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3 years ago

(CO 2) A random selection from a deck of cards selects one card. What is the probability of selecting a spade

Mathematics

1 answer:

lora16 [44]3 years ago

3 0

Answer:

$P(Spade) = 0.25$

Step-by-step explanation:

Given.

In a deck, there are:

$Cards = 52$

$Spade= 13$

Required

Determine the probability that a selected card is a spade

This is represented as: P(Spade) and it is calculated using:

$P(Spade) = \frac{Spade}{Cards}$

Substitute 13 for Spade and 52 for Cards

$P(Spade) = \frac{13}{52}$

$P(Spade) = 0.25$

Hence, the required probability is 0.25

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3 years ago

Can someone help me please

FinnZ [79.3K]

Answer:

1/9, 3, 27

(A / the first option listed)

Step-by-step explanation:

Here, we are given an expression with an "x" in it, and we are to evaluate possible x values, which means we plug them in as the x in the expression

first,

x = -2

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when we raise something to a negative power, we simply find the positive version of that power, and put a 1 over it:

3² = 3 · 3 = 9

so, 3⁻² = $\frac{1}{9}$ (1/9)

next,

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now, here we have something raised to the power of 1, which is always itself, so 3¹ = 3

(think of ^to the power of as how many times we multiply that number, so 3² = 3 · 3, 3¹ = 3)

and finally,

x = 3

3ˣ → 3³

here, we can multiply 3 by itself 3 times:

3 · 3 · 3 = 27

so, 3³ = 27

[important note: when we raise something to a power, we are not multiplying the base number by that number, but this is a common mistake. but the exponent is how many times we are multiplying that number, which is confusing at first}

So, our answers in order are:

1/9, 3, 27 (the first option listed)

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2 years ago

a red hot ar balloon is 100 feet off the ground and is rising at a rate of 8 feet per second. a green hot air balloon is 160 fee

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Answer:

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Step-by-step explanation:

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3 years ago

Find two power series solutions of the given differential equation about the ordinary point x = 0. y'' + xy = 0

nalin [4]

Answer:

First we write y and its derivatives as power series:

y=∑n=0∞anxn⟹y′=∑n=1∞nanxn−1⟹y′′=∑n=2∞n(n−1)anxn−2

Next, plug into differential equation:

(x+2)y′′+xy′−y=0

(x+2)∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

x∑n=2∞n(n−1)anxn−2+2∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

Move constants inside of summations:

∑n=2∞x⋅n(n−1)anxn−2+∑n=2∞2⋅n(n−1)anxn−2+∑n=1∞x⋅nanxn−1−∑n=0∞anxn=0

∑n=2∞n(n−1)anxn−1+∑n=2∞2n(n−1)anxn−2+∑n=1∞nanxn−∑n=0∞anxn=0

Change limits so that the exponents for x are the same in each summation:

∑n=1∞(n+1)nan+1xn+∑n=0∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−∑n=0∞anxn=0

Pull out any terms from sums, so that each sum starts at same lower limit (n=1)

∑n=1∞(n+1)nan+1xn+4a2+∑n=1∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−a0−∑n=1∞anxn=0

Combine all sums into a single sum:

4a2−a0+∑n=1∞(2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an)xn=0

Now we must set each coefficient, including constant term =0 :

4a2−a0=0⟹4a2=a0

2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an=0

We would usually let a0 and a1 be arbitrary constants. Then all other constants can be expressed in terms of these two constants, giving us two linearly independent solutions. However, since a0=4a2 , I’ll choose a1 and a2 as the two arbitrary constants. We can still express all other constants in terms of a1 and/or a2 .

an+2=−(n+1)nan+1+(n−1)an2(n+2)(n+1)

a3=−(2⋅1)a2+0a12(3⋅2)=−16a2=−13!a2

a4=−(3⋅2)a3+1a22(4⋅3)=0=04!a2

a5=−(4⋅3)a4+2a32(5⋅4)=15!a2

a6=−(5⋅4)a5+3a42(6⋅5)=−26!a2

We see a pattern emerging here:

an=(−1)(n+1)n−4n!a2

This can be proven by mathematical induction. In fact, this is true for all n≥0 , except for n=1 , since a1 is an arbitrary constant independent of a0 (and therefore independent of a2 ).

Plugging back into original power series for y , we get:

y=a0+a1x+a2x2+a3x3+a4x4+a5x5+⋯

y=4a2+a1x+a2x2−13!a2x3+04!a2x4+15!a2x5−⋯

y=a1x+a2(4+x2−13!x3+04!x4+15!x5−⋯)

Notice that the expression following constant a2 is =4+ a power series (starting at n=2 ). However, if we had the appropriate x -term, we would have a power series starting at n=0 . Since the other independent solution is simply y1=x, then we can let a1=c1−3c2, a2=c2 , and we get:

y=(c1−3c2)x+c2(4+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(4−3x+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(−0−40!+0−31!x−2−42!x2+3−43!x3−4−44!x4+5−45!x5−⋯)

y=c1x+c2∑n=0∞(−1)n+1n−4n!xn

Learn more about constants here:

brainly.com/question/11443401

#SPJ4

6 0

1 year ago

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DochEvi [55]

Answer:

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Step-by-step explanation:

hopes this helps:)

4 0

3 years ago