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Snowcat [4.5K]
2 years ago
12

(HELP ME) I really need this done

Mathematics
2 answers:
Tatiana [17]2 years ago
6 0
Hey, here’s your answer :)

RoseWind [281]2 years ago
4 0

Answer:

56.88

god bless all my dudes!!!

p.s. plz give crown, I need it to level up.

thx

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What quadrant is the<br> ordered pair (2,-3) in?
Neko [114]

Answer:

quadrant 4

Step-by-step explanation:

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2 years ago
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Help, I'm doing rhis problem and need help checking if I'm right.
umka2103 [35]
To me it looks like your right
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3 years ago
Which pair of complex numbers has a real-number product? (1 + 2i)(8i) (1 + 2i)(2 – 5i) (1 + 2i)(1 – 2i) (1 + 2i)(4i)
kupik [55]

Answer:

(1+2i)(1-2i)

Step-by-step explanation:


5 0
3 years ago
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1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n
Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 70% of fatalities involve an intoxicated driver, hence p = 0.7.
  • A sample of 15 fatalities is taken, hence n = 15.

The probability is:

P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

Hence

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

P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061

P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186

P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700

P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916

P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305

P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047

Then:

P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0
2 years ago
Solve for a.<br><br> a + (-1.3) = -4.5 <br><br> Enter your answer in the box.
Vlada [557]

Answer:

a=(-3.2)

Step-by-step explanation:

a+(-1.3)=(-4.5)=a-1.3=(-4.5)

a=(-3.2)

Please mark as Brainliest! :)

Have a nice day.

3 0
2 years ago
Read 2 more answers
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