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3 years ago

Which statement is true?

Mathematics

1 answer:

Sedbober [7]3 years ago

7 0

Answer:

This is the right one that's true

$\frac{2}{5 } = 0.4$

$8.6 = 0.086\%$

You might be interested in

Let f(x) = cxe−x2 if x ≥ 0 and f(x) = 0 if x < 0.

sergij07 [2.7K]

(a) If f(x) is to be a proper density function, then its integral over the given support must evaulate to 1:

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx = \int_0^\infty cxe^{-x^2}\,\mathrm dx=1$

For the integral, substitute u = x ² and du = 2x dx. Then as x → 0, u → 0; as x → ∞, u → ∞:

$\displaystyle\frac12\int_0^\infty ce^{-u}\,\mathrm du=\frac c2\left(\lim_{u\to\infty}(-e^{-u})-(-1)\right)=1$

which reduces to

c / 2 (0 + 1) = 1 → c = 2

(b) Find the probability P(1 < X < 3) by integrating the density function over [1, 3] (I'll omit the steps because it's the same process as in (a)):

$\displaystyle\int_1^3 2xe^{-x^2}\,\mathrm dx = \boxed{\frac{e^8-1}{e^9}} \approx 0.3678$

5 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPP

Rus_ich [418]

Answer:

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wa=d

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= 2412512.1.4.21.5421.3.12.542.3.215.42.4.2143.51.3

Step-by-step explanation:

8 0

4 years ago

The sequence$$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,2,\dots$$consists of $1$'s separated by blocks of $2$'s with $n$ $2$'s i

kicyunya [14]

Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

block 3 - 1, 2, 2, 2 - length 4

block 4 - 1, 2, 2, 2, 2 - length 5

and so on.

Recall the formula for the sum of consecutive positive integers,

$\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2$

Now,

$1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016$

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

$\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224$

numbers, and their sum is

$\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400$

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of $1+9\times2=19$ .

So, the sum of the first 1234 terms in the sequence is 2419.

8 0

2 years ago

Help please.........

WINSTONCH [101]

Lesser slope

other stuff bla bla

6 0

3 years ago

Limit of x^2-81/x+9 As x goes toward -9

Semmy [17]

Hello,

Use the factoration

a^2 - b^2 = (a - b)(a + b)

Then,

x^2 - 81 = x^2 - 9^2

x^2 - 9^2 = ( x - 9).(x + 9)

Then,

Lim (x^2- 81) /(x+9)

= Lim (x -9)(x+9)/(x+9)

Simplity x + 9

Lim (x -9)

Now replace x = -9

Lim ( -9 -9)

Lim -18 = -18
_______________

The second method without using factorization would be to calculate the limit by the hospital rule.

Lim f(x)/g(x) = lim f(x)'/g(x)'

Where,

f(x)' and g(x)' are the derivates.

Let f(x) = x^2 -81

f(x)' = 2x + 0
f(x)' = 2x

Let g(x) = x +9

g(x)' = 1 + 0
g(x)' = 1

Then the Lim stay:

Lim (x^2 -81)/(x+9) = Lim 2x /1

Now replace x = -9

Lim 2×-9 = Lim -18

= -18

7 0

3 years ago