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3 years ago

Given A(-2,1) and B(3,-4)

1 answer:

4 0

$\bf f(x)=-2x^2-5x-7\\\\ -------------------------------\\\\ \cfrac{f(x+h)-f(x)}{h} \\\\\\ \cfrac{[-2(x+h)^2-5(x+h)-7]~~-~~[-2x^2-5x-7]}{h} \\\\\\ \cfrac{[-2(x^2+2xh+h^2)-5(x+h)-7]~~-~~[-2x^2-5x-7]}{h} \\\\\\ \cfrac{\underline{-2x^2}-4xh-2h^2\underline{-5x}-5h\underline{-7}\underline{+2x^2+5x+7}}{h} \\\\\\ \cfrac{-4xh-2h^2-5h}{h}\implies \cfrac{\underline{h}(-4x-2h-5)}{\underline{h}}\implies -4x-2h-5$

$\bf -------------------------------\\\\
~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad 
B(\stackrel{x_2}{3}~,~\stackrel{y_2}{-4})\qquad \qquad 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AB=\sqrt{[3-(-2)]^2+[-4-1]^2}\implies AB=\sqrt{(3+2)^2+(-4-1)^2}
\\\\\\
AB=\sqrt{25+25}\implies AB=\sqrt{2\cdot 25}
\\\\\\
AB=\sqrt{2\cdot 5^2}\implies AB=5\sqrt{2}\\\\
-------------------------------\\\\
$

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad 
B(\stackrel{x_2}{3}~,~\stackrel{y_2}{-4})\qquad \qquad 
\left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left(\cfrac{3-2}{2}~~,~~\cfrac{-4+1}{2} \right)\implies \left(\frac{1}{2}~~,~~-\frac{3}{2} \right)\implies \left(\frac{1}{2}~~,~~-1\frac{1}{2} \right)$

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4 years ago

Which expression represents "eight more than one-tenth a"?

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"eight more than one-tenth a" = 8 + 1/10 a or 1/10 a + 8

Answer

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3 years ago

Read 2 more answers

The probability that a single radar station will detect an enemy plane is 0.65.

taurus [48]

Answer:

a) We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

b) If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

Step-by-step explanation:

For each station, there are two two possible outcomes. Either they detected the enemy plane, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

The probability that a single radar station will detect an enemy plane is 0.65. This means that $n = 0.65$ .

(a) How many such stations are required to be 98% certain that an enemy plane flying over will be detected by at least one station?

This is the value of n for which $P(X = 0) \leq 0.02$ .

n = 1.

$P(X = 0) = C_{1,0}.(0.65)^{0}.(0.35)^{1} = 0.35$

n = 2

$P(X = 0) = C_{2,0}.(0.65)^{0}.(0.35)^{2} = 0.1225$

n = 3

$P(X = 0) = C_{3,0}.(0.65)^{0}.(0.35)^{3} = 0.0429$

n = 4

$P(X = 0) = C_{4,0}.(0.65)^{0}.(0.35)^{4} = 0.015$

We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

(b) If seven stations are in use, what is the expected number of stations that will detect an enemy plane?

The expected number of sucesses of a binomial variable is given by:

$E(x) = np$

So when $n = 7$

$E(x) = 7*(0.65) = 4.55$

If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

6 0

3 years ago

To turn 2⁄5 into a decimal, what operation must you perform?

natali 33 [55]

D. 2 ÷ 5.......................

4 0

3 years ago

The inequality 9/5C + 32 <-40 can be used to find Celsius temperatures that are less than -40 Fahrenheit. What is the solutio

Blababa [14]

The solution of inequality is C < -40

Solution:

Given that the inequality that is used to find the celsius temperature that are less than -40 fahrenheit

To find: solution of inequality

Given inequality is:

$\frac{9}{5}C + 32 < -40$

So we have to solve the above inequality for C

$\rightarrow \frac{9}{5}C + 32 < -40$

Add -32 on both sides

$\rightarrow \frac{9}{5}C + 32 -32 < -40 - 32\\\\\rightarrow \frac{9}{5}C < -72$

Multiply both the sides by $\frac{5}{9}$

$\rightarrow\frac{9}{5}C \times \frac{5}{9} < -72 \times \frac{5}{9}\\\\\rightarrow C < -8 \times 5\\\\\rightarrow C < -40$

Thus the solution of inequality is C < -40

3 0

4 years ago