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bulgar [2K]
3 years ago
9

You spin the spinner shown. A spinner is divided into 8 equal regions. Each region is labeled with a numeral from 1 to 8. How ma

ny possible outcomes are there? There are possible outcomes.
Mathematics
1 answer:
Anastasy [175]3 years ago
8 0

Answer:

Therefore,

There are 8 possible outcomes.

Step-by-step explanation:

Given:

A spinner is divided into 8 equal regions.

While Spinner spins it can stop at

Numeral 1 or

Numeral 2 or

Numeral 3 or

Numeral 4 or

Numeral 5 or

Numeral 6 or

Numeral 7 or

Numeral 8 or

So in All, there are 8 Possible outcomes Also Known as Sample Space

Sample Space:

Total number of favorable outcomes.

Denote by "S"

S = { 1 , 2 , 3 , 4 ,5 , 6, 7 ,  8 }

Possible outcomes = n(S) = 8

Therefore,

There are 8 possible outcomes.

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Round $56.29 to the nearest dollar.
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Step-by-step explanation:

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Consider the sequence 3, -9, 27, -81, .. Find the 14th term of the sequence.
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Answer:

let 'a' be the first term, 'd' be the common difference between all the terms of the sequence

Step-by-step explanation:

therefore, a = 3,

and, d = -9 -3

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hence the 14th term would be,

=> a + 13d

=> 3 + 13( -12 )

=> 3 - 156

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Every 2 minutes, the height of a glider decreases by 8 feet. The initial height was 47 feet.
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Step-by-step explanation:


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A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

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3 years ago
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