All categories

2 years ago

Prove that f(x) = x^3 – 1000x^2 + x – 1 is ω(x^3) and o(x^3).

1 answer:

5 0

F(x) = x 3 − 1000x^2 + x − 1

> x3 − 1000x^ 2

= (x − 1000)x^2

> (.9x)x^2

= .9x^3

Therefore, f(x) is Ω(x^3 ) with C = .9, k = 10, 000. Also, for all x > 0:

f(x) = x^3 − 1000x^2 + x − 1

< x^3 + 1000x^3 + x^3 + x^3

= 1002x^3

Therefore, f(x) is O(x^3 ) with C = 1002, k = 1.

You might be interested in

{(–9, 1), (–5, 2), (0, 7), (1, 3), (6, –10)} Josiah claims that the ordered pair (6, –10) can be replaced with any ordered pair

Scrat [10]

B. (–9, –6)
C. (–5, 1)
D. (1, 12)

Remember that no two x's can be the same in a function.

6 0

2 years ago

What is the cube root of -729a9b6

Harman [31]

ANSWER

$\sqrt[3]{- 729{a}^{9} {b}^{6} } = - 9 {a}^{3} {b}^{2}$

EXPLANATION

We want to find the cube root of

$- 729 {a}^{9} {b}^{6}$

We express this symbolically as:

$\sqrt[3]{- 729 {a}^{9} {b}^{6} }$

The expression under the radical called the radicand.

We need to express this radical in exponential form using the property,

${x}^{ \frac{m}{n} } = \sqrt[n]{ {x}^{m} }$

Applying this rule gives us:

$\sqrt[3]{- 729 {a}^{9} {b}^{6} } = ({- 729 {a}^{9} {b}^{6}})^{ \frac{1}{3} }$

$\sqrt[3]{- 729{a}^{9} {b}^{6} } = ({- {9}^{3} {a}^{9} {b}^{6}})^{ \frac{1}{3} }$

Recall that

$({a}^{m} )^{n} = {a}^{mn}$

We apply this rule on the RHS to get,

$\sqrt[3]{- 729{a}^{9} {b}^{6} } = ({- {9}^{3 \times { \frac{1}{3} } } {a}^{9 \times { \frac{1}{3} } } {b}^{6 \times { \frac{1}{3} } }})$

This simplifies to

$\sqrt[3]{- 729{a}^{9} {b}^{6} } = - 9 {a}^{3} {b}^{2}$

3 0

2 years ago

Read 2 more answers

What is the area of a rectangle with length 1/12 ft and width 3/4 ft

Mila [183]

Answer:

3/48 or reduced 1/16

Step-by-step explanation:

1/3 x 3/4 = 3/48

Reduced:

3/48 ÷ 3/3 = 1/16

Only put the reduced fraction if it asks.

Hope this helps :)

7 0

2 years ago

A construction company is considering submitting bids for contracts of three different projects. The company estimates that it h

julsineya [31]

Answer:

a. $P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\$

b. $E(x) = 0.3$

c. $S(x)=0.5196$

d. $E=5,000$

Step-by-step explanation:

The probability that the company won x bids follows a binomial distribution because we have n identical and independent experiments with a probability p of success and (1-p) of fail.

So, the PMF of X is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\$

Where p is 0.1 and it is the chance of winning. Additionally, n is 3 and it is the number of bids. So the PMF of X is:

$P(x)=\frac{3!}{x!(3-x)!}*0.1^{x}*(1-0.1)^{n-x}\\$

For binomial distribution:

$E(x)=np\\S(x)=\sqrt{np(1-p)}$

Therefore, the company can expect to win 0.3 bids and it is calculated as:

$E(x) = np = 3*0.1 = 0.3$

Additionally, the standard deviation of the number of bids won is:

$S(x)=\sqrt{np(1-p)}=\sqrt{3(0.1)(1-0.1)}=0.5196$

Finally, the probability to won 1, 2 or 3 bids is equal to:

$P(1)=\frac{3!}{1!(3-1)!}*0.1^{1}*(1-0.1)^{3-1}=0.243\\P(2)=\frac{3!}{2!(3-2)!}*0.1^{2}*(1-0.1)^{3-2}=0.027\\P(3)=\frac{3!}{3!(3-3)!}*0.1^{3}*(1-0.1)^{3-3}=0.001$

So, the expected profit for the company is equal to:

$E=-10,000+50,000(0.243)+100,000(0.027)+150,000(0.001)\\E=5,000$

Because there is a probability of 0.243 to win one bid and it will produce 50,000 of income, there is a probability of 0.027 to win 2 bids and it will produce 100,000 of income and there is a probability of 0.001 to win 3 bids and it will produce 150,000 of income.

5 0

2 years ago

The following chart shows a store's records of sales of stuffed toys over two months.

Svetllana [295]

Answer:

Hello, the correct answer to this question is letter D) 20.7%

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers