Please, need this urgently.Graph the function f(x) = 18(0.8)x.

an exponential graph decreasing from the left and crossing the y axis at 0 comma 21
an exponential graph decreasing from the left and crossing the y axis at 0 comma 5
an exponential graph decreasing from the left and crossing the y axis at 0 comma 18
an exponential graph decreasing from the left and crossing the y axis at 0 comma 8

Answers

Answer 1
Answer:

Answer:

i think the answer is:

D - an exponential graph decreasing from the left and crossing the y axis at 0 comma 8

Answer 2
Answer:

Answer:

the answer is D

Step-by-step explanation:

i took the test


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-5(4x + 7) <25
Answer here

Answers

Answer:

X>-3

Step-by-step explanation:

Hope I helped

Do u want the steps ?

Answer: x > -3

Step-by-step explanation: -5(4x + 7) <25

                                              -20x - 35 < 25

                                              -20x -35 + 35 < 25 + 35

                                              -20x/20 < 60/20

                                              x > -3

What is the inequality solution for y+2<4

Answers

Answer:

y<2

Step-by-step explanation:

Subtract 2 on both sides

Answer:

y<2

Step-by-step explanation:

MATH 1325 – EXAM 4 NAME: ______________________________ SHOW ALL WORK. ANSWERS WITHOUT WORK WILL RECEIVE NO CREDIT. YOU MUST USE A PENCIL. READ ALL DIRECTIONS. POINTS WILL BE DEDUCTED FOR FAILURE TO FOLLOW DIRECTIONS. TRUE/FALSE – WRITE THE WORD THAT BEST DESCRIBES THE GIVEN STATEMENT BY WRITING EITHER "TRUE" OR "FALSE" IN THE SPACE PROVIDED TO THE LEFT OF THE PROBLEM. __________ 1. THE ABSOLUTE MAXIMUM OF A FUNCTION ALWAYS OCCURS WHERE THE DERIVATIVE HAS A CRITICAL FUNCTION. __________ 2. IMPLICIT DIFFERENTIATION CAN BE USED TO FIND dy dx WHEN x IS DEFINED IN TERMS OF y . __________ 3. IN A RELATED RATES PROBLEM, THERE CAN BE MORE THAN TWO QUANTITIES THAT VARY WITH TIME. __________ 4. A CONTINUOUS FUNCTION ON AN OPEN INTERVAL DOES NOT HAVE AN ABSOLUTE MAXIMUM OR MINIMUM. __________ 5. IN A RELATED RATES PROBLEM, ALL DERIVATIVES ARE WITH RESPECT TO TIME. MULTIPLE CHOICE – CHOOSE THE ONE ALTERNATIVE THAT BEST COMPLETES THE STATEMENT OR ANSWERS THE QUESTION BY CIRCLING THE CORRECT LETTER. 6. FIND THE MAXIMUM ABSOLUTE EXTREMUM AS WELL AS ALL VALUES OF x WHERE IT OCCURS ON THE SPECIFIED DOMAIN

Answers

Answer: Please see explanation column for answers. Also check number 6, its question is incomplete.  i used an assumed function, incase its not the same function with the one omitted, just follow steps

Step-by-step explanation: Questions 1-5 do not need any step by step explanation, its purely straight forward but Question 6 involves step by step explanation but  is not a complete question, though i used an assumed function.

FALSE   ---> 1. THE ABSOLUTE MAXIMUM OF A FUNCTION ALWAYS OCCURS WHERE THE DERIVATIVE HAS A CRITICAL FUNCTION. ___TRUE_____-->__ 2. IMPLICIT DIFFERENTIATION CAN BE USED TO FIND dy/dx WHEN x IS DEFINED IN TERMS OF y .

TRUE__--->3. IN A RELATED RATES PROBLEM, THERE CAN BE MORE THAN TWO QUANTITIES THAT VARY WITH TIME.

_FALSE  ---> 4. A CONTINUOUS FUNCTION ON AN OPEN INTERVAL DOES NOT HAVE AN ABSOLUTE MAXIMUM OR MINIMUM.

