3e x + 2 : (a) Make the substitution u = e x to convert Z f(x) dx into an integral in u (HINT: The ea

MrRa [10]

Answer:

Step-by-step explanation:

Given;

$f(x)=\frac{e^{2x}}{e^{2x}+3e^x+2}$

substitute $u=e^x\\du=e^x dx\\\\\int\frac{e^{2x}}{e^{2x}+3e^x+2}dx=\int\frac{e^x\dot e^x}{e^x^{2x}+3e^x+2}dx\\\\=\int\frac{udu}{u^2+3u+2}$

Apply partial fraction in (a), we get;

$\frac{u}{u^2+3u+2}=\frac{2}{u+2}-\frac{1}{u+1}\\\\\therefore u^2+3u+2\\=u^2+2u+u+2\\=u(u+2)+1(u+1)\\=(u+2)(u+1)\\\\Now\,\int\frac{u}{u^2+3u+2}\,du=\int\frac{2}{u+2}du-\int\frac{1}{u+1}du\\\\=2ln|u+2|-ln|u+1|+c\\=2ln|e^x+2|-lm|e^x+1|+c$

where C is an arbitrary constant

8 0

2 years ago

Bones Brothers & Associates prepare individual tax returns. Over prior years, Bones Brothers have maintained careful records

madreJ [45]

Answer:

For this case we have the following info related to the time to prepare a return

$\mu =90 , \sigma =14$

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard deviation would be:

$\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2$

And the best answer would be

b. 2 minutes

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

For this case we have the following info related to the time to prepare a return

$\mu =90 , \sigma =14$

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard deviation would be:

$\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2$

And the best answer would be

b. 2 minutes

3 0

2 years ago

A web server hosting company advertises 99.999 percent guaranteed network uptime. (a) how many independent network servers would

miv72 [106K]

The solution for this problem would be:
Given that there is 99.999%.

Let denote n as the network servers and p as the reliability of each server.
So the probability that the network uptime = 1 - (1 - p)^n

Therefore, (1-p) ^n = 0.00001

a. x= log(1-.99999)÷log(1-.97)= 3.2833 is the answer

1-(1-.97)^3= 0.99999 + 0.0001 = 1

b. x = log(1-.99999)÷log(1-.88) = 5.43 is the answer

1-(1-.88)^3= 0.99 + 0.0001 = approx 1

4 0

2 years ago

Game 1 gives you 200 points every time you score. Game 2 doubles your points every time you score (2, 4, 8, and so on). In which

nignag [31]

Its depends on how many times you score during the game

in second game the number of points increases as a geometric sequence
with common ratio 2

so for example if you score ten times in first game you get 2000 points

if you score 10 times in Game 2 you score 2 * (2^10) - 1 = 2046 points

so playing 10 or more games is best with Game 2. Any less plays favours gane 1.

6 0

2 years ago

Of all the Sunny Club members in a particular city, 25% prefer swimming on weekends and 75% prefer swimming on weekdays. It is f

Ivan

P(f | weekend) = p(f & weekend)/p(weekend)
.. = 10%/25%
.. = 2/5 = 0.4

3 0

2 years ago

Read 2 more answers