Simplyfiy3 3/4 x 1 1/5

Help! It says that I need to determine If both equations represent lines which are parallel, perpendicular, or none...

12345 [234]

4x + 2y = 8 (1)
 8x + 4y = -4y (2)

A) Two lines are parallel if they have the same gradient
- putting both equations into the gradient- intercept form ( y = mx + c where m is the gradient)
(1) 4x + 2y = 8
 2y = 8 - 4x
y = -2x + 4

 (2) 8x + 4y = -4y
 8x = -4y - 4y
y = $\frac{-8x}{-8}$
y = -x
Thus the gradient of the two equations are different and as such the two lines are not parallel

B) When two lines are perpendicular, the product of their gradient is -1
$m_{1} * m_{2} = p$
p = (-2) * (-1)
p = 2
 ∴ the two lines are not perpendicular either.

Thus these lines are SKEWED LINES

4 0

2 years ago

Find f(4) please, I need help

mixer [17]

Answer:

f(4) = 1

Step-by-step explanation:

f(4) = what?

So if we know that x, aka 4, is technicially f is greater then 2, so you can plug in x as 4. So that means 5(4) - 19 = 20 - 19 = 1, so f(4) = 1

7 0

2 years ago

What is 100,034 - 9,348

pochemuha

The answer is 90,686

7 0

2 years ago

Read 2 more answers

six men completed a task in 24 hours and 48 minutes. how long would it take four men to complete the same task

Komok [63]

Answer:

37 hours and 12 minutes

Step-by-step explanation:

six men completed a task in 24 hours and 48 minutes which means (24 x 60) + 48 = 1488 minutes

x is the number of minutes it would take 4 men to complete

6(1488) = 4x

8928 = 4x

x = 2232

x = 2220 + 12 or 37 hours and 12 minutes

8 0

1 year 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