19c10x42c4/61c14 what would that equal to?

PLS HELP THIS IS LAST OF MY POINTS AND RUNNING OUT OF TIME. ANSWER ONLY IF U KNOW MATH.

ipn [44]

Answer:

using y=mx +c

c=0

m=y2-y2/x2-x1

m=16-0/2-0

m=8

y=8x+0

y=8x

hope that helped

if yes give me brainliest and if no draw my attention by hitting the comment section

7 0

2 years ago

Read 2 more answers

What are the x intercepts of the quadratic function? Y=-x^2+2x-1

saw5 [17]

Answer:

x = 1

Step-by-step explanation:

Given in the question the equation

y = -x² + 2x - 1

To find the x-intercept, substitute in 0 for y

0 = -x² + 2x - 1

To find value of x use quadratic equation

x = -b ± √b²-4ac / 2a

here a = -1

b = 2

c = -1

x1 = -2 + √2²-4(-1)(-1) / 2(-1)

= -2 + 0 / -2

= -2 / -2

= 1

x2 = -2 - √2²-4(-1)(-1) / 2(-1)

= -2 - 0 / -2

= -2 / -2

= 1

8 0

3 years ago

A car insurance company has determined that 8% of all drivers were involved in a car accident last year. Among the 15 drivers li

svetlana [45]

Answer:

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

Step-by-step explanation:

For each driver surveyed, there are only two possible outcomes. Either they were involved in a car accident last year, or they were not. This means that we solve this problem using binomial probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem

15 drivers are randomly selected, so $n = 15$ .

A success consists in finding a driver that was involved in an accident. A car insurance company has determined that 8% of all drivers were involved in a car accident last year. This means that $\pi = 0.08$ .

What is the probability of getting 3 or more who were involved in a car accident last year?

This is $P(X \geq 3)$ .

Either less than 3 were involved in a car accident, or 3 or more were. Each one has it's probabilities. The sum of these probabilities is decimal 1. So:

$P(X < 3) + P(X \geq 3) = 1$