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1 year ago

A customer's stock value seems to be rising exponentially. The equation for

Mathematics

1 answer:

pashok25 [27]1 year ago

4 0

Considering the given regression equation, the number of weeks that will pass before the value of the stock reaches $600 is of:

B. 8.3.

<h3>What is the regression equation?</h3>

The value of the stock y after x weeks is given as follows:

log(y) = 0.3x + 0.296.

We have to solve for x when y = 600, hence:

log(600) = 0.3x + 0.296

0.3x = 2.7782 - 0.296

x = (2.7782 - 0.296)/0.3

x = 8.3 weeks.

More can be learned about regression equations at brainly.com/question/25987747

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Which statement is true? A. 2x – 1 = 13 and x = 6 B. 2x – 1 = 13 and x = 7 C. 2x + 1 = 13 and x = 7 D. 2x – 1 = 13 and x = 11

alexandr1967 [171]

Answer:

B. 2x – 1 = 13 and x = 7

Step-by-step explanation:

We are given 4 equations and a solution for each. We have to tell which of the given solution satisfies the given equation.

Option A.

2x -1 = 13 and x = 6

Using this value in the equation, we get:

2(6) -1 = 13

12 - 1 = 13

11 = 13, which is not true. Hence this option is not valid

Option B.

2x - 1 = 13 and x = 7

Using the value in the equation, we get:

2(7) - 1 =13

14 - 1 =13

13 = 13, which is true. Hence this option is valid.

Option C.

2x + 1 =13 and x = 7

Using the value in the equation, we get:

2(7) + 1 = 13

15 = 13, which is not true. So this option is not valid

Option D.

2x - 1 = 13 and x = 11

Using this value in the equation, we get:

2(11) - 1 = 13

21 = 13, which is not true. Hence this option is not valid.

6 0

3 years ago

PLEASE HELP IMMEDIATELY!!! WILL MARK AS BRAINLIEST

Levart [38]

Answer:

Step-by-step explanation:

5 0

2 years ago

For a set of five whole numbers, the mean is 4, the mode is 1, and the median is 5. What are the five numbers? Plz help me plz

solniwko [45]

Mean is the average, mode is the number that is repeated most often and the median is the middle number. We know there are 5 numbers so, 5 is the middle as it is the median #. The mode has to appear more than once or 5 would also be a mode.
1 1 5 _ _
M = 1+1+5+x+y /5
4 = 1+1+5+x+y/ 5
4 = 7+x+y / 5
Average of 5 numbers is 4 therefore 4 times 5 = 20
20-7 = 13, what two numbers that are more than 5 equal 13? Only 6 &7.
4 = 1+1+5+6+7 / 5

3 0

3 years ago

At one point the average price of regular unleaded gasoline was $3.41 per gallon. Assume that the standard deviation price per

Aleonysh [2.5K]

Answer:

a) $1-\frac{1}{k^2} =1- \frac{1}{2^2}= 1-0.25 = 0.75$

So we expected about 75% within two deviations from the mean

b) $1-\frac{1}{k^2} =1- \frac{1}{1.5^2}= 1-0.4444 = 0.556$

So we expected about 55.6% within 1.5 deviations from the mean

And the limits are:

$Lower = 3.41 -1.5*0.09 = 3.275$

$Upper = 3.41 +1.5*0.09 = 3.545$

c) We can calculate how many deviations we are within the mean with the limits with this formula:

$z =\frac{x-\mu}{\sigma}$

And using the lower limit we got:

$z = \frac{3.05-3.41}{0.09}=-4$

And with the upper limit we got:

$z = \frac{3.77-3.41}{0.09}=4$

So then the value of k =4 and the percentage is given by:

$1-\frac{1}{k^2} =1- \frac{1}{4^2}= 1-0.0625 = 0.9375$

Step-by-step explanation:

Previous concepts and Data given

$\mu =3.41$ reprsent the population mean

$\sigma=0.09$ represent the population standard deviation

The Chebyshev's Theorem states that for any dataset

• We have at least 75% of all the data within two deviations from the mean.

• We have at least 88.9% of all the data within three deviations from the mean.

• We have at least 93.8% of all the data within four deviations from the mean.

Or in general words "For any set of data (either population or sample) and for any constant k greater than 1, the proportion of the data that must lie within k standard deviations on either side of the mean is at least: $1-\frac{1}{k^2}$

Part a

For this case we can find the percentage required replaincg k =2 and we got:

$1-\frac{1}{k^2} =1- \frac{1}{2^2}= 1-0.25 = 0.75$

So we expected about 75% within two deviations from the mean

Part b

For this case we can find the percentage required replaincg k =2 and we got:

$1-\frac{1}{k^2} =1- \frac{1}{1.5^2}= 1-0.4444 = 0.556$

So we expected about 55.6% within 1.5 deviations from the mean

And the limits are:

$Lower = 3.41 -1.5*0.09 = 3.275$

$Upper = 3.41 +1.5*0.09 = 3.545$

Part c

We can calculate how many deviations we are within the mean with the limits with this formula:

$z =\frac{x-\mu}{\sigma}$

And using the lower limit we got:

$z = \frac{3.05-3.41}{0.09}=-4$

And with the upper limit we got:

$z = \frac{3.77-3.41}{0.09}=4$

So then the value of k =4 and the percentage is given by:

$1-\frac{1}{k^2} =1- \frac{1}{4^2}= 1-0.0625 = 0.9375$

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3 years ago

What is the highest point the squirrel reaches?

Yuliya22 [10]

?? Is there a picture so I can go based off that

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3 years ago