All categories

3 years ago

How do you draw a line for this

Mathematics

2 answers:

m_a_m_a [10]3 years ago

8 0

I'd do a little more than 77minutes for the 6and 7for if you could have it on 77and but you could have it in your head

Studentka2010 [4]3 years ago

3 0

Put 0 in for x and you get 0. The first point to graph is (0,0)
Put 1 in for x and you get -5. The second point is (1,-5)
Third point is put 2 in for x and you get -10. The last point is (2,-10)
Plot and connect the points

You might be interested in

What is the missing number in the number sentence?

MatroZZZ [7]

Answer:

the missing number is 47............

8 0

2 years ago

Read 2 more answers

What are the slope and the y-intercept of the linear function that is represented by the table?

Hunter-Best [27]

Answer:

Slope is -2/5 and y-intercept is -1/3

Step-by-step explanation:

To get the slope, we pick any two points and apply the slope formula

The slope formula is;

m = (y2-y1)/(x2-x1)

To get the slope, we select any two points

That will be;

(-3/4, -1/30) and (1/4,-13/30)

Substituting these in the slope formula, we have:

m = (-13/30 + 1/30)/(1/4 + 3/4)

m = 12/30 = -2/5

Generally, the equation of a straight line is;

y = mx + b

m is slope and b is the y-intercept

We already have the slope, to get the y-intercept, we select any of the points

-3/5 = -2/5(2/3) + b

= -3/5 + 4/15 = b

b = (-9 + 4)/15 = -5/15 = -1/3

Slope is -2/5 and y-intercept is -1/3

6 0

3 years ago

Read 2 more answers

in a random sample of 45 endangered species, 12 are categorized as vulnerable, 11 are categorized as endangered, and 22 are cate

horrorfan [7]

The probability that the randomly selected sample is vulnerable or critically endangered is equal to 0.76

The probability can be determined using the following formula;

probability = number of desired events or outcomes ÷ total number of events or outcomes

The total is known to be 45, the favorable or desired outcomes can be calculated as follows;

number of desired outcomes = number of vulnerable species + number of critically endangered species

number of desired outcomes = 12 + 22

number of desired outcomes = 34

Now we calculate the probability by using the above formula;

probability = 34 ÷ 45

probability = 0.76

Therefore, 0.76 is the probability that the randomly selected species is vulnerable or critically endangered.

Although a part of your question is missing, you might be referring to this question:

In a random sample of 45 endangered species, 12 are categorized as vulnerable, 11 are categorized as endangered, and 22 are categorized as critically endangered. what is the probability that a randomly selected species from this sample is vulnerable or critically endangered? group of answer choices

11/45

0.24

0.83

0.76

12/45

To learn more about probability, click here:

brainly.com/question/24756209

#SPJ4

8 0

1 year ago

Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and

Naddika [18.5K]

Answer:

Using a 90% confidence level

A. A sample size of 68 should be used.

B. A sample size of 98 should be used.

Step-by-step explanation:

I think there was a small typing mistake and the confidence level was left out. I will use a 90% confidence level.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

A. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 72 seconds, what sample size should be used?

We have the standard deviation in minutes, so the margin of error should be in minutes.

72 seconds is 72/60 = 1.2 minutes.

So we need a sample size of n, and n is found when M = 1.2. We have that $\sigma = 6$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$1.2 = 1.645*\frac{6}{\sqrt{n}}$

$1.2\sqrt{n} = 6*1.645$

$\sqrt{n} = \frac{6*1.645}{1.2}$

$(\sqrt{n})^{2} = (\frac{6*1.645}{1.2})^{2}$

$n = 67.65$

Rounding up.

A sample size of 68 should be used.

B. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 1 minute, what sample size should be used?

Same logic as above, just use M = 1.

$M = z*\frac{\sigma}{\sqrt{n}}$

$1 = 1.645*\frac{6}{\sqrt{n}}$