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Charra [1.4K]
2 years ago
15

Melissa has three different positive integers. She adds their reciprocals together and gets a sum of $1$. What is the product of

her integers? Melissa has three different positive integers. She adds their reciprocals together and gets a sum of $1$. What is the product of her integers?
Mathematics
2 answers:
ryzh [129]2 years ago
6 0

Answer:

36

Step-by-step explanation:

Let the three positive integers be x, y, and z. Then

1/x + 1/y + 1/z = 1

Assume x = 2.

Then 1/x = ½ and 1/y + 1/x = 1/2

Divide the second portion (1/y + 1/z) into three parts.

3/6 = 1/6 + (1/6 +1/6)

Combine two of the fractions.

1/2 = 1/6 + 2/6

1/2 = 1/6 + 1/3

1/2 + 1/3 + 1/6 = 1

The integers are 2, 3, and 6.

2 × 3 × 6 = 36

The product of Melissa’s integers is 36.

Whitepunk [10]2 years ago
4 0

If the three integers are a,b,c, then we have

\dfrac1a+\dfrac1b+\dfrac1c=1

We can combine the fractions on the left side:

\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=\dfrac{bc+ac+ab}{abc}=1

\implies abc=bc+ac+ab

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denpristay [2]

Answer:

-73, -34, 0.5, 1.2, 23

Step-by-step explanation:

In Mathematics, the set of numbers are arranged in two ways. It means that the numbers can be arranged either in ascending order or descending order. If the numbers are arranged from the least to the greatest, then it is called ascending order. In this form, the numbers are in increasing order. The first number should be lesser than the second number.

If the numbers are arranged from the greatest to the least, then it is called descending order. In this form, the numbers are in decreasing form. The first number should be greater than the second number.

Also, read: Descending Order

Standard Form

<em>The standard form to represent the least to the greatest arrangement </em>of numbers is given by:

a < b < c < d  < …..

Here,  

a, b, c, d represent the numbers

Example: 2 < 5 < 7 < 8

5 0
2 years ago
Which ordered pairs are solutions to the inequality 2y−x≤−6 ?
Vinil7 [7]
Given inequality: 2y−x ≤ −6

Option-1 : (-3,0)
2×0 - (-3) = 0 + 3 = 3 > -6
Not satisfied

Option-2 : (6,1)
2×1 - 6 = 2 - 6 = -4 > -6
Not satisfied

Option-3 : (1, -4)
2×(-4) - 1 = -8 - 1 = -9 < -6
Satisfied.
Thus, (1, -4) is a solution.

Option-4 : (0, -3)
2×(-3) - 0 = -6 - 0 = -6 = -6
Satisfied.
Thus, (0, -3) is a solution.

Option-5 : (2, -2)
2×(-2) - 2 = -4 - 2 = -6 = -6
Satisfied.
Thus, (2, -2) is a solution.

Solutions are: (1, -4), (0, -3) , (2, -2)

4 0
2 years ago
A lake contains 4 distinct types of fish. Suppose that each fish caught is equally likely to be any one of these types. Let Y de
strojnjashka [21]

Answer:

a) P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

a= μ-3.16*σ , b= μ+3.16*σ

b) P(Y≥ μ+3*σ ) ≥ 0.90

b= μ+3*σ

Step-by-step explanation:

from Chebyshev's inequality for Y

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

where

Y =  the number of fish that need be caught to obtain at least one of each type

μ = expected value of Y

σ = standard deviation of Y

P(| Y - μ|≤ k*σ ) = probability that Y is within k standard deviations from the mean

k= parameter

thus for

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

P{a≤Y≤b} ≥ 0.90 →  1-1/k² = 0.90 → k = 3.16

then

P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

using one-sided Chebyshev inequality (Cantelli's inequality)

P(Y- μ≥ λ) ≥ 1- σ²/(σ²+λ²)

P{Y≥b} ≥ 0.90  →  1- σ²/(σ²+λ²)=  1- 1/(1+(λ/σ)²)=0.90 → 3= λ/σ → λ= 3*σ

then for

P(Y≥ μ+3*σ ) ≥ 0.90

5 0
2 years ago
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Answer:

Step-by-step explanation:

$27.63 / 2 = $13.82

$13.82 * 0.20 = $2.76

$13.82 + $2.76 = $16.58

6 0
1 year ago
A group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are list
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Answer:

Step-by-step explanation:

Given that a group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are listed below.

Data set is as ollows:

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Mean = 62.27Variance =333.35std dev = 18.258std error = 4.714

H0: mu = 60 sec

Ha: mu not equals 60 sec

Mean diff = 2.27

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Test statistic = 2.27/SE =0.4815

p value =0.6376

Since p>0.05, we accept null hypothesis

i.e. there is statistical evidence to say that students are reasonably good at estimating one​ minute

8 0
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