What is the slope of the line below (-1,6) (2,-3)

-2/3x+7(3/7-3/7x) -12 divided by 6*3

Anna71 [15]

Answer:

x = (-27)/11

Step-by-step explanation:

Solve for x:

(-11 x)/54 - 1/2 = 0

Put each term in (-11 x)/54 - 1/2 over the common denominator 54: (-11 x)/54 - 1/2 = (-27)/54 - (11 x)/54:

(-27)/54 - (11 x)/54 = 0

(-27)/54 - (11 x)/54 = (-11 x - 27)/54:

(-11 x - 27)/54 = 0

Multiply both sides of (-11 x - 27)/54 = 0 by 54:

(54 (-11 x - 27))/54 = 54×0

(54 (-11 x - 27))/54 = 54/54×(-11 x - 27) = -11 x - 27:

-11 x - 27 = 54×0

0×54 = 0:

-11 x - 27 = 0

Add 27 to both sides:

(27 - 27) - 11 x = 27

27 - 27 = 0:

-11 x = 27

Divide both sides of -11 x = 27 by -11:

(-11 x)/(-11) = 27/(-11)

(-11)/(-11) = 1:

x = 27/(-11)

Multiply numerator and denominator of 27/(-11) by -1:

Answer:x = (-27)/11

5 0

3 years ago

Three coins are flipped,what is the Probability of getting one HEAD and two TAILs?

diamong [38]

Answer:

3/8

Step-by-step explanation:

So, three outcomes give two heads and a tail (HHT, HTH, THH). Therefore, the probability of getting two heads and a tail is 3/8.

4 0

3 years ago

Read 2 more answers

Simplify x²-4xy+4y²/x²+xy-6y²÷x²+3xy/x²+6xy+9y²

LenaWriter [7]

Answer:

x-2y/x

Step-by-step explanation:

x²-4xy+4y²/x²+xy-6y²÷x²+3xy/x²+6xy+9y² = x-2y/x

3 0

2 years ago

The rule is y=− 3/4 x+ 1 /4

Effectus [21]

The answer is C because y=-3/4 which is a negative.

7 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