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kondor19780726 [428]
3 years ago
14

Christopher is a graphic designer who creates business websites. It takes him 2.4 hours to complete one website page. He finds o

ut about a new software program that will cut his time in half for completing one page, but it will take him 15 hours to learn the new program. Which equation can be used to find the number of website pages, x, that Christopher needs to create so that his time spent using the new program will be the same as his current time? How many website pages would Christopher need to create in order to save time using the new software program?
Mathematics
2 answers:
const2013 [10]3 years ago
4 0
The equation that the question asked for would be:
1.2x + 15 = 2.4x

Now, let's solve the second part of the problem.
1.2x + 15 = 2.4x
1.2x - 2.4x + 15 = 2.4x - 2.4x
-1.2x + 15 = 0
-1.2x + 15 - 15 = 0 - 15
-1.2x = -15
-1.2x/-1.2 = -15/-1.2
x = 12.5
So, he needs to create 12 and a half pages for his time spent using the new program to be the same as his current time.

So, any value of x higher than 12.5 would save him time using the new software.
jeka943 years ago
3 0

Answer:

15 + 1.2x = 2.4x

13 pages

Step-by-step explanation:

Let x be the number of pages he needs to create,

∵ In the original programm the time taken by him in each page = 2.4 hours,

So, the total original time = 2.4x

Now, In the new program,

The learning time = 15 hours,

While the time per each page = \frac{2.4}{2} = 1.2 hours,

Thus, the total new time to create x pages = 15 + 1.2x

According to the question,

15 + 1.2x = 2.4x

Which is the required equation,

Subtract 15 on both sides,

1.2x = 2.4x - 15

Subtract both sides by 2.4x

-1.2x = - 15

x = 12.5 ≈ 13

Hence, he needs to create 13 pages.

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Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati
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Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly​ selected,  the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c.   D. Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X \sim N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})

P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})

P(X < 76) = P(Z< \dfrac{3}{12.5})

P(X < 76) = P(Z< 0.24)

From the standard normal distribution tables,

P(X < 76) = 0.5948

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b.  If 25 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})

P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})

P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})

P( \overline X < 76) = P(Z< 1.2)

From the standard normal distribution tables,

P(\overline X < 76) = 0.8849

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

In order to determine the probability in part (b);  the  normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

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