72. Which best describes the polynomial -5x^3?Third degree binomial


First degree trinomial


Third degree monomial


Third degree binomial

Answers

Answer 1
Answer:

-5x^3?

This is a third degree monomial

It has one term so it is a monomial

the exponent is to the power of 3 so it is third degree

Answer 2
Answer:

Answer:

Third degree monomial.

Step-by-step explanation:

because it only has one number and is to the third degree


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A cut piece of ribbon is 21 inches long. The piece of ribbon is 15% of the original length of the ribbon.How long was the original length of ribbon?
Victor collects data on the price of a dozen eggs at 8 different stores.median: $ 1.55Find the lower quartile and upper quartile ofthe data set.lower quartile: $upper quartile: S?$1.39 $1.40 $1.44 $1.50 $1.60 $1.63 $1.65 $1.80
Use Euler's formula to find the missing number vertices 22 edges 34 faces?a.10b.12c.14d.16

A certain city's population is 120,000 and decreases 1.4% per year for 15 years.Is this exponential growth or decay? Growth
What is the rate of growth or decay?
What was the initial amount? 120000
What is the function?
What is the population after 10 years? Round to the nearest whole number.

Answers

Answer:

  • Decay Problem.
  • Decay rate,  r = 0.014
  • Initial Amount =120,000
  • P(t)=120000(0.986)^t
  • P(10)=104,220

Step-by-step explanation:

The exponential function for growth/decay is given as:

P(t)=P_0(1 \pm r)^t, where:\nP_0$ is the Initial Population\nr is the growth/decay rate\nt is time

In this problem:

The city's initial population is 120,000 and it decreases by 1.4% per year.

  • Since the population decreases, it is a Decay Problem.
  • Decay rate, r=1.4% =0.014
  • Initial Amount =120,000

Therefore, the function is:

P(t)=120000(1 - 0.014)^t\nP(t)=120000(0.986)^t

When t=10 years

P(10)=120000(0.986)^10\n=104219.8\n\approx 104220 $ (to the nearest whole number)

A manufacturer claims that only 1% of their computers are defective, but in a sample of 600 3% were found to be defective. If the 1% claim were true there would be less than 1 chance in 1000 of getting this number of defects in the sample. Is there statistically significant evidence against the manufacturer's claim? Why or why not?No, because the difference between a 1% and a 3% defect rate is insignificant.

Yes, because the source of the data was unbiased.

Yes, because the results are unlikely to occur by chance.

No, because the sample size was too small to reach a conclusion.

Answers

Answer:

Step-by-step explanation:

Here population parameter p= 1% = 0.01

But sample proportion = 0.03

Sample size = n=600

Std error of the sample = \sqrt{(pq)/(n) } =\sqrt{(0.1(0.9))/(600) } \n=0.01225\n

Let us assume significance level as 5%

For proportion z critical for 95% is 1.96

Margin of error = 1.96(std error) = 0.024

Conf interval for proportion lower bound = 0.01-0.024 =- 0.014

Upper bound = 0.01+0.024 = 0.124

Thus conf interval (-0.014, 0.124)

Our sample proportion is 0.03 which does not lie within this interval.

Hence we conclude that

Yes, because the results are unlikely to occur by chance.

Answer:

Yes, because the results are unlikely to occur by chance.

Step-by-step explanation:

We have to remember that when dealing with statistics the larger the sample we are taking, the more the results will tend to the statistical reality, for example, if we flip a coin, the chances or statistics are 50%-50% but if we only toss it two times, theres a singnificant chance that it could be 100% tails, the more we continue to toss the coin, the closer we will get to the 50-50, here we have a really large sample of 600 computers, where 3% of them were defective, so we can assure that it wasn´t by chance, because an increase of 2% on the percetange of the defective devices from the ideal to the reality is not by chance.

Determine whether the improper integral converges or diverges, and find the value of each that converges.∫^0_-[infinity] 5e^60x dx

Answers

Answer:

The improper integral converges.

\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12)

General Formulas and Concepts:
Calculus

Limit

Limit Rule [Variable Direct Substitution]:                                                         \displaystyle \lim_(x \to c) x = c

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                       \displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals

Integration Rule [Reverse Power Rule]:                                                           \displaystyle \int {x^n} \, dx = (x^(n + 1))/(n + 1) + C

Integration Rule [Fundamental Theorem of Calculus 1]:                                 \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Integration Property [Multiplied Constant]:                                                     \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

Integration Method: U-Substitution

Improper Integral:                                                                                             \displaystyle \int\limits^(\infty)_a {f(x)} \, dx = \lim_(b \to \infty) \int\limits^b_a {f(x)} \, dx

Step-by-step explanation:

Step 1: Define

Identify.

\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx

Step 2: Integrate Pt. 1

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:             \displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = 5 \int\limits^0_(- \infty) {e^(60x)} \, dx
  2. [Integral] Rewrite [Improper Integral]:                                                     \displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) 5 \int\limits^0_(a) {e^(60x)} \, dx

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

  1. Set u:                                                                                                         \displaystyle u = 60x
  2. [u] Differentiate [Derivative Properties and Rules]:                                 \displaystyle du = 60 \ dx
  3. [Bounds] Swap:                                                                                         \displaystyle \left \{ {{x = 0 \rightarrow u = 0} \atop {x = a \rightarrow u = 60a}} \right.

