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lbvjy [14]
3 years ago
8

Determine the transformations that produce the graph of the functions g(x)=0.2log(x+14)+10 and h(x)=5log(x+14)−10 from the paren

t function f(x)=logx. Then compare the similarities and differences between the two functions, including the domain and range.
Mathematics
1 answer:
liubo4ka [24]3 years ago
4 0

Let's focus on g(x) for now.

0.2 = 1/5

f(x) = log(x) is the parent function

0.2*f(x) = 0.2*log(x) .... vertical compression by factor of 5

0.2*f(x+14) = 0.2*log(x+14) .... shift 14 units to the left

0.2*f(x+14)+10 = 0.2*log(x+14)+10 .... shift 10 units up

g(x) = 0.2*log(x+14)+10

The transformations to go from f(x) to g(x) are:

  1. vertical compression by factor of 5
  2. shift 14 units to the left
  3. shift 10 units up

The domain of g(x) is x > -14 to ensure that x+14 > 0. This is so we avoid applying a log to 0 or a negative value. The parent function domain was x > 0. Note how the domain shifted 14 units to the left (step 2)

======================================================

Now onto h(x)

f(x) = log(x)

5f(x) = 5log(x) .... vertical stretch by factor of 5

5f(x+14) = 5log(x+14) ... shift 14 units left, just like earlier

5f(x+14)-10 = 5log(x+14)-10 ... shift 10 units down

h(x) = 5log(x+14)-10

The transformations to go from f(x) to h(x) are:

  1. vertical stretch by factor of 5
  2. shift 14 units to the left (same as g(x))
  3. shift 10 units down (instead of up compared to g(x))

This time h(x) is taller compared to f(x), while g(x) is more vertically squished compared to f(x). Both involve being shifted 14 units to the left, so both g(x) and h(x) have the same domain. Vertical shifting and compressing/stretching does not affect the domain.

The range of f(x), g(x) and h(x) is the set of all real numbers.

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108 identical books have a mass of 30 kg. Find
julia-pushkina [17]

Answer:

The mass of 150 books = 41.7kg

The number of books that have mass 20kg = 72 books

Step-by-step explanation:

(I) Number of identical books = 108

Mass of  books = 30 kg

Mass of 150 books = ?

Let the mass of 150 books = x kg

To find the mass of 150 books we will use ratio and proportion:

108 books: 30kg = 150 books: x kg

You can also write it like this because ratio can be written in division:

108/30 = 150/x

By cross multiplication:

108x = 150*30

108 x= 4500

Divide both sides by 108

108 x/108 = 4500/108

x= 41.7

Thus the mass of 150 books = 41.7kg

( ii ) The number of books that have a mass of 20 kg = ?

Let assume that the number of books of mass 20 kg = y books

We will again use ratio and proportion method.

108 books: 30kg = y: 20kg

We can write it as:

108/30 = y/20

By cross multiplication:

108 *20 = 30y

2160= 30y

Divide both the terms by 30

2160/30 = 30y/30

72 = y

Therefore the number of books that have mass 20kg = 72 books ....

8 0
4 years ago
How do I solve this?
aleksandr82 [10.1K]
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3 years ago
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3 years ago
According to an​ airline, flights on a certain route are on time ​% of the time. Suppose flights are randomly selected and the n
neonofarm [45]

Answer:

(a) Explained below.

(b) 0.0294

(c) 0.0173

(d) 0.09827

(e) 0.0452

Step-by-step explanation:

The complete question is:

According to an​ airline, flights on a certain route are on time 80​% of the time. Suppose 25 flights are randomly selected and the number of​ on-time flights is recorded.

​(a) Explain why this is a binomial experiment.

​(b) Find and interpret the probability that exactly 16 flights are on time. ​

(c) Find and interpret the probability that fewer than 16 flights are on time.

​(d) Find and interpret the probability that at least 16 flights are on time.

​(e) Find and interpret the probability that between 14 and ​16 flights, inclusive, are on time.

Solution:

(a)

Let the random variable <em>X</em> be defined as the number of​ on-time flights.

A Binomial experiment has the following properties:

  • There are a fixed number of trials (n).
  • Each trial are independent of the others.
  • Each trial has only two outcomes: Success and Failure
  • Each trial has the same probability of success (p).

If a random variable <em>X</em> is used in an experiment and the experiment has all the above mentioned properties, then the random variable X is known as a binomial random variable.

All of these properties can be confirmed for the random variable <em>X</em>.

Thus, this is a binomial experiment.

(b)

Compute the probability that exactly 16 flights are on time as follows:

P(X=16)={25\choose 16}(0.80)^{16}(0.20)^{25-16}

        =2042975\times 0.0281475\times 0.000000512\\=0.029442375072\\\approx 0.0294

Thus, the probability that exactly 16 flights are on time is 0.0294.

(c)

Compute the probability that fewer than 16 flights are on time as follows:

P(X

                 =0.0000+0.0000+....+0.011777\\=0.0173

Thus, the probability that fewer than 16 flights are on time is 0.0173.

(d)

Compute the probability that at least 16 flights are on time as follows:

P(X\geq 16)=1-P(X

                 =1-0.0173\\=0.9827

(e)

Compute the probability that between 14 and 16 ​flights, inclusive, are on time as follows:

P(14\leq X\leq 16)=\sum\limits^{16}_{x=14}{{25\choose x}(0.80)^{x}(0.20)^{25-x}}

                          =0.004+0.0118+0.0294\\=0.0452

Thus, the probability that between 14 and 16 ​flights, inclusive, are on time is 0.0452.

8 0
3 years ago
The sum of two integers is 33. The larger is 6 more than twice the smaller. Find the two
grandymaker [24]

Answer:

24 and 9

Step-by-step explanation:

Hi there!

Let x be equal to the larger integer.

Let y be equal to the smaller integer.

<u>1) Construct equations</u>

  1. x+y=33 (The sum of two integers is 33)
  2. x=6+2y (The larger is 6 more than twice the smaller)

<u>2) Solve for one of the integer</u>

Isolate x in the first equation

x+y=33\\x=33-y

Plug the first equation into the second

33-y=6+2y

Combine like terms

33-6=2y+y\\27=3y\\9=y

Therefore, the smaller integer is 9.

<u />

<u>3) Solve for the other integer</u>

x+y=33

Plug in y (9)

x+9=33\\x=33-9\\x=24

Therefore, the larger integer is 24.

I hope this helps!

4 0
3 years ago
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