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Helga [31]
4 years ago
15

Find n(A) for the set. A = {4, 6, 8,1 0, 12}

Mathematics
1 answer:
slega [8]4 years ago
4 0

Answer:

n(A) = 5

Step-by-step explanation:

The notation "n(A)" means the "number of elements in the set A".

Note, that this has to be "unique" elements in the set. If there is five 6's, we will count as one...

Looking at the set given, there are 4, 6, 8, 10, 12 -- 5 elements. All are unique, no repetitions. Hence n(A) = 5

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The answer to your questions is 3.12
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4 years ago
A company has 2 machines that manufacture widgets. An older machine manufactures 25% defective widgets, while the new machine ma
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Answer:

73.90%

Step-by-step explanation:

Let Event D=Defective,  D' = Non Defective

Let Event N=New Machine, N' = Old Machine

From the given information:

P(D|N')=0.25\\P(D|N)=0.09\\P(N)=0.7\\P(N')=0.3

We are required to calculate the probability that a widget was manufactured by the new machine given that it is non defective.  

i.e. P(N|D')

P(D'|N')=1-P(D|N')=1-0.25=0.75\\P(D'|N)=1-P(D|N)=1-0.09=0.91

Using Baye's Law of conditional Probability

P(N|D')=\dfrac{P(D'|N)P(N)}{P(D'|N)P(N)+P(D'|N')P(N')} \\=\dfrac{0.91*0.7}{0.91*0.7+0.75*0.3}\\ =0.73897\\\approx 0.7390

Therefore given that a selected widget is non-defective, the probability that it was manufactured by the new machine is 73.9%.

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3 years ago
In a standard Normal distribution, what percentage of observations lie between z = 0.37 and z = 1.65?
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Answer:

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Step-by-step explanation:

7 0
3 years ago
What is the approximate volume of a cone with a height of 6 mm and radius of 18 mm?
Verdich [7]
Given:
radius = 18mm
height = 6mm
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volume of a cone is computed by multiplying pi, radius raised to the 2nd power, and height, which has been divided by 3.

v = πr²h/3

v = 3.14 * (18mm)² x 6mm/3
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<span>volume rounded to the nearest tenth would be 2034.70 mm³</span>
6 0
3 years ago
Read 2 more answers
50 POINTS 10 QUESTIONS! <br><br> Solve each quadratic equation using factoring<br><br> PLEASE HELP!!
Tanzania [10]

Answer:

Quadratic Equation | Factoring

Solve each quadratic equation using factoring.

1) v² + 5v + 6 = 0

doing middle term factorisation

v²+(3+2)v+6=0

v²+3v+2v+6=0

v(v+3)+2(v+3)=0

(v+3)(v+2)=0

either

<u>v=-3</u>

<u>or</u>

<u>v=-2</u>

2) g² - 3g = 4

keeping all terms in one side

g²-3g-4=0

doing middle term factorisation

g²-(4-1)g-4=0

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g(g-4)+1(g-4)=0

(g-4)(g+1)=0

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<u>g=4</u>

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w(w+4)=0

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<u>w=0</u>

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4) s² - 8s + 12 = 0

doing middle term factorisation

s²-(6+2)+12=0

s²-6s-2s+12=0

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(s-6)(s-2)=0

either

<u>s=6</u>

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doing middle term factorisation

x²+(7-5)x-35=0

x²+7x-5x-35=0

x(x+7)-5(x+7)=0

(x+7)(x-5)=0

either

<u>x=-7</u>

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6) r(r + 2) = 99

opening bracket

r²+2r=99

keeping all terms in one side

r²+2r-99=0

r²+(11-9)r-99=0

r²+11r-9r-99=0

r(r+11)-9(r+11)=0

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opening bracket

k²-4k=-3

keeping all terms in one side

k²-4k+3=0

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8)t²+ 3t + 2 = 0

doing middle term factorisation

t²+(2+1)t+2=0

t²+2t+t+2=0

t(t+2)+1(t+2)=0

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either

<u>t</u><u>=</u><u>-</u><u>2</u>

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doing square root in both side

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<u>m=±9</u>

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<u>m</u><u>=</u><u>9</u>

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<u>m</u><u>=</u><u>-</u><u>9</u>

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6 0
3 years ago
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