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3 years ago

Graph the ordered-pair solutions of y=lx-1 when x=-5, -3,,0,1,2.

1 answer:

34kurt3 years ago

4 0

Y = x - 1
y = -5 - 1
y = -6
(x, y) = (-5, -6)

y = x - 1
y = -3 - 1
y = -4
(x, y) = (-3, -4)

y = x - 1
y = 0 - 1
y = -1
(x, y) = (0, -1)

y = x - 1
y = 1 - 1
y = 0
(x, y) = (1, 0)

y = x - 1
y = 2 - 1
y = 1
(x, y) = (2, 1)

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4 people can dig a hole in 3 hours. How long does it take 9 people.

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Answer: 1 hour 20 minutes

<u>Step-by-step explanation:</u>

Let's figure out how long it would take 1 person to dig the hole and then divide that number by 9 (since they are sharing the workload).

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3 years ago

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Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

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2 years ago

Answer by today!!! x^2 + (2y) ÷ (2w) + 3z w = 2, x = 5, y = 8, z = 3

inysia [295]

Answer:

4.1

Step-by-step explanation:

5^2 + (2*8) / (2*2) + 3*3

25 + 16 / 4 + 6

41 / 10

4.1

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2 years ago

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statuscvo [17]

Answer:

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3 years ago

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Hoochie [10]

Answer:

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Step-by-step explanation:

We can do it by long division method the required quotient is 3x-5 and r=0

not 3x-1 , r=1

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Now after solving we will get

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