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alekssr [168]
2 years ago
6

The table below represents the closing prices of stock ABC for the last five days. What is the r-value of the linear regression

that fits these data?
DAY VALUE
1 24.31
2 23.31
3 23.01
4 23.16
5 23.27

A. -0.69771
B. 0.56941
C. -0.68427
D. 0.89215
Mathematics
2 answers:
Mekhanik [1.2K]2 years ago
7 0

Answer:

C. -0.68427

Step-by-step explanation:

Enter the data into a graphing calculator.

First decide which is the independent and which is the dependent variable.  Since the value of the stock depends on the day, value will be dependent and day will be independent.

This means that the days will be entered into the first list and the values will be entered into the second list.

Entering these and running the linear regression, we find the r-value to be -0.68427 .

svlad2 [7]2 years ago
4 0
I think it may b either c or b I'm not sure
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There are 6 different ways:

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3 0
2 years ago
Suppose the Sunglasses Hut Company has a profit function given by P ( q ) = − 0.02 q 2 + 5 q − 30 , where q is the number of tho
Andrew [12]

Answer:

A) Pairs of sunglasses should be sold to maximize profits is 100 thousand.

B) The actual maximum profits that can be expected is 180 thousands dollars.

Step-by-step explanation:

Given : Suppose the Sunglasses Hut Company has a profit function given by P(q)=-0.02q^2+4q-20

where, q is the number of thousands of pairs of sunglasses sold and produced and P(q) is the total profit, in thousands of dollars, from selling and producing q pairs of sunglasses.

To find :

A) How many pairs of sunglasses (in thousands) should be sold to maximize profits?

Profit is P(q)=-0.02q^2+4q-20

Compare with general quadratic equation ax^2+bx+c

a=-0.02, b=4, c=-20

The maximum value of quadratic function is at x=-\frac{b}{2a}

So, maximum profit occur at q is given by,

q=-\frac{4}{2(-0.02)}

q=\frac{4}{0.04}

q=100

i.e. 100 thousand pair of sunglasses.

Pairs of sunglasses should be sold to maximize profits is 100 thousand.

B) What are the actual maximum profits (in thousands) that can be expected?

Maximum profit occurs at q=100 thousand pairs is given by,

P(100)=-0.02(100)^2+4(100)-20

P(100)=-0.02\times 10000+400-20

P(100)=180

The actual maximum profits that can be expected is 180 thousands dollars.

7 0
1 year ago
The Bureau of the Census reports that the median family income for all families in the United States during the year 2003 was $4
miv72 [106K]

Answer:

{GGGG, GGGN, GGNG, GNGG, NGGG, GGNN, GNGN, GNNG, NGGN, NGNG, NNGG, GNNN, NGNN, NNGN. NNNG, NNNN}

Step-by-step explanation:

4 families are surveyed and each one reveals whether its income exceeded $43,318 in 2003. Each family has two possibilities: Having an income greater than 43,318 (G), or is not greater than $43,318 (N). We have 2⁴ = 16 different outcomes.

The possible outcomes are: GGGG, GGGN, GGNG, GNGG, NGGG, GGNN, GNGN, GNNG, NGGN, NGNG, NNGG, GNNN, NGNN, NNGN. NNNG, NNNN.

Therefore the sample space is {GGGG, GGGN, GGNG, GNGG, NGGG, GGNN, GNGN, GNNG, NGGN, NGNG, NNGG, GNNN, NGNN, NNGN. NNNG, NNNN}

8 0
1 year ago
A population of 950 bacteria grows continuously at a rate of 4.75% per day. (a) Write an exponential function, N(t), that repres
krek1111 [17]

Answer: a) N(t) = 950e^0.0475t b) 1020

Step-by-step explanation:

a)

N(t) = 950e^0.0475t.

N/B — Base is e (exponential) because growth is continuous.

b)

N = {36/24} =(approximately) 1020

4 0
2 years ago
Read 2 more answers
Last year Beth's annual salary was $38,350. This year she received a promotion and now earns $46,462 annually. She is paid biwee
kenny6666 [7]

Answer:

Quite simple. There are 52 weeks in a year. She gets paid 26 times, every other week.

38,350 / 26 = her salary from last year per check.

46,462 / 26 = her new salary per check

(46,462 - 38,350) / 26 = the difference per week

8 0
1 year ago
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