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Aliun [14]
3 years ago
8

One card is selected at random from a deck of cards. Determine the probability that the card selected is a 10. The probability t

hat the card selected is a 10 is
Mathematics
2 answers:
Alinara [238K]3 years ago
7 0

Answer:

Four out of fifty-two(4/52), or one out of thirteen (1/13)

Step-by-step explanation:

There are 52 cards in standard set of cards.

There are 13 ranks of cards, ranging from Ace(1) to King.

There are four cards of each rank. example) Four aces, four jacks etc

From this information, we can then deduce that the probability of selecting a 10 is 4/52 because there are four tens out of fifty-two cards. Simplified, one out of thirteen.

\frac{4}{52} =\frac{1}{13} \\

Daniel [21]3 years ago
6 0

Answer:

1:13

Step-by-step explanation:

A standard deck has 52 cards not including the joker. each suit has one 10. there are total four "10" in a deck of playing cards. 4/52 = 1/13 or 1:13

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If AB = AD + DB and AD = AC, then AB = AC + DB because of the
weqwewe [10]
Substitution property
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3 years ago
A car costs £20000. Work out the price after a 5% discount.
Mandarinka [93]

Answer:

£19000

Step-by-step explanation:

Discount cost = 5% of 20000 = 5/100 x 20000 = 1000

SO, cost after discount = 20000 - 1000 = 19000

hope this helps

6 0
3 years ago
Jonathan has collected of 400 marbles. Blue marbles make up 17% of his collection.
Inessa05 [86]
Hello there!

Based from this question above, what this is asking is to find how much does 17% goes into 400 marbles.

We want to know how many marbles are in this 400 marbles. (This would be the key point)

We do (17% x 400).

By putting the "%" symbol this allows us the understand that we are trying to find out how many blue marbles they're going to be.

By multiplying them both together, we got our answer as (68) marbles.

Your correct answer would he 68.

I hope this helps you!
3 0
3 years ago
Which of the values shown are potential roots of f(x)=3x^3-13x^2-3x+45? Select all that apply
GrogVix [38]
<h3>f(x)=3x³-13x²-3x+45</h3><h3>f(x)=3x³-9x²-4x²+12x-15x+45</h3><h3>f(x)=3x²(x-3)-4x(x-3)-15(x-3)</h3><h3>f(x)=(x-3)(3x²-4x-15)</h3><h3>f(x)=(x-3)(x-3)(3x+5)</h3><h2>f(x)=(x-3)²(3x+5)</h2>

<h3><u>Roots:</u></h3><h3>x=3</h3><h3>x=-5/3</h3>
5 0
2 years ago
Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.
Cloud [144]

If F_Y(y) is the cumulative distribution function for Y, then

F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)

Then the probability density function for Y is f_Y(y)={F_Y}'(y):

f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}

The nth moment of Y is

E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy

Let u=\ln y, so that \mathrm du=\frac{\mathrm dy}y and y^n=e^{nu}:

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du

Complete the square in the exponent:

nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du

But \frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2} is exactly the PDF of a normal distribution with mean n and variance 1; in other words, the 0th moment of a random variable U\sim N(n,1):

E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1

so we end up with

E[Y^n]=e^{\frac12n^2}

3 0
2 years ago
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