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3 years ago

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Find the midpoint of the line segment joining the points (10,7) and (negative 2,negative 7).

1 answer:

Basile [38]3 years ago

8 0

The midpoint of the line segment joining the points (10,7) and (negative 2,negative 7) is (4, 0)

<h3><u>Solution:</u></h3>

Given that line segment joining the points (10,7) and (negative 2, negative 7)

Thus the two points are (10, 7) and (-2, -7)

The midpoint of two points $(x_1, y_1)$ and $(x_2, y_2)$ is given as:

$M(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

$\text {Here } x_{1}=10 \text { and } y_{1}=7 \text { and } x_{2}=-2 \text { and } y_{2}=-7$

Substituting the values in formula, we get

$\begin{array}{l}{M(x, y)=\left(\frac{10+(-2)}{2}, \frac{7+(-7)}{2}\right)} \\\\ {M(x, y)=\left(\frac{10-2}{2}, \frac{7-7}{2}\right)} \\\\ {M(x, y)=(4,0)}\end{array}$

Thus the midpoint of line segment is (4, 0)

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2 years ago

Sarah,Leah,Haley and Kim bought 5 bottles of water to share among themselves. They divided each bottle of water into 4 equal por

Irina18 [472]

Answer:

Sarah have $\frac{5}{4}$ bottle of water.

Step-by-step explanation:

We are given that,

Total number of bottles = 5

As, each bottle is divided into 4 portions.

The amount of water per person = $\frac{1}{4}$

Further, Sarah took 1 portion from each bottle and there are 5 bottles.

So, the amount of water Sarah have = $\frac{1}{4}\times 5$ = $\frac{5}{4}$

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3 years ago

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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