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3 years ago

The audience thinks 30% of Brandon's jokes are funny. What fraction of Brandon's jokes does the audience think are funny?

Mathematics

2 answers:

horsena [70]3 years ago

8 0

Hi!

30 % represented as a fractions is 30/100.

There is simpler form which is 3/10.

Hope this helps!

expeople1 [14]3 years ago

4 0

30% as a fraction would be 30/100

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ε = {x: 2 x 30, x is an integer}, M = {even numbers}, P = {prime numbers}, T = {odd numbers} Find: I) MUP ii) M - T iii) P(MT) i

mariarad [96]

Answer:

Step-by-step explanation:

ε = {x: 2≤ x ≤30}

M = { even numbers} = {2,4,6,8,10,12,14,16,18,20,22,24,26,28,30}

P = { prime numbers} = {2,3,5,7,11,13,17,19,23,29}

T = {odd numbers} = {3, 5, 7, 9, 11,13,15,17,19,21,23,25,27,29}

1. M ∪ P = {2,3,4,5,6,7,8,10,11,12,13,14,16,17,18,19,20,22,23,24,26,28,29,30}

2. M - T =

n(M) - n(T)

15- 14 = 1

3. P(MT)

(MT) = M ∩ T = 0

P ∪ (M ∩ T ) = {2,3,5,7,11,13,17,19,23,29}

4. P' = not a prime number

T' = not odd number = M

P' ∪(M∩T')

P' ∪ {2,4,6,8,10,12,14,16,18,20,22,24,26,28,30}

= {2,4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30}

6 0

3 years ago

Round each number to the place of the underlined digit the 4 is underlined in 324,650

Otrada [13]

Answer: my bad 324,000 isnt the answer. Its 325,000

Step-by-step explanation:

4 0

3 years ago

I need help with #2 please I don’t understand

kipiarov [429]

1. Square
2. Circle
3. Triangle (inside of a square)
4. Pentagon
5. Rectangle

Hope this helps :)

6 0

3 years ago

Read 2 more answers

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago

Which polynomial is in standard form? A) 26x^5+12x^7-8x^3+6x

Bess [88]

It's C. In order to be in standard form you have to have all your exponents in descending order. They don't all have to be there in order, the ones that are just have to go from highest to lowest.

7 0

3 years ago

Read 2 more answers