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Triss [41]
3 years ago
13

Select the outlier in the data set.

Mathematics
1 answer:
Katarina [22]3 years ago
6 0

Answer:

86

Step-by-step explanation:

21   <u>3</u>6

<em>5</em>2   <em>4</em>0

<u>3</u>2   <em><u>86</u></em>

<u>3</u>3   <u>3</u>8

28   <u>3</u>4

<u>3</u>0   <em>1</em>9

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Given f(x) = 4 - 3x, what is the value of f(-5)?
Studentka2010 [4]

Answer:

Step-by-step explanation:

If f= 4 and x=3 then you times 4 times 5 equals 20

3 0
3 years ago
Which number is equal to 10 Superscript negative 3? –1,000 –30 0.001 0.003
Karolina [17]

<u>Given</u>:

The given expression is 10^{-3}

We need to determine the equivalent number.

<u>Equivalent expression:</u>

To determine the equivalent number, let us solve the expression 10^{-3}

Thus, let us apply the exponent rule a^{-b}=\frac{1}{a^{b}} , we get;

10^{-3}=\frac{1}{10^{3}}

Also, the term 10³ can be written as 10^{3}=1000

Thus, we get;

\frac{1}{1000}

Dividing, we get;

0.001

Hence, the equivalent number is 0.001

Hence, Option c is the correct answer.

5 0
3 years ago
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Select which of the following are surds
Airida [17]
36 , 100, are surfs
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2 years ago
If all 108 people take 2 biscuits, how many biscuits will you need?
grigory [225]

Answer:

216

Step-by-step explanation:

108 x 2 = 216

7 0
4 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
3 years ago
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