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4 years ago

Write an equation of a line with the given slope and y-intercept. m = 4/5 , b = 2 y = 2x + y = x – 2 y = x + 2 y = x + 2

Mathematics

1 answer:

xxTIMURxx [149]4 years ago

7 0

Answer:

$y = \frac{4}{5} x + 2$

Step-by-step explanation:

Put the slope value given as the coefficient of x.

It is the "m" in the formula y = mx + b

I'm not sure what the other equations in the question are there for. Maybe to change from slope-intercept form to standard form?

y = 2x +

y = x – 2 becomes y -x = -2 or x - y = 2

y = x + 2 becomes y -x = 2 or x - y = -2

Some examples of the graphs are in the attachment

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What is the third quartile of this data set ? 14,18,20,21,25,32,40,44,48

kotykmax [81]

Step-by-step explanation:

Step 1 :

Given data set - 14,18,20,21,25,32,40,44,48

We need to get the 3rd quartile

Step 2 :

To find the quartile's , we have to divide the data set into 4 equal quarters.

The 3rd cut will give the 3rd quartile

Step 3:

The median of this data set is at 25, there are 4 numbers below and above it.

Now we take the 2nd half of the set , after the median and take the median of it.

There are 4 numbers in the 2nd half 32,40,44,48. Cutting this into 2 , we have the cut to fall between 40 and 44. Hence taking the average of this will give the 3rd quartile.

The average of 40 and 44 is 42 .

Hence the 3rd quartile is 42.

7 0

4 years ago

The Potter Valley basketball team did not record how many baskets each player made during the last game. Jenny remembers that sh

lesya692 [45]

Answer: The approximate values that every player make are,

Tammy = grace = 5.43 baskets

Jenny = 16.30 Baskets

Alex = 11.43 Baskets

Joan = 1.43

Step-by-step explanation:

Let Grace made x baskets.

Then According to the question,

Jenny made baskets = 3x

Alexis made basket = x + 6

Joan made basket = x - 4

Tammy made baskets = x

Since, Altogether the five players made 40 baskets.

⇒ x+ 3x + x + 6 + x - 4 + x = 40

⇒ 7x + 2 = 40

⇒ 7x = 38

⇒ x = 38 / 7 = 5.42857142857 ≈ 5.43

Thus, Grace made baskets= Tammy made baskets = x = 5.43

Jenny made baskets = 3 × x = 16.2857142857 ≈ 16.30

Alexis made basket = x + 6 = 11.4285714286 ≈ 11.43

Joan made basket = x - 4 = 1.42857142857

≈ 1.43

5 0

3 years ago

Find the sample mean on the outputs. Does the sample mean provide some evidence that the mean of the population of all possible

soldi70 [24.7K]

I know it and it is 7

8 0

3 years ago

Please help will give brainliest

Gelneren [198K]

Answer:

Step-by-step explanation:

It is B because to find the volume you multiply all three values together, which gets you 68.448. I multiplied 9.2x3.1, which got me 28.52, and then I multiplied 28.52 by 2.4 which got me 68.448.

6 0

3 years ago

A teacher categorized students by whether they regularly completed their homework (or not) and whether they passed the course (o

ella [17]

Probabilities are used to determine the chances of an event.

Let P represents the event that a student passes the course, and C represents the event that a student completes the assignment

(a) The probability a student passed the course given that they completed their homework?

From the table, we have:

$\mathbf{n(P\ n\ C) = 25}$

$\mathbf{n(C) = 30}$

So, the required probability is:

$\mathbf{Pr = \frac{n(P\ n\ C)}{n(C)}}$

This gives

$\mathbf{Pr = \frac{25}{30}}$

$\mathbf{Pr = 0.8333}$

Express as percentage

$\mathbf{Pr = 83.33\%}$

Hence, the probability a student passed the course given that they completed their homework is 83.33%

(b) The probability a student passed the course given that they didn't complete their homework

From the table, we have:

$\mathbf{n(P\ n\ C'= 7}$

$\mathbf{n(C') = 25}$

So, the required probability is:

$\mathbf{Pr = \frac{n(P\ n\ C')}{n(C')}}$

This gives

$\mathbf{Pr = \frac{7}{25}}$

$\mathbf{Pr = 0.28}$

Express as percentage

$\mathbf{Pr = 28\%}$

Hence, the probability a student passed the course given that they didn't complete their homework is 28%

(c) The probability a student completed their homework given that passed the course?

From the table, we have:

$\mathbf{n(P\ n\ C) = 25}$

$\mathbf{n(P) = 32}$

So, the required probability is:

$\mathbf{Pr = \frac{n(P\ n\ C)}{n(P)}}$

This gives

$\mathbf{Pr = \frac{25}{32}}$

$\mathbf{Pr = 0.78125}$

Express as percentage

$\mathbf{Pr = 78.125\%}$

Approximate

$\mathbf{Pr = 78.13\%}$

Hence, the probability a student completed their homework given that they passed the course is 78.13%

(d) The probability a student didn't complete their homework given that they didn't pass the course

From the table, we have:

$\mathbf{n(P'\ n\ C'= 18}$

$\mathbf{n(P) = 23}$

So, the required probability is:

$\mathbf{Pr = \frac{n(P'\ n\ C')}{n(P')}}$

This gives

$\mathbf{Pr = \frac{18}{23}}$

$\mathbf{Pr = 0.7826}$

Express as percentage

$\mathbf{Pr = 78.26\%}$

Hence, the probability a student didn't complete their homework given that they didn't pass the course is 78.26%