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3 years ago

I WILL MARK YOU THE HIGHEST IF RIGHT NEED ASAP!!! DONT ANSWER IF YOU DONT KNOW

Mathematics

1 answer:

Len [333]3 years ago

4 0

Sin 30= x/8
X= sin 30*8
X=4

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Which is a correct first step in solving 5 - 2x < 8x - 3?

charle [14.2K]

Answer:

5 < 10x-3

Step-by-step explanation:

5 - 2x < 8x - 3

Add 2x to each side

5 - 2x+2x < 8x - 3+2x

5 < 10x-3

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3 years ago

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URGENT!!!!!

hichkok12 [17]

Answer:

Step-by-step explanation:

1. Cross Multiply. $\frac{5}{6} =\frac{30}{x}$

2. Solve. $\frac{5}{6} =\frac{30}{x} =36$

3. Your answer is 36

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Kelly has four times as many songs on her music player as Lou Tiffany has six times as many songs on your music player as Lou.al

Ivahew [28]

Kelly: 4y Lou: y Tiffany: 6y

4y + y + 6y = 11y

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Kelly: 4x62 = 248 songs
Lou: 1x63 = 62 songs
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3 years ago

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Leni [432]

hi buddy i can help you with this

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3 years ago

Military radar and missile detection systems are designed to warn a coutry of an enemy attack. Assume that a particular detectio

Tom [10]

Answer:

0.96 = 96% probability that at least one of them detect an enemy attack.

Step-by-step explanation:

For each radar, there are only two possible outcomes. Either it detects the attack, or it does not. The missiles are operated independently, which means that we use the binomial probability distribution to solve this question.

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 combinations 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.

Assume that a particular detection system has a 0.80 probability of detecting a missile attack.

This means that $p = 0.8$

If two military radars are installed in two different areas and they operate independently, the probability that at least one of them detect an enemy attack is

This is $P(X \geq 1)$ when $n = 2$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{2,0}.(0.8)^{0}.(0.2)^{2} = 0.04$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.04 = 0.96$

0.96 = 96% probability that at least one of them detect an enemy attack.

6 0

3 years ago