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3 years ago

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Mathematics

1 answer:

Jet001 [13]3 years ago

7 0

Answer:

Step-by-step explanation:

Let the number of cockles be c and that of winkles be w.

From the first relation,

w = 2c ..........(I)

From the second relation, 6 winkles are accidentally split.

This in equation form means the following:

c/w- 6 = 3/5

By cross multiplication,

5c = 3(w - 6)...........(ii)

We know w = 2c , let’s substitute this into the second equation.

5c = 3(2c - 6)

5c = 6c - 18

Collecting like times, c = 18

And w = 2c = 2(18) = 36

Hence, there were initially 18 cockles and 36 winkles

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The weights of cars passing over a bridge have a mean of 3,550 pounds and standard deviation of 870 pounds. Assume that the weig

Annette [7]

Answer:

0.24315

Step-by-step explanation:

Using the z score formula to solve this question

z = (x - μ) / σ,

Such that:

x = raw score

μ = population mean

σ = population standard deviation.

From the question:

x = 3000

μ = 3550

σ = 870

z = (3000 - 3550) / 870

z = -550/870

z = -0.6962

Using the z score table as well as probability calculator(as requested in the question to find the z score)

The probability of having less than 3000 is obtained as:

P(x<3000) = 0.24315

7 0

3 years ago

Consider the function f given by f(x)=x*(e^(-x^2)) for all real numbers x.

NISA [10]

Answer:

$\frac{\sqrt{\pi}}{4}$

Step-by-step explanation:

You are going to integrate the following function:

$g(x)=x*f(x)=x*xe^{-x^2}=x^2e^{-x^2}$ (1)

furthermore, you know that:

$\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}$

lets call to this integral, the integral Io.

for a general form of I you have In:

$I_n=\int_0^{\infty}x^ne^{-ax^2}dx$

furthermore you use the fact that:

$I_n=-\frac{\partial I_{n-2}}{\partial a}$

by using this last expression in an iterative way you obtain the following:

$\int_0^{\infty}x^{2s}e^{-ax^2}dx=\frac{(2s-1)!!}{2^{s+1}a^s}\sqrt{\frac{\pi}{a}}$ (2)

with n=2s a even number

for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:

$\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}$

5 0

3 years ago

Read 2 more answers

The following table shows a proportional relationship between mmm and nnn.

Marina CMI [18]

Answer:

70707mmm = 111nnn

Step-by-step explanation:

Using the basic equation of a line:

y = mx + c;. where m is slope and c is intercept on y axis.

Let mmm = y and nnn = x

(i) 333 = 212121m + c

(ii) 555 = 353535m + c

Making c the subject of the formula in both (i) and (ii)

c = 333 - 212121m = 555 - 353535m

353535m -212121m = 555 - 333

141414m = 222

m = 111/70707

Substitute in (i) above

c = 333 -333 = 0

Hence; y = 111/70707x + c

Finally, mmm = 111/70707 nnn

i.e. 70707mmm = 111nnn

Hope this helps.

5 0

3 years ago

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Find the radius of convergence, r, of the series. ? n2xn 7 · 14 · 21 · ? · (7n) n = 1

defon

I'm guessing the series is supposed to be

$\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}$

By the ratio test, the series converges if the following limit is less than 1.

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|$

The first $n$ terms in the numerator's denominator cancel with the denominator's denominator:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|$

$|x^n|$ also cancels out and the remaining factor of $|x|$ can be pulled out of the limit (as it doesn't depend on $n$ ).

$\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0$

which means the series converges everywhere (independently of $x$ ), and so the radius of convergence is infinite.

3 0

3 years ago

True or false: is ST parallel to PR? (photo above)

Deffense [45]

Answer:

true

Step-by-step explanation:

5 0

3 years ago

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