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Alla [95]
3 years ago
14

What is the simplified version of 3\sqrt{135}

Mathematics
2 answers:
Aleonysh [2.5K]3 years ago
7 0

Answer:

The simplified version of \sqrt[3]{135} is 3\sqrt[3]{5}.

Step-by-step explanation:

The given expression is

\sqrt[3]{135}

According to the property of radical expression.

\sqrt[n]{x}=(x)^{\frac{1}{n}}

Using this property we get

\sqrt[3]{135}=(135)^{\frac{1}{3}}

\sqrt[3]{135}=(27\times 5)^{\frac{1}{3}}

\sqrt[3]{135}=(3^3\times 5)^{\frac{1}{3}}

\sqrt[3]{135}=(3^3)^{\frac{1}{3}}\times (5)^{\frac{1}{3}}      [\because (ab)^x=a^xb^x]

\sqrt[3]{135}=3\times \sqrt[3]{5}     [\because \sqrt[n]{x}=(x)^{\frac{1}{n}}]

\sqrt[3]{135}=3\sqrt[3]{5}

Therefore the simplified version of \sqrt[3]{135} is 3\sqrt[3]{5}.

ohaa [14]3 years ago
4 0

9 \sqrt{15}
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Answer with Step-by-step explanation:

a) {n}^{2}  + 3 \\  \\ n = 1 \\  {1}^{2}  + 3 \\ 1 + 3 \\  = 4 \\  \\ n = 2 \\  {2}^{2}  + 3 \\ 4 + 3 \\  = 7 \\  \\ n = 3 \\  {3}^{2}  + 3 \\ 9 + 3 \\  = 12 \\  \\ n = 4 \\  {4}^{2}  + 3 \\ 16 + 3 \\  = 19 \\  \\ n = 10 \\  {10}^{2}  + 3 \\ 100 + 3  \\  = 103

In this sequence,

1st term = 4

2nd term = 7

3rd term = 12

4th term = 19

10th term = 103

b)2 {n}^{2}  \\  \\ n = 1 \\ 2 \times  {1}^{2}  \\ 2 \times 1 \\  = 2 \\  \\ n = 2 \\ 2 \times  {2}^{2}  \\ 2 \times 4 \\  = 8 \\  \\ n = 3 \\ 2 \times  {3}^{2}  \\ 2 \times 9 \\  = 18 \\  \\ n = 4 \\ 2 \times  {4}^{2}  \\ 2 \times 16 \\ =   32 \\  \\ n = 10 \\ 2 \times  {10}^{2}  \\ 2 \times 100 \\  = 200

in this sequence,

1st term = 2

2nd term =8

3rd term = 18

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3 years ago
Of 500 people surveyed, 460 liked the new flavor of Mintie Toothpaste. Suppose the manufacturer operates on the standard that ov
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3 years ago
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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!!!

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