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2 years ago

I need help on how to write an EQUATION of how to solve for X 2x<15 simpliefied answer anyone that can help tysm

Mathematics

1 answer:

goblinko [34]2 years ago

6 0

Answer:

2x < 15

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Simplify the expression<br> (-1-3i)(-6-i)

Ivan

(-1-3i)(-6-i)
=6+i+18i+3i^2
=3i^2+19i+6. Hope it help!

7 0

2 years ago

Solve the system by substitution. y=2x^2-3x-1 y=x-3

padilas [110]

The answer is B(1,-2). This is the only point where the two lines intercept. & if you plug the numbers into the equations they fit.

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2 years ago

Read 2 more answers

You play a game on your phone. For every correct move you make, you receive 50 points. For every second that it takes you to com

Vladimir79 [104]

Answer:

2005 points

Step-by-step explanation:

1. 35x50=750

2. 2 min. 45 sec.=(-165 sec.)

3. 500x5=2500

4. 2500-495=2005

Hope this helped!!:)

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

2 years ago

Find the length of the third side to the nearest tenth.<br> Hurry need help asap!!

gavmur [86]

Answer:

2.8

Step-by-step explanation:

6 0

2 years ago