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4 years ago

Use the graph below for this question:

Mathematics

1 answer:

skelet666 [1.2K]4 years ago

5 0

The average rate of change of a function f(x) on the interval [a,b] is given by:

f(b) - f(a)

-------------

b-a

In this specific case, a=2 and b = 3.

f(2) = -3 and f(3) = -1.

Thus, the average rate of change here of f(x) on the interval [2,3] is:

-1 - (- 3) 2

------------- = ------------- = 2 (answer)

3-2 1

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On December 31, 2050, General Industries Corp. (GIC) has just established two joint

siniylev [52]

The combined value available to GIC in terms of the present value as of

December 2050 is $5,050,413.22

What are year 1 cash flows to GIC?

GIC is entitled to 50% of the cash inflows of the joint venture both in UK and South Korea in years 1 and 2

Annual cash flow=(1,000,000*1.02*50%)+(6,000,000,000/1250*50%)

Annual cash flow=$2,910,000

Now the present value of years 1 and 2 cash flows can be determined using the present value formula of a single cash flow provided below:

PV=FV/(1+r)^N

PV=present value=unknown

FV=future cash flows in years 1 and 2

r=discount rate=10%

N=years of cash flow,1 for year and 2 for year 2

PV=2,910,000/(1+10%)^1+2,910,000/(1+10%)^2

PV=$5,050,413.22

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2 years ago

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Schach [20]

0 is greater than -15

8 0

4 years ago

Read 2 more answers

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qwelly [4]

Answer:

2.765 Miles is how much he ran

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4 years ago

Can you please do 1 to 10

ehidna [41]

1) 4(x-6)<2
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4 years ago

A college student organization wants to start a nightclub for students under the age of 21. To assess support for this proposal,

eduard

Answer:

a) Sample size = 822

b) Margin of error = 0.03407

Step-by-step explanation:

We are given the following in the question:

p = 74% = 0.74

a) Sample size is required to obtain margin of error of 0.03

Formula:

$\text{Margin of error} = z_{\text{statistic}}\times \sqrt{\dfrac{p(1-p)}{n}}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get,

$0.03 = 1.96\times \sqrt{\dfrac{0.74(1-0.74)}{n}}\\\\n = (\dfrac{1.96}{0.03})^2(0.74)(1-0.74)\\\\n = 821.24 \approx 822$

Thus, the sample size must be approximately 822 to obtain a 95% confidence interval with an approximate margin of error of 0.03

b) Margin of error for the 95% confidence interval

p = 54% = 0.54

Formula:

$\text{Margin of error} = z_{\text{statistic}}\times \sqrt{\dfrac{p(1-p)}{n}}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get,

$\text{Margin of error} = 1.96\times \sqrt{\dfrac{0.54(1-0.54)}{822}}\\\\=0.03407$

The margin of error now will be 0.03407.

5 0

3 years ago