Step 4: Integrate Pt. 3

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:             \displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) \int\limits^0_(a) {60e^(60x)} \, dx
  2. [Integral] Apply Integration Method [U-Substitution]:                             \displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) \int\limits^0_(60a) {e^(u)} \, du
  3. [Integral] Apply Exponential Integration:                                                 \displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) e^u \bigg| \limits^0_(60a)
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:       \displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1 - e^(60a))/(12)
  5. [Limit] Evaluate [Limit Rule - Variable Direct Substitution]:                     \displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1 - e^(60(-\infty)))/(12)
  6. Rewrite:                                                                                                     \displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12) - (1)/(12e^(60(\infty)))
  7. Simplify:                                                                                                     \displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12)

∴ the improper integral equals\displaystyle \bold{(1)/(12)}  and is convergent.

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Learn more about improper integrals: brainly.com/question/14413972

Learn more about calculus: brainly.com/question/23558817

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Topic: AP Calculus BC (Calculus I + II)

Unit: Integration

Answer:

\int_(-\infty)^0 5 e^(60x) dx = (1)/(12)[e^0 -0]= (1)/(12)  

Step-by-step explanation:

Assuming this integral:

\int_(-\infty)^0 5 e^(60x) dx

We can do this as the first step:

5 \int_(-\infty)^0 e^(60x) dx

Now we can solve the integral and we got:

5 (e^(60x))/(60) \Big|_(-\infty)^0

\int_(-\infty)^0 5 e^(60x) dx = (e^(60x))/(12)\Big|_(-\infty)^0 = (1)/(12) [e^(60*0) -e^(-\infty)]

\int_(-\infty)^0 5 e^(60x) dx = (1)/(12)[e^0 -0]= (1)/(12)  

So then we see that the integral on this case converges amd the values is 1/12 on this case.

Which number line represents the solutions to |–2x| = 4?

Answers

Answer:

the answer is C

Step-by-step explanation:

Answer:

I want to say its b

Calculate had a net income of 5 million dollars in 2010, while a small competing company, Computate, had anet income of 2 millions dollars. The management of Calculate develops a business plan for future growth
that projects an increase in net income of 0.5 million per year, while the management of Computate
develops a plan aimed at increasing its net income kshy15% each year.
a. Create standard mathematical model (table, graph, or equations) for the projected net income for the
next 10 years for both companies. Make sure that each model is accurate and labeled properly so that it
represents the situation
b. If both companies were able to meet their net income growth goals, which company would you choose
to invest in? Why?
c. When, if ever, would your projections suggest that the two companies have the same net income? How
did you find this? Will they ever have the same net income again?

Answers

9514 1404 393

Answer:

  a) see the attached spreadsheet (table)

  b) Calculate, for a 10-year horizon; Computate for a longer horizon.

  c) Year 13; no

Step-by-step explanation:

a) The attached table shows net income projections for the two companies. Calculate's increases by 0.5 million each year; Computate's increases by 15% each year. The result is rounded to the nearest dollar.

__

b) After year 4, Computate's net income is increasing by more than 0.5 million per year, so its growth is faster and getting faster yet. However, in the first 10 years, Calculate's net income remains higher than that of Computate. If we presume that some percentage of net income is returned to investors, then Calculate may provide a better return on investment.

The scenario given here is only interested in the first 10 years. However, beyond that time frame (see part C), we find that Computate's income growth far exceeds that of Calculate.

__

c) Extending the table through year 13, we see that Computate's net income exceeds Calculate's in that year. It continues to remain higher as long as the model remains valid.

Final answer:

To create a mathematical model for the next 10 years' projected net income for Calculate and Computate, use the given growth rates. Compare the projected net incomes to decide which company to invest in. Find the year when the two companies have the same net income.

Explanation:

To create a mathematical model for the projected net income for the next 10 years for both companies, we can use equations. Let's start with Calculate:

Net Income(Calculate) = 5 + 0.5*year

For Computate, the net income growth rate is 15%, so the equation would be:

Net Income(Computate) = 2 * (1 + 0.15)^year



To compare the two companies' projected net income, we can create a table or graph using the equations. By comparing the values, we can determine which company would be a better choice for investment. To find when the two companies have the same net income, we can set the two equations equal to each other and solve for the year.

Learn more about Projected net income here:

brainly.com/question/33816331

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More on the Leaning Tower of Pisa. Refer to the previous exercise. (a) In 1918 the lean was 2.9071 meters. (The coded value is 71.) Using the least-squares equation for the years 1975 to 1987, calculate a predicted value for the lean in 1918. (Note that you must use the coded value 18 for year.)

Answers

Answer:

2.9106

Step-by-step explanation:

According to the information of the problem

Year 75   76   77   78    79    80    81      82 83 84 85 86 87

Lean 642 644 656  667   673  688 696  698 713 717 725 742 757

If you use a linear regressor calculator you find that approximately

y = 9.318 x - 61.123

so you just find x = 18  and then the predicted value would be 106mm

therefore the predicted value for the lean in 1918 was 2.9106