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3 years ago

Which line is a linear model for the data? Please Help Me!!!!

Mathematics

2 answers:

podryga [215]3 years ago

7 0

, this is best fit. But the best fit for this graph is variable b

Alekssandra [29.7K]3 years ago

5 0

The answer would be letter b. The line starts at the first dot and goes towards the last with an arrow indicating it goes on.

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Include parentheses in your answer.<br> (x3 + y3) ÷ (x - y)

-Dominant- [34]

I'm assuming you mean $\frac{x^3+y^3}{x-y}$

we can factor that sum of perfect cubes from $x^2+y^3$ into $(x+y)(x^2-xy+y^2)$

so therefor $\frac{x^3+y^3}{x-y}=\frac{(x+y)(x^2-xy+y^2)}{x-y}$
that is simpliest form

8 0

3 years ago

Match each set of numbers with their least common multiple.

mylen [45]

Answer:

27, 12, and 36 = 108

22 and 4 = 44

14 and 12 = 84

25 and 8 =200

9, 8, and 7 = 504

32, 24, and 18 = 288

Step-by-step explanation: You are welcome.

5 0

2 years ago

your sister can get $30 per day for selling clothes at the market. but she has to $60 at first for renting a stand. you can get

ella [17]

Day Sis you
1 -30 20
2 0 40
3 30 60
4 60 80
5 90 100
6 120 120

So day 6.

8 0

3 years ago

Read 2 more answers

Rectangular pyramid:volume 448 inches cubed, base edge 12 inches.;base length 8 inches

saw5 [17]

Whats your question?

4 0

3 years ago

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by

Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}$

$(\text{Average velocity})_{0.1\ s}=696\ ft/s$

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

5 0

3 years ago