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2 years ago

PLS HELP BRAINLIEST ! What is the conjugate of -8-2i?

Mathematics

1 answer:

AlladinOne [14]2 years ago

7 0

Answer:

it is already conjucated i belive

Step-by-step explanation:

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Need help ASAP pls thx

Mazyrski [523]

$\\ \sf\longmapsto RElative\:Frequency(Green)$

$\\ \sf\longmapsto RF(Green)=\dfrac{28}{14+28+28}$

$\\ \sf\longmapsto RF(Green)=\dfrac{28}{70}$

$\\ \sf\longmapsto RF(Green)=\dfrac{2}{5}$

5 0

2 years ago

Read 2 more answers

Please need solution .. Thank you

hjlf

anong edad mo na Kasi 13 na ko

6 0

2 years ago

What is 2x3-4x5 and 3x6=

lukranit [14]

Answer:

4 I believe

Step-by-step explanation:

sorry if im wrong, (didn't know if the and was to add or multiply,sorry!)

3 0

3 years ago

Which equation can be used to solve for x in the parallelogram below?

densk [106]

Answer:

B. (3x - 12)° = (2x + 8)°

Step-by-step explanation:

This equation can be used to solve x.

3x - 12 = 2x + 8

3x - 2x = 12 + 8

x = 20°

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Verification:

3(20) - 12 = 2(20) + 8

60 - 12 = 40 + 8

48° = 48°

6 0

2 years ago

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