____TRUE__--->____ 5. IN A RELATED RATES PROBLEM, ALL DERIVATIVES ARE WITH RESPECT TO TIME.

6. FIND THE MAXIMUM ABSOLUTE EXTREMUM AS WELL AS ALL VALUES OF x WHERE IT OCCURS ON THE SPECIFIED DOMAIN

----Incomplete question Please.

But assuming the function---- f(x)= x³ -3x+1

 for (E)=(0,3)

step 1= let us use the power rule to find derivative of   f(x)= x^3 -3x+1

we will have f¹ (x) = 3x² -3

The critical values occurs when  3x² -3 = 0

which makes x=⁺₋1

As can be seen 3x² -3 = 0

                         3x²=3

                          x²=3/3=1

                       x= ⁺₋1

step 2=Now x= -1 cannot be considered because it is not in the interval  of the critical values (0,3)

therefore we consider x=1

step 3=The absolute extremes occurs at x=0, x=1, x=3 forf(x)= x³ -3x+1

when x=0,  f(0)= 0³-3(0)+ 1= 1

         x=1    f(1)=1³-3(1) +1=  -1

         x=3    f(3)= 3³ -3(3)+1= 19

Absolute minimum at x=1 has absolute value of-1

Absolute maximum of x=3 has absolute value of 19

Find the zeros of f(x) =x^3+2x^2-3x

Answers

Answer: 0, 1, -3

Step-by-step explanation:

f(x) = x^3 +2x^2 -3x

f(x) = x(x^2 +2x-3)

f(x) = x(x-1)(x+3)

to find zeroes, plug into f(x) and solve for each value of x. doing this gives 0, 1, and -3

What are the answers?

Answers

send me the question bc cant seeeeeeeee

In a survey of 1016 ?adults, a polling agency? asked, "When you? retire, do you think you will have enough money to live comfortably or not. Of the 1016 ?surveyed, 535 stated that they were worried about having enough money to live comfortably in retirement. Construct a 99?% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement.A. There is a 99?% probability that the true proportion of worried adults is between ___ and ___.

B. 99?% of the population lies in the interval between ___ and ___.

C. There is 99?% confidence that the proportion of worried adults is between ___ and ___.

Answers

Answer:

C. There is 99% confidence that the proportion of worried adults is between 0.487 and 0.567

Step-by-step explanation:

1) Data given and notation  

n=1016 represent the random sample taken    

X=535 represent the people stated that they were worried about having enough money to live comfortably in retirement

\hat p=(535)/(1016)=0.527 estimated proportion of people stated that they were worried about having enough money to live comfortably in retirement

\alpha=0.01 represent the significance level

Confidence =0.99 or 99%

z would represent the statistic

p= population proportion of people stated that they were worried about having enough money to live comfortably in retirement

2) Confidence interval

The confidence interval would be given by this formula

\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_(\alpha/2)=2.58

And replacing into the confidence interval formula we got:

0.527 - 2.58 \sqrt{(0.527(1-0.527))/(1016)}=0.487

0.527 + 2.58 \sqrt{(0.527(1-0.527))/(1016)}=0.567

And the 99% confidence interval would be given (0.487;0.567).

There is 99% confidence that the proportion of worried adults is between 0.487 and 0.567

Final answer:

To build a 99% confidence interval, we first calculate our sample proportion by dividing the number of such instances by the total sample size. Next, we determine the standard error of the proportion, then our margin of error by multiplying the standard error by the Z value of the selected confidence level. Lastly, we determine the confidence interval by adding and subtracting the margin of error from the sample proportion.

Explanation:

To construct a 99% confidence interval for the proportion of adults worried about having enough money to live comfortably in retirement, we will utilize statistical methods and proportions. First, we must calculate the sample proportion. The sample proportion (p) is equal to 535 (the number who are worried) divided by 1016 (the total number of adults surveyed).

Then, we find the standard error of the proportion which we get by multiplying the square root of ((p*(1-p))/n) where n is the number of adults sampled. The margin of error is found using the Z value corresponding to the desired confidence level, in this case, 99%. Multiply the standard error by this Z value. Lastly, we construct the confidence interval by taking the sample proportion (p) ± the margin of error.

The result will give you the 99% confidence interval - meaning we are 99% confident that the true proportion of adults who are worried about having enough money to live comfortably in retirement lies within this interval.

Learn more about Confidence Interval here:

brainly.com/question/34700241

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