If () = − 10, find (20)

Simplify the expression. Write the answer using scientific notation. 0.7(3.6x10^-2)

kobusy [5.1K]

The answer would be 2.52×10^-2.

5 0

3 years ago

Jan and two of her friends raised money by having a bake sale. They split the proceeds. If each person received $32.19, how much

Dimas [21]

You need to multiply the answer by how many people got that amount, Jan +her 2 friends would be 3 people so
32.19 x 3 = 96.57

5 0

3 years ago

Read 2 more answers

Are the equations sometimes, always, or never true?

AURORKA [14]

15 + 2x - 4 = 9x + 11 - 7x
2x + 11 = 2x + 11
always true

2x + 3(4x - 1) = 2(5x + 3) + 4x
2x + 12x - 3 = 10x + 6 + 4x
14x - 3 = 14x + 6
never true

4 0

3 years ago

The graph shows a probability distribution. What is P(X<5)

Thepotemich [5.8K]

Answer:

P(x < 5) = 0.70

Step-by-step explanation:

Note: The area under a probability "curve" must be = to 1.

Finding the sub-area representing x < 5 immediately yields the desired probability.

Draw a dashed, vertical line through x = 5. The resulting area, on the left, is a trapezoid. The area of a trapezoid is equal to:

(average length)·(width, which here is:

2 + 5

----------- · 0.02 = (7/2)(0.2) = 0.70

2

Thus, P(x < 5) = 0.70

7 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