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slamgirl [31]
3 years ago
14

he initial number of views for a blog was 20. The number of views is growing exponentially at a rate of 20% per week. What is th

e number of views expected to be four weeks from now? Round to the nearest whole number. Enter your answer in the box.
Mathematics
2 answers:
sergeinik [125]3 years ago
5 0
<h2>Answer</h2>

41 views

<h2>Explanation</h2>

To solve this, we are going to use the exponential function:

f(t)=a(1+b)^t

f(t) is the final number of views after t weeks

a is the number of views

b is the growing rate in decimal form

t is the time in weeks

We know from our problem that the number initial views was 20 and the growing rate is 20%, so a=20 and r=\frac{20}{100} =0.2. Let's replace the values in our exponential function:

f(t)=a(1+b)^t

f(t)=20(1+0.2)^t

f(t)=20(1.2)^t

Now to find the number of views expected four weeks from now, we just need to replace t with 4, so t=4

f(t)=20(1.2)^t

f(4)=20(1.2)^4

f(t)=41.472

And rounded to the nearest whole number:

f(t)=41

Goryan [66]3 years ago
3 0
The answer to this problem is about 42 views.
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Write the word sentence as an inequality: A number y plus 2 is greater than -5​
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5 0
4 years ago
Suppose that an airline overbooks seats on their flights. In particular, it sells 300 tickets for a flight when there are only 2
vladimir1956 [14]

Using the <u>normal approximation to the binomial</u>, it is found that there is a 0.994 = 99.4% probability that we will have enough seats for everyone who shows up.

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.  
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • The binomial distribution is the probability of <u>x successes on n trials</u>, with <u>p probability</u> of a success on each trial. It can be approximated to the normal distribution with \mu = np, \sigma = \sqrt{np(1-p)}.

In this problem:

  • 15% do not show up, so 100 - 15 = 85% show up, which means that p = 0.85.
  • 300 tickets are sold, hence n = 300.

The mean and the standard deviation are given by:

\mu = np = 300(0.85) = 255

\sigma = \sqrt{np(1-p)} = \sqrt{300(0.85)(0.15)} = 6.185

The probability that we will have enough seats for everyone who shows up is the probability of at most <u>270 people showing up</u>, which, using continuity correction, is P(X \leq 270 + 0.5) = P(X \leq 270.5), which is the <u>p-value of Z when X = 270.5</u>.

Z = \frac{X - \mu}{\sigma}

Z = \frac{270.5 - 255}{6.185}

Z = 2.51

Z = 2.51 has a p-value of 0.994.

0.994 = 99.4% probability that we will have enough seats for everyone who shows up.

A similar problem is given at brainly.com/question/24261244

8 0
3 years ago
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