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Ad libitum [116K]
4 years ago
12

Point R is on line segment QS. Given QS= 5x-2, QR= 3x-6, and RS=4x-2, determine the numerical length of RS.

Mathematics
1 answer:
pashok25 [27]4 years ago
8 0

Answer:

The numerical length of RS is 10 units

Step-by-step explanation:

* <em>Lets explain How to solve the problem</em>

- Point R is on line segment QS

- The length of QS is (5x - 2)

- The length of QR is (3x - 6)

- The length of RS is (4x - 2)

- <em>The length of QS is the sum of the lengths of QR and RS</em>

∵ QS = QR + RS

∴ (5x - 2) = (3x - 6) + (4x - 2)

- Simplify the right hand side by adding like terms

∴ 5x - 2 = (3x + 4x) + (-6 + -2)

∴ 5x - 2 = 7x + -8 ⇒ (+)(-) = (-)

∴ 5x - 2 = 7x - 8

- Add 8 for both sides

∴ 5x + 6 = 7x

- Subtract 5x from both sides

∴ 6 = 2x

- Divide both sides by 2

∴ x = 3

- <em>To find the length of RS substitute the value of x in its expression</em>

∵ RS = 4x - 2

∵ x = 3

∴ RS = 4(3) - 2 = 12 - 2 = 10

∴ The numerical length of RS is 10 units

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Answer:

The answer to the question is

The probability of winning second prize (that is, picking five of the six winning numbers) with a 6/44 lottery, as played in Connecticut, Missouri, Oregon, and Virginia is 3.49905×10⁻⁵≡ 0.00003 to five decimal places.

Step-by-step explanation:

The probability of winning the second prize or picking five of the six winning numbers) with a 6/44 lottery is given by

Number of 5 sets of numbers in 44 = ₄₄C₅ = 1086008 ways

Number of 5 set of winning numbers in 44 = 1

Number of ways of picking the last number to make it 6 numbers is given by

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Therefore, there are 38 ways from 1086008 of selecting the 5 second place winning numbers

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4 years ago
The amount dispensed by a soft-drink dispensing machine has a normal distribution with mean μ and standard deviation 0.191 ounce
nirvana33 [79]

Answer:

The value of mean is  \mu = 15.64

Step-by-step explanation:

The normal distribution of the soft-drink dispensing machine can be stated statistically like

                   X ~ N (\mu ,0.191^2)

The standard form representation is

           P(X

     Now we need to obtain \mu in such a way that

                   P(X>16) = 0.01

   In standard form

       Since

                 P(Z>16 -\frac{\mu}{0.191} ) = 0.03

   This means

              P(Z

Now looking at the z-table for probability of 0.99 we obtain 1.88

    i.e

            16 - \frac{\mu}{0.191} = 1.88

Making \mu\ the \ subject

        16 - \mu = 1.88 *0.191

       \mu = 16 -(1.88*0.191)

          =15.64

   

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

Answer

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