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Darya [45]
3 years ago
7

Alysha is two years younger than Bryce the sum of their ages is 28 how old is Alysha

Mathematics
2 answers:
IrinaK [193]3 years ago
8 0

Alright, lets get started.

Suppose the age of Alysha = x

As given in question, Alysha is 2 years younger than Bryce.

So age of Bryce will be = x+2

As per given in question the sum of their ages is 28, so

x + x +2 = 28

2x + 2 = 28

Subtracting 2 fom both sides

2x + 2 - 2 = 28 - 2

2x = 26

Dividing 2 in both sides

x = 13

Means age of Alysha will be = 13

Means age of Bryce will be = 13 +2 = 15

Alysha and Bryce ages are 13 and 15 years respectively. : Answer

Hope it will help :)


vova2212 [387]3 years ago
5 0
Alysha is 13 yrs old while Bryce is 15
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If Juan drives 24 miles in 3 hours,how many miles does Juan drive in 1 hour
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PLEASE HELP SUPER EASY!!!
Effectus [21]

A1. 12 i.e option D

A2. 3n-7 i.e option A

A3. -6n+20 i.e option D

A4. -70 i.e option C

Step-by-step explanation:

aₙ = a₁ + (n - 1) × d  

aₙ = the nᵗʰ term in the sequence

a₁ = the first term in the sequence

d = the common difference between terms

Using the above formula to solve the first part, we have :

  • -8 = a₁ + (2-1) × 5
  • -13 = a₁
  • a₆ = -13 + (6-1) × 5
  • a₆ = 12

For the second part, we have :

  • aₙ = -4 + (n-1)×3
  • aₙ = -4 + 3n -3
  • aₙ = 3n-7

For the third part, we have :

  • a₁=14 ; d=-6
  • aₙ = 14 + (n-1)×(-6)
  • aₙ = -6n + 20

For the fourth part, we have :

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7 0
3 years ago
Thetime(inhours)requiredtorepairamachineis an exponentially distributed random variable with parameter λ = 1 . What is 2 (a) the
mihalych1998 [28]

Answer:

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Step-by-step explanation:

Let's start by defining the random variable T.

T : ''The time (in hours) required to repair a machine''

T ~ exp (λ)

T ~ exp (1)

The probability density function for the exponential distribution is

(In the equation I replaced λ = L)

f(x)=Le^{-Lx}

With L > 0 and x ≥ 0

In this exercise λ = 1 ⇒

f(x)=e^{-x}

For a)

P(T>2)

P(T>2)=1-P(T\leq 2)

P(T>2)=1-\int\limits^2_0 {e^{-x} } \, dx

P(T>2)=1-(-e^{-2}+1)

P(T>2)=e^{-2}=0.1353

For b)

P(T\geq 10/T>9)

The event (T ≥ 10 / T > 9) is equivalent to the event T ≥ 1 so they have the same probability of occur

P(T\geq 10/T>9)=P(T\geq 1)

P(T\geq 1)=1-P(T

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