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Ymorist [56]
3 years ago
14

Assume that when adults with smartphones are randomly​ selected,41 ​% use them in meetings or classes. If 5 adult smartphone use

rs are randomly​ selected, find the probability that exactly 2 of them use their smartphones in meetings or classes.
Mathematics
1 answer:
kotegsom [21]3 years ago
3 0

Answer:

0.345

Step-by-step explanation:

Use binomial probability:

P = nCr pʳ qⁿ⁻ʳ

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1−p).

n = 5, r = 2, p = 0.41, and q = 0.59.

P = ₅C₂ (0.41)² (0.59)⁵⁻²

P = 0.345

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Multiplying a trinomial by a trinomial follows the same steps as multiplying a binomial by a trinomial. Determine the degree and
FinnZ [79.3K]

Answer: Degree of polynomial (highest degree) =4

Maximum possible terms =9

Number of terms in the product = 5

Step-by-step explanation:

A trinomial is a polynomial with 3 terms.

The given product of trinomial: (x^2 + x + 2)(x^2 - 2x + 3)

By using distributive property: a(b+c+d)= ab+ac+ad

(x^2 + x + 2)(x^2 - 2x + 3)=(x^2 + x + 2) x^2+(x^2 + x + 2) (-2x)+(x^2 + x + 2)(3)\\\\=x^2(x^2)+x(x^2)+2(x^2)+x^2 (-2x)+x (-2x)+2 (-2x)+x^2 (3)+x (3)+2 (3)\\\\\\=x^4+x^3+2x^2-2x^3-2x^2-4x+3x^2+3x+6

Maximum possible terms =9

Combine like terms

x^4+x^3-2x^3+3x^2-4x+3x+6\\\\=x^4-x^3+3x^2-x+6

Hence, \left(x^2\:+\:x\:+\:2\right)\left(x^2\:-\:2x\:+\:3\right)=x^4-x^3+3x^2-x+6

Degree of polynomial (highest degree) =4

Number of terms = 5

6 0
3 years ago
A system has two failure modes. One failure mode, due to external conditions, has a constant failure rate of 0.07 failures per y
nadya68 [22]

Answer:

0.9177

Step-by-step explanation:

let us first represent the two failure modes with respect to time as follows

R₁(t) for external conditions

R₂(t) for wear out condition ( Wiebull )

Now,

R1(t) = e^{-nt} .....1

where t = time in years = 1,

n = failure rate constant = 0.07

Also,

R2(t)=e^{-(\frac{t}{Q} )^{B} }......2

where t = time in years = 1

where Q = characteristic life in years = 10

and B = the shape parameter = 1.8

Substituting values into equation 1

R1(t) = e^{-(0.07)(1)} \\\\R1(t) = e^{-0.07}

Substituting values into equation 2

R2(t)=e^{-(\frac{1}{10} )^{1.8} }\\\\R2(t)=e^{-(0.1)}^{1.8} }\\\\R2(t)=e^{-0.0158}

let the <em>system reliability </em>for a design life of one year be Rs(t)

hence,

Rs(t) = R1(t) * R2(t)

t = 1

Rs(1) = [e^{-0.07} ] * [e^{-0.0158} ] = 0.917713

Rs(1) = 0.9177 (approx to four decimal places)

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Answer:

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