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Nataly [62]
3 years ago
9

Find the equation of the line with a slope of 2/5 and containing the point (10,6).

Mathematics
1 answer:
Kipish [7]3 years ago
4 0

Use the poin-slope form of the equation of a line:-


y - y1 = m(x - x1) (where m = slope and the point is (x1, y1))


Plugging in the given values:-


y - 6 = 2/5(x - 10)


multiplying through by 5:-


5y - 30 = 2(x - 10)

5y - 30 = 2x - 20

5y = 2x + 10


In standard form the equation is


2x - 5y = -10 (answer)



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salantis [7]

Answer:

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Step-by-step explanation:

The area can be figured as the difference of the enclosing rectangle area ((3x-2)×(4x-1)) and the area of the white space at the lower left (horizontal dimension x+1).

The vertical dimension of the white space at lower left is The difference of the given vertical dimensions:

  (4x -1) -(x -2) = 3x +1

So, the area of the figure is ...

  (3x -2)(4x -1) -(x +1)(3x +1)

  = (12x² -3x -8x +2) -(3x² +x +3x +1) . . . . "FOIL" each product

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<em>Comment on products of binomials</em>

The product of any polynomial with any other can be found using the distributive property repeatedly. When both polynomials are <em>binomials</em>, a mnemonic can help you make sure all product terms are included.

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8 0
3 years ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.7
Zinaida [17]

Answer:

(a) The probability that the sample average sediment density is at most 3.00 is 0.092.

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(b) The sample size must be at least 77.

Step-by-step explanation:

The random variable <em>X</em> ca be defined as the sediment density (g/cm) of a specimen from a certain region.

The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 2.7 and standard deviation, <em>σ</em> = 0.75.

(a)

A random sample of <em>n</em> = 25 specimens is selected.

Compute the probability that the sample average sediment density is at most 3.00 as follows:

Apply continuity correction:

P(\bar X\leq 3.00)=P(\bar X

                    =P(\bar X

                    =P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}

                    =P(Z

Thus, the probability that the sample average sediment density is at most 3.00 is 0.092.

Compute the probability that the sample average sediment density is between 2.70 and 3.00 as follows:

P(2.70

                                =P(0

*Use a <em>z</em>-table.

Thus, the probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b)

It is provided that:

P(\bar X\leq 3.00)\geq 0.99

P(\bar X

P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}

P(Z

The value of <em>z</em> for the probability above is

<em>z</em> ≥ 2.33

Compute the value of <em>n</em> as follows:

\frac{|2.50-2.70|}{0.75/\sqrt{n}}\geq 2.33

\frac{|-0.20|}{2.33}\geq \frac{0.75}{\sqrt{n}}

\sqrt{n}\geq \frac{0.75\times 2.33}{|-0.20|}\\\sqrt{n}\geq 8.7375\\n\geq 76.3439\\\approx n\geq 77

Thus, the sample size must be at least 77.

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AveGali [126]

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8 0
3 years ago
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