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3 years ago

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There are five candidates for class president. McKayla and Jacob have the same chance of winning. Daniel has a 20% chance of win

ning, and Samantha and Maria are both half likely to win as Daniel. Create a table of probabilities for the sample space.

1 answer:

Marysya12 [62]3 years ago

3 0

Answer:

Mckayla 30%

Jacob 30%

Daniel 20%

Samantha 10%

Maria 10%

Step-by-step explanation: First Divide 20% into half to get samantha and marias percent then add those together then subtract what you get from that to 100% and then divide into half for Mckayla and Jacobs percent

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Oscar uses place value to divide 63.81 by 9 his steps are shiwn below what mistake does oscar make

Ksenya-84 [330]

Check the picture below.

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3 years ago

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A class of 32 students is organised in 33 teams every team consists of 3 students and there are no identical teams . show that t

Tresset [83]

Answer:

Step-by-step explanation:

Let's start by making up as many teams as we can with the 32 student. Given that each team is different, we can make 10 teams of 3 each. (we still have 23 more teams to make).

The last two people make a team of only 2. No matter which student from the 30 other students is picked, the team of two and the one the student is coming from will have one student in common. Though there are more borrowings that take place (many more), the results remain as stated. At least 2 teams will have 1 person in common.

The method is called the pigeon hole method.

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3 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

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3 years ago

What is the solution to<br><br> −(6m+8) =4(17−m)

Sedbober [7]

Answer:

$\boxed{m = -38}$

Step-by-step explanation:

$-(6m+8) =4(17-m)$

→ Expand brackets

$-6m-8=68-4m$

→ Add 8 to both sides to collect the whole numbers

$-6m = 76-4m$

→ Add -4m to both sides to collect the m terms

$-2m=76$

→ Divide both sides by -10 to isolate m

$m=-38$

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3 years ago

How to calculate roots when the equation is non quadratic?

likoan [24]

Solve like a linear algebra equation
$y = mx + b$
use for slope
$m = y1 - y2 \div x1 - x2$
the b is intersept

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3 years ago