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2 years ago

Plz Help.......

Mathematics

1 answer:

lana66690 [7]2 years ago

6 0

Answer:

For product A, the product is increasing, for the bigger the number you plug into x (due to the fact that the numbers become bigger because of the time: year 1, year 2, etc)

Product A is 82% change rate, while

product B is 983.45/4 = 245.8625, 1756.16/3 = 585.3867

Product B is 245.8625/585.3867

product B is 42% change rate

Product A change rate is higher than Product B by 40%

Step-by-step explanation:

Hope this helped! :)

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An online photo services charge $20 to make a photo book with 16 pages each extra page cost a dollar 75 the cost to ship the com

Alisiya [41]

Answer:

The expression to determine the total cost in dollars is $y=25+0.75x$ .

Step-by-step explanation:

Given:

Service charge to make a photo book = $20

Cost per extra page = $0.75

Cost to ship the complete photo book = $5

We need to write expression to determine the total cost in dollars.

Solution:

Let the number of extra pages be denoted by 'x'.

Also Let the Total cost be denoted by 'y'.

Now we can say that;

Total cost will be equal to sum of Service charge to make a photo book and Cost per extra page multiplied by number of extra pages and Cost to ship the complete photo book.

framing in equation form we get;

$y=20+0.75x+5\\\\y=25+0.75x$

Hence the expression to determine the total cost in dollars is $y=25+0.75x$ .

3 0

2 years ago

Write 0.09 as a simplified fraction

Illusion [34]

9/100, let me know if this helps.

4 0

2 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

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

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

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

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

3 years ago

The distance between City A and City B is 250 miles. A length of 2.3 feet represents this distance on a certain wall map. City C

masya89 [10]

The distance between city C and city D is 450 miles.

<u>Solution:</u>

Given, The distance between City A and City B is 250 miles.

A length of 2.3 feet represents this distance on a certain wall map.

City C and City D are 4.14 feet apart on this map.

We have to find what is the actual distance between City C and City D?

Now, distance between a and b is 250 miles ⇒ 2.3 feet on map

Then, distance between c and d be n miles ⇒ 4.14 feet on map.

Now, by chris cross method.

250 x 4.14 = 2.3 x n miles.

2.3n = 1035

n = 450

Hence, the distance between city C and city D is 450 miles.

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3 years ago

The ratio of boys to girls at a daycare is 5 to 4 which of these cannot be the total number of kids at the daycare ?

Andre45 [30]

I believe the answer is D

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2 years ago