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Nadya [2.5K]
2 years ago
13

There were s students at a summer camp in 2010. In 2011, there were half as many students as the previous year. In 2012, there w

ere 54 fewer students than in 2010. What expression would represent the total number of students at the summer camp for all three years?
55s
Five-halves s minus 54
3s – 54
Five-halves s + 54
Mathematics
1 answer:
valentina_108 [34]2 years ago
8 0

The expression that would represent the total number of students at the summer camp for all three years is (b) 5.5s - 54

<h3>How to determine the expression?</h3>

Let the number of students in 2010 be s.

So, we have:

  • Year 2011 = 1/2s i.e. half as many students in 2010
  • Year 2012 = s - 54 i.e. 54 students less than 2010

The total number of students is:

Total = s + 1/2s + s - 54

Evaluate

Total = 9s/2 - 54

Rewrite s:

Total = 5.5s - 54

Hence, the expression that would represent the total number of students at the summer camp for all three years is (b) 5.5s - 54

Read more about expressions at:

brainly.com/question/723406

#SPJ1

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3 years ago
Let x1, x2, and x3 represent the times necessary to perform three successive repair tasks at a certain service facility. suppose
kirza4 [7]

Answer:

P(T₀ < 200) = 0.99856

P(150 < T₀ < 200) = 0.99856

Step-by-step explanation:

The expected values for each of the tasks is μ₁ = 60, μ₂ = 60, μ₃ = 60

The variances for each of the 3 tasks

σ₁² = 15, σ₂² = 15, σ₃² = 15

calculate P(T₀ < 200) and P(150 < T₀ < 200)

When independent distributions are combined, the combined mean and combined variance are given through the relation

Combined mean = Σ λᵢμᵢ

(summing all of the distributions in the manner that they are combined)

Combined variance = Σ λᵢ²σᵢ²

(summing all of the distributions in the manner that they are combined)

Distribution of total time taken for the 3 successive tasks

= X₁ + X₂ + X₃

Expected value = Combined Mean = μ₁ + μ₂ + μ₃ = 60 + 60 + 60 = 180

Combined Variance = 1²σ₁² + 1²σ₂² + 1²σ₃²

= (1² × 15) + (1² × 15) + (1² × 15)

= 45

standard deviation of the combined distribution = √(variance) = √45 = 6.708

Since each of the distributions are said to be normal, the combined distribution too, is normal.

P(T₀ < 200)

We first standardize 200

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (200 - 180)/6.708 = 2.98

The required probability

= P(T₀ < 200) = P(z < 2.98)

We'll use data from the normal probability table for these probabilities

P(T₀ < 200) = P(z < 2.98) = 0.99856

b) P(150 < T₀ < 200)

We first standardize 150 and 200

For 150

z = (x - μ)/σ = (150 - 180)/6.708 = -4.47

For 200

z = (x - μ)/σ = (200 - 180)/6.708 = 2.98

The required probability

= P(150 < T₀ < 200) = P(-4.47 < T₀ < 2.98)

We'll use data from the normal probability table for these probabilities

P(150 < T₀ < 200) = P(-4.47 < T₀ < 2.98)

= P(z < 2.98) - P(z < -4.47)

= 0.99856 - 0.0000 = 0.99856

Hope this Helps!!!

6 0
3 years ago
Help me with this please!!!!!!!!!!!!!!!!!!!!!
Andrej [43]

Answer:

y = 2 units

Step-by-step explanation:

We can solve this in 2 ways, both ways would give the same answer:

<u>#1) Sin(45) = </u>\frac{\sqrt{2} }{y}<u />

=> Sin(45) * y = \sqrt{2}

=> y = \frac{\sqrt{2} }{Sin(45)}

=> y = 2

<u>#2) Cos(45) = </u>\frac{\sqrt{2} }{y}<u />

=> Cos(45) * y = \sqrt{2}

=> y = \frac{\sqrt{2} }{Cos(45)}

=> y = 2

Hope this helps!

8 0
3 years ago
Read 2 more answers
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