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4 years ago

Please help!!! (-x/2)+(1/2x)=(-4/x)

1 answer:

7 0

Are you solving this for x? You can't because the left side is eliminated by subtracting (1/2)x from (1/2)x. That leaves nothing on the left. $0= \frac{-4}{x}$
you can't solve that for x because when you multiply both sides by x, 0*x = 0 and $0 \neq 4$
I think you made a mistake in your typing or something...unless the answer was no solution!

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Question 7 of 10

Afina-wow [57]

Answer:

B. x = 5/2

Step-by-step explanation:

quadratic formula is:

( $x=-b$ ± $\sqrt{b^2-4ac}$ )/2a

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in this equation, a = 1, b = -5, c = 25/4

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2 years ago

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.

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3 years ago

Synthetic Division <br> -2 | 1 2 -3 1

Tom [10]

Answer:

Step-by-step explanation:

I am assuming that the division problem looks like this:

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Going off that assumption, we will work this problem. The first thing you always do in the execution of synthetic division is to bring down the first number. Then multiply that number by the one "outside", which is -2, then put that number up under the next number in the line:

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-2

Now add the 2 and -2 and bring that down as a 0 and multiply the -2 times the 0:

-2| 1 2 -3 1

-2 0

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Now add -3 and 0 to get -3 and multiply that -3 times the -2 and put the product up under the next numbe in line;

-2| 1 2 -3 1

-2 0 6

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Now add the 1 and the 6 to get the remainder:

Read more on Brainly.com - brainly.com/question/12992907#readmore

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3 years ago

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alina1380 [7]

Answer:

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