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3 years ago

Please help I give 20 points please no fake answers =/

Mathematics

2 answers:

Licemer1 [7]3 years ago

5 0

Answer to the question is b 1 4/5

Svetach [21]3 years ago

5 0

Answer:

1 4/5

Step-by-step explanation:

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I only need number one. <br> Which variable should be the dependent variable?

Natali [406]

The dependent variable is y, that represents the years

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3 years ago

Nancy's Cupcakes recorded how many cupcakes it recently sold of each flavor.

Alexeev081 [22]

Answer:

pumpkin because it got the most sold

7 0

3 years ago

Write down the first even multiple of 7

Fiesta28 [93]

Answer:

Step-by-step explanation:

multiples are the list of number that can be multiplied by a certain number.

list the multiples

make a chart

1 × 7 = 7 (odd)

2 × 7 = 14 (even)

the first even multiple is 14 (2×7)

6 0

3 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

3 years ago

What is 3x-9-5x=-7 if you solve for x

deff fn [24]

3x-9-5x = -7

combine like terms -9-2x = -7

add 9 to both sides

-2x = 2

divide both sides by -2

x - 2/-2 = -1

x = -1

6 0

3 years ago