2+2= x?? i dont understand this

Plz help I have no clue how to do 1 and 2

Elan Coil [88]

1) y= 1/2x+2 and I have no idea about 2

3 0

3 years ago

A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 2

Shalnov [3]

Answer:

The expected winnings for a person buying 1 ticket is -0.2.

Step-by-step explanation:

Given : A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 25 cents each, find the expected winnings for a person buying 1 ticket.

To find : What are the expected winnings?

Solution :

There are one first prize, 2 second prize and 20 third prizes.

Probability of getting first prize is $\frac{1}{20000}$

Probability of getting second prize is $\frac{2}{20000}$

Probability of getting third prize is $\frac{20}{20000}$

A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each.

So, The value of prizes is

$\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10$

If 20000 tickets are sold at 25 cents each i.e. $0.25.

Remaining tickets = 20000-1-2-20=19977

Probability of getting remaining tickets is $\frac{19977}{20000}$

The expected value is

$E=\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10-\frac{19977}{20000}\times 0.25$

$E=\frac{1000+600+200-4994.25}{20000}$

$E=\frac{-3194.25}{20000}$

$E=-0.159$

Therefore, The expected winnings for a person buying 1 ticket is -0.2.

3 0

2 years ago

Suppose Upper F Superscript prime Baseline left-parenthesis x right-parenthesis equals 3 x Superscript 2 Baseline plus 7 and Upp

Sedaia [141]

It looks like you're given

F'(x) = 3x² + 7

and

F (0) = 5

and you're asked to find F(b) for the values of b in the list {0, 0.1, 0.2, 0.5, 2.0}.

The first is done for you, F (0) = 5.

For the remaining b, you can solve for F(x) exactly by using the fundamental theorem of calculus:

$F(x)=F(0)+\displaystyle\int_0^x F'(t)\,\mathrm dt$

$F(x)=5+\displaystyle\int_0^x(3t^2+7)\,\mathrm dt$

$F(x)=5+(t^3+7t)\bigg|_0^x$

$F(x)=5+x^3+7x$

Then F (0.1) = 5.701, F (0.2) = 6.408, F (0.5) = 8.625, and F (2.0) = 27.

On the other hand, if you're expected to approximate F at the given b, you can use the linear approximation to F(x) around x = 0, which is

F(x) ≈ L(x) = F (0) + F' (0) (x - 0) = 5 + 7x

Then F (0) = 5, F (0.1) ≈ 5.7, F (0.2) ≈ 6.4, F (0.5) ≈ 8.5, and F (2.0) ≈ 19. Notice how the error gets larger the further away b gets from 0.

A better numerical method would be Euler's method. Given F'(x), we iteratively use the linear approximation at successive points to get closer approximations to the actual values of F(x).

Let y(x) = F(x). Starting with x₀ = 0 and y₀ = F(x₀) = 5, we have

x₁ = x₀ + 0.1 = 0.1

y₁ = y₀ + F'(x₀) (x₁ - x₀) = 5 + 7 (0.1 - 0) → F (0.1) ≈ 5.7

x₂ = x₁ + 0.1 = 0.2

y₂ = y₁ + F'(x₁) (x₂ - x₁) = 5.7 + 7.03 (0.2 - 0.1) → F (0.2) ≈ 6.403

x₃ = x₂ + 0.3 = 0.5

y₃ = y₂ + F'(x₂) (x₃ - x₂) = 6.403 + 7.12 (0.5 - 0.2) → F (0.5) ≈ 8.539

x₄ = x₃ + 1.5 = 2.0

y₄ = y₃ + F'(x₃) (x₄ - x₃) = 8.539 + 7.75 (2.0 - 0.5) → F (2.0) ≈ 20.164

4 0

2 years ago

A person invests 10000 dollars in a bank. The bank pays 4.5% interest compounded

Greeley [361]

The person would have to leave the money in the bank for 7.8 years for it to reach 13,500 dollars.

Step-by-step explanation:

Step 1; First we must calculate how much interest is generated for a single year. The annual interest rate is 4.5% i.e. 4.5% of 10,000 dollars which equals 0.045 × 10,000 = 450 dollars a year. As the years pass, more and more will be put into the account due to interest.

Step 2; For there to be 13,500 dollars in the bank account we need to calculate how much money is added due to interest.

The money needed to be added through interest = 13,500 - 10,000 = 3,500 dollars.

So we need to determine how long it will take for the bank to add 3,500 dollars by adding 450 dollars a year.

The number of years to reach 13,500 dollars = $\frac{3,500}{450}$ = 7.777 years. By rounding this value to the nearest tenth, we get 7.8 years.