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erastova [34]
3 years ago
5

How many ways are possible to choose 3 days out of February? NOTE: Use 28 days for the number of days in February. A) 3276 B) 37

8 C) 20475 D) 2925
Mathematics
1 answer:
Charra [1.4K]3 years ago
5 0

Answer:

A) 3,276 ways.

Step-by-step explanation:

In this case, choosing February 1, February 12, and February 20 would be the same thing as choosing February 12, February 1, and February 20. So, since order does not matter, we will use a combination to solve the question.

The formula for combinations is...

n! / [r!(n - r)!], where n = the number of days in February (28) and r = the number of days you are choosing (3).

28! / [3! * (28 - 3)!]

= 28! / (6 * 25!)

= (28 * 27 * 26) / 6

= (14 * 9 * 26) / 1

= 14 * 9 * 26

= 126 * 26

= A) 3,276.

Hope this helps!

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Answer:

y = 3/4 x -1

Step-by-step explanation:

The equation for a line with the slope and the intercept is y = mx+b  where m is the slope and b is the y intercept

y = 3/4 x -1


5 0
3 years ago
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Wanda pays $19.50 per month for her gym membership. She can take yoga classes at her gym that
PSYCHO15rus [73]

Answer: She took 6 yoga classes

Step-by-step explanation: So first

Subtract

$43.50 - $19.50 = $24

Carry then Divide

$24 / 8 = 6 Classes

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6 0
3 years ago
Sam plans to walk his dog a distance of a mileHe walks 3/8 of a mile and stop and get a bottle of waterThen he walks 1/8 of a mi
DedPeter [7]

Answer:

1/2 mile

Step-by-step explanation:

3/8+1/8=4/8 or 1/2 simplified

1-1/2=1/2

same has to walk 1/2 mile more to walk his dog a full mile

3 0
3 years ago
Which of the equations below could be the equation of this parabola?
nirvana33 [79]

Answer:

 y=-4x^2  is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

y=-4x^2

\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}

\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)

As

\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)

\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}

y=-4x^2

\mathrm{The\:parabola\:params\:are:}

a=-4,\:m=0,\:n=0

x_v=\frac{m+n}{2}

x_v=\frac{0+0}{2}

x_v=0

\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=-4\cdot \:0^2

y_v=0

Therefore, the parabola vertex is

\left(0,\:0\right)

\mathrm{If}\:a

\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}

a=-4

\mathrm{Maximum}\space\left(0,\:0\right)

so,

\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)

Therefore,  y=-4x^2  is the equation of this parabola. The graph is also attached.

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