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3 years ago

Which expression is equivalent to 60x^20y^24/30x^10y^12 ?

Mathematics

2 answers:

olganol [36]3 years ago

8 0

Answer:

2x^(10)y^(12)

Hope This Helps! Have A Nice Day!!

mariarad [96]3 years ago

3 0

Answer:

The correct answer is 2X¹⁰Y¹²

Step-by-step explanation:

Points to remember

identities

Xᵃ * Xᵇ = X⁽ᵃ⁺ᵇ⁾

Xᵃ/Xᵇ = X⁽ᵃ⁻ᵇ⁾

To find the equivalent to given expression

It is given that,

60X²⁰Y²⁴/30X¹⁰Y¹²

Using identities we can write,

60X²⁰Y²⁴/30X¹⁰Y¹² = (60/30) * (X²⁰/X¹⁰) * (Y²⁴/y¹²)

= 2 * X⁽²⁰ ⁻ ¹⁰⁾ * Y⁽²⁴ ⁻ ¹²⁾

= 2 * X¹⁰ * Y¹²

= 2X¹⁰Y¹²

The correct answer is 2X¹⁰Y¹²

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A sprinkler system inside an officebuilding has 2 types of not-very-reliable activation devices,A1 and A2, which operate indepen

Charra [1.4K]

Answer:

(a) The probability that both sprinklers operate correctly is 0.18.

(b) The probability that at least one sprinkler operates correctly is 0.72.

(d) The probability that neither sprinkler operates correctly is 0.28.

(e) The probability that no more than one sprinkler operates correctly is 0.82.

Step-by-step explanation:

The probability of the two sprinklers operating is:

P (A₁) = 0.60

P (A₂) = 0.30

The two sprinklers operate independently.

(a)

If events X and Y are independent than, $P(X\cap Y)=P(X)\times P(Y)$ .

Compute the probability that both sprinklers operate correctly as follows:

$P(A_{1}\cap A_{2})=P(A_{1})\times P(A_{2}) = 0.60 \times0.30=0.18$

Thus, the probability that both sprinklers operate correctly is 0.18.

(b)

Compute the probability that at least one sprinkler operates correctly as follows:

P (At least 1 operates correctly) = 1 - P (None operates correctly)

$=1-P(A^{c}_{1}\cap A^{c}_{2} )\\= 1 - P(A^{c}_{1})P(A^{c}_{2})\\= 1 -[1-P(A_{1})][1-P(A_{2})]\\= 1 - [1-0.60][1-0.30]\\=1-(0.40\times0.70)\\=0.72$

Thus, the probability that at least one sprinkler operates correctly is 0.72.

(c)

Compute the probability that exactly one sprinkler fails to operate as follows:

P (Exactly One fails to operates) = P (A₁ works but A₂ does not) +

P (A₁ does not works but A₂ works)

$=P(A_{1}\cap A^{c}_{2})+P(A^{c}_{1}\cap A_{2})\\=[P(A_{1})P(A^{c}_{2})]+[P(A^{c}_{1})P(A_{2})]\\=[P(A_{1})(1-P(A_{2}))]+[(1-P(A_{1}))P(A_{2})]\\=[0.60(1-0.30)]+[(1-0.60)0.30]\\=0.54$

Thus, the probability that exactly one sprinkler fails to operate is 0.54.

(d)

Compute the probability that neither sprinkler operates correctly as follows:

P (Neither operate correctly) = P (A₁ and A₂ does not works)

$=P(A^{c}_{1}\cap A^{c}_{2} )\\= P(A^{c}_{1})P(A^{c}_{2})\\= [1-P(A_{1})][1-P(A_{2})]\\= [1-0.60][1-0.30]\\=(0.40\times0.70)\\=0.28$

Thus, the probability that neither sprinkler operates correctly is 0.28.

(e)

Compute the probability that no more than one sprinkler operates correctly as follows:

P (No more than 1 operates correctly) = 1 - P (Both operates correctly)

$=1-P(A_{1}\cap A_{2})\\= 1-0.18\\=0.82$

Thus, the probability that no more than one sprinkler operates correctly is 0.82.

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The correct answer is D
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