Which statement is true about the local minimum of the graphed function over the function

Suppose that you draw two cards from a deck. After drawing the first card, you do not put the first card back in the deck. What

AVprozaik [17]

Answer:

The correct answer is 0.05882.

Step-by-step explanation:

A deck of cards have 52 cards, 13 cards of each suit.

We are drawing two cards without replacement.

We need to find the probability of getting a diamond as the first card.

Favorable outcomes are 13 and total number of outcomes are 52.

Thus this probability is $\frac{13}{52} = \frac{1}{4}$ .

Now for the next draw we again want to pick a diamond card.

Favorable outcomes are 12 and total number of leftover cards are 51.

Thus this probability is $\frac{12}{51}$ .

Now the probability that both cards are diamonds is $\frac{1}{4}$ × $\frac{12}{51}$ = $\frac{3}{51} = \frac{1}{17}$ = 0.0588235 ≈ 0.05882

4 0

3 years ago

Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

4 0

2 years ago

Solve the system of equations by graphing on your own paper. What is the y

xenn [34]

Answer: The coordinate of the solution is (0.25,-0.5)

Step-by-step explanation:

the solution is (0.25,-0.5) because when you graph them on a graph, they both intercept at that point.

3 0

2 years ago

Latasha can read 30 pages of economics in an hour. She can also read 20 pages of sociology in an hour. She spends 4 hours per da

stiv31 [10]

Answer:

Her opportunity cost of reading 10 pages of sociology is 15 pages of Economics

Her opportunity cost of reading 18 pages of economics is 12 pages of sociology.

Step-by-step explanation:

For economics, Latasha reads 1 page in 2 mins (gotten by dividing the number of pages by the time taken to read it i.e 30 pages in 1 hr or 60 mins)

Her sociology is read 1 page in 3 mins (20 pages read in 60 mins)

To read 10 pages of sociology will take 30 mins. She will read 15 pages of Economics in the same time which is the opportunity cost.

In the same vein, It will take her 36 mins to read 18 pages of Economics at the same rate. This is the opportunity cost of reading 12 pages of Sociology (which she reads at 1 page in 3 mins).

8 0

3 years ago

Find the unit rate. 7 square foot in - hour

notka56 [123]

i need more info. Is this the whole question?

6 0

2 years ago