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2 years ago

Please helpim not great at math.

Mathematics

2 answers:

choli [55]2 years ago

4 0

2^4 = 16

3^3 = 27

Explanation:

Basile [38]2 years ago

3 0

Answer:

1)16

2)27

Step-by-step explanation:

1) is 2*2*2*2 which is 16

2) is 3*3*3 which is 27

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A and B are independent events. P(A) = 0.50 and P(B) = 0.30. What is PA and B)?

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2 years ago

Not sure about this please help!!!

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8 0

2 years ago

Describe any domain restrictions that make sense for each of these scenarios: The price of a stock fluctuates several times thro

mrs_skeptik [129]

Answer:

a) Restriction: days of the year (a discrete time)

b) The time the ball is on the air: t = 0, until the balls touchs the ground (any real value from t = 0 to t when the ball hits the ground)

c) Any positive real value: x ≥ 0

Explanation:

The domain is the set of elements that can be inputs of the function.

1. The price of a stock fluctuates several times throughout the year as a function of time.

The input variable is the time through out the year.

If you assume that the fluctuaion is recorded once a day, and not every second, the logical assumption is that the time is a day of the year.

Thus, the restricion is the days of the year.

You could represent them in different forms, like DD/MM/YY, or even as a number from 1 to 365 if it covers just one year.

2. A ball is hit into the air by a baseball player, rises until it reaches its highest point, and then falls into left field.

The domain is the time the ball is on the air, which may be any real value from t = 0, until the balls touchs the ground.

It is not restricted to integer numbers.

3. The volume of a spherical balloon increases as the balloon is blown up and decreases as air is let out of it.

The inputs of the function are amounts of air, which may be measured as number of moles.

The number of moles can be any real value equal or greater than zero.

Thus, the domain is the positive real values x ≥ 0.

8 0

2 years ago

Read 2 more answers

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

2 years ago