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2 years ago

Find the coordinates of the missing endpoint if B is the midpoint of AC. C(-5,4), B(-2,5) I need the steps and answer pls and ty

1 answer:

3 0

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ x}}\quad ,&{{ y}})\quad 
% (c,d)
C&({{ -5}}\quad ,&{{ 4}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)$

$\bf \left( \cfrac{-5+x}{2}~,~\cfrac{4+y}{2} \right)=\stackrel{B}{(-2,5)}\implies 
\begin{cases}
\cfrac{-5+x}{2}=-2\\\\
-5+x=-4\\
\boxed{x=1}\\
----------\\
\cfrac{4+y}{2}=5\\\\
4+y=10\\
\boxed{y=6}
\end{cases}$

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Solve each system by substitution<br><br> 2x - 3y = -1<br> X + 4y = -6

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Answer:

(- 2, - 1 )

Step-by-step explanation:

Given the 2 equations

2x - 3y = - 1 → (1)

x + 4y = - 6 → (2)

Rearrange (2) expressing x in terms of y by subtracting 4y from both sides

x = - 6 - 4y → (3)

Substitute x = - 6 - 4y into (1)

2(- 6 - 4y) - 3y = - 1 ← distribute and simplify left side

- 12 - 8y - 3y = - 1

- 12 - 11y = - 1 ( add 12 to both sides )

- 11y = 11 ( divide both sides by - 11 )

y = - 1

Substitute y = - 1 into (3) for corresponding value of x

x = - 6 - 4(- 1) = - 6 + 4 = - 2

Solution is (- 2, - 1 )

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2 years ago

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2 years ago

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muminat

Answer:

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Step-by-step explanation:

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2 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

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2 years ago

A man surveyed a sample of 36 high school students and asked, "How many days in the past week have you consumed an alcoholic b

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Answer: