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100 PTS AND BRAINLIEST WRONG ANSWERS REPORTED

The function p(w)=230(1.1)^W represents the number of specialty items produced at the old factory w
weeks after a change in management. The graph represents the number of specialty items produced at the
new factory during the same time period.

(a) During Week 0, how many more specialty items were produced at the old factory than at the new
factory? Explain.
(b) Find and compare the growth rates in the weekly number of specialty items produced at each factory.
Show your work.


100 PTS AND BRAINLIEST WRONG ANSWERS REPORTED The Function Pw23011W Represents The Number Of Specialty Items Produced At The Old Factory W Weeks After A Change class=

Sagot :

Answer:

40; 0.1 and 0.16

Step-by-step explanation:

a. We know that p(w)=230(1.1)^W represents the number of specialty items produced at the old factory.

This equation is in the form of an exponential growth function:

f(x)=a(1+r)^x

where a is the initial value, r is the growth rate, and x is the number of time intervals.

That means that for p(w)=230(1.1)^W:

230 is the initial value, and 0.1 is the growth rate. (1.1-1)

Therefore, for week 0, there were 230 specialty items produced at the old factory.

We can also plug in 0 for w (0 time intervals) to get the same answer

That would look like:

p(0)=230(1.1)^0

p(0)=230(1)

p(0)=230

On the other hand, for the new factory, if you go find 0 on the x-axis, you'll notice that 190 specialty items were produced in week 0.

230-190=40

This means that our answer for a is that 40 more specialty items were produced at the old factory than the new factory.

_____

b. Our first step for this question would be to find the equation for the graph.

We can immediately recognize that this is an exponential growth function, similar to the other equation.

As a reminder, exponential growth functions are in the form: f(x)=a(1+r)^x

We can put it in terms of production and weeks:

p(w)=a(1+r)^w

We know that the initial value is 190 by the graph.

p(w)=190(1+r)^w

Now we just need to find the growth rate:

We can do this by choosing a value (other than the first point) and plugging it in.  Let's use (1, 220) which is in the form (w, p)

Plug it in to the equation:

220=190(1+r)^1

220=190(1+r)

Distribute

220=190+190r

Subtract 190 from both sides

30=190r

Divide 190 from both sides

r=0.1578947368 (round how your teacher asks)

Next, all we have to do is find the growth rate for the old factory:

p(w)=230(1.1)^W

1.1=1+r

Therefore, r=0.1

The growth rate at the old factory is 0.1.

When comparing them, we'll realize that the growth rate at the new factory is about 0.06 higher.

#1

For old factory

  • p(w)=230(1.1)^w

So

  • week =0

[tex]\\ \rm\dashrightarrow p(0)=230(1.1)^0=230(1)=230[/tex]

For nEW factory

  • On graph y intercept is 190

Difference:-

  • 230-190=40

40 items were more produced

#2

For old factory

  • Turn it to p(1+r)^t form where r is rate of change

[tex]\\ \rm\dashrightarrow 230(1+0.1)^t[/tex]

  • Growth rate is 0.1

For new factory

  • (0,190)
  • (1,220)

Find slope:-

[tex]\\ \rm\dashrightarrow m=\dfrac{220-190}{1}=30[/tex]

Find growth rate

  • m/b(b is y intercept)
  • 30/190
  • 3/19
  • 0.16

New factory has greater growth rate