All categories

3 years ago

Look at the graph shown:

Mathematics

1 answer:

NNADVOKAT [17]3 years ago

4 0

We are going to base the answer off of the y=mx+b form where m represents slope and b represents the y-intercept.

First, we know that b must be -1 because the line hits the y-axis when y is -1
So this leaves us with either the first or second choice

Then we choose two points to find the slope. I am using (0,-1) and (1,2). If you count the boxes, the y-value increases by 3 as the x-value increases by 1. So the slope is 3/1 or 3.

So then we know the answer is y=3x-1

Hope this helps

You might be interested in

The weights of the fish in a certain lake are normally distributed with a mean of 12 lb and a standard deviation of 12. If 16 fi

erastova [34]

Answer:

If 4 fish are randomly selected, what is the probability that the mean weight will be between 9.6 and 15.6 lb? Round your answer to four decimal places.

6 0

2 years ago

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species

kotegsom [21]

Answer:

$P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}$

Step-by-step explanation:

The logistic equation is the following one:

$P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}$

In which P(t) is the size of the population after t years, K is the carrying capacity of the population, r is the decimal growth rate of the population and P(0) is the initial population of the lake.

In this problem, we have that:

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 2,000. This means that $P(0) = 80, K = 2000$ .

The number of fish tripled in the first year. This means that $P(1) = 3P(0) = 3(80) = 240$ .

Using the equation for P(1), that is, P(t) when $t = 1$ , we find the value of r.

$P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}$

$240 = \frac{2000*80e^{r}}{2000 + 80(e^{r} - 1)}$

$280*(2000 + 80(e^{r} - 1)) = 160000e^{r}$

$280*(2000 + 80e^{r} - 80) = 160000e^{r}$

$280*(1920 + 80e^{r}) = 160000e^{r}$

$537600 + 22400e^{r} = 160000e^{r}$

$137600e^{r} = 537600$

$e^{r} = \frac{537600}{137600}$

$e^{r} = 3.91$

Applying ln to both sides.

$\ln{e^{r}} = \ln{3.91}$

$r = 1.36$

This means that the expression for the size of the population after t years is:

$P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}$

4 0

2 years ago

Six tenths as a decimal

vampirchik [111]

0.6 is the answer you're looking for.

4 0

3 years ago

Read 2 more answers

Can I have some help

jok3333 [9.3K]

18/20=27/x
18x=27*20
x=(27*20)/18=30

Answer: 30

6 0

3 years ago

Asher pays $156 every 6 weeks for piano lessons. what is the price for per year for piano lesson?

scoundrel [369]

Price for every 6 weeks= $ 156
Price for 1 week = 156/6 = $ 26

Assuming there are approximately 52 weeks in a year
Price for lessons per year = 26*52
= $1352

7 0

3 years ago