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alexira [117]
3 years ago
14

Find the Slope using the Slope Formula 1. (8.6).(2-8)​

Mathematics
1 answer:
mojhsa [17]3 years ago
5 0

Answer: 2.3 repeating

Please mark me as Brainliest

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Solve for t.<br><br> t/-3.2 &lt; 5
Black_prince [1.1K]

Step-by-step explanation:

t/-3,2 < 5

t > 5 × -3,2

t >-16

4 0
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Is there negative correlation, or no correlation between the day and the sales amount?
FrozenT [24]

Answer:

A: no correlation

Step-by-step explanation:

Hopefully this helps!

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3 years ago
What is the value for ƒ(x) = 42x - 100 when x = 2 ?
Anton [14]
F(2)=42(2)-100
f(2)= -16
f(2)/-16= -8
3 0
3 years ago
Students sometimes suspect that studying more isn't worth it for "harder" math and science classes: either you understand it, or
Zolol [24]

ANSWER: The statistical procedure that should be performed is REGRESSION.

Step-by-step explanation: Regression is a statistical procedure that is used to estimate the relationship between an independent variable and a dependent variable using their mean values.

The independent variable in this case is the hours each student spend in studying, while the dependent variable is the students grade.

Since the researcher wants to determine if the hours a student spend in studying maths and science has any significant effect on their grades. The researcher should use regression, because it will show if the two variables are related and how it relates, by showing how far the points are from the trend lines of the graph.

6 0
4 years ago
A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are
skelet666 [1.2K]

A) The maximum revenue is 450000$

B) The maximum profit is 216000$ when 2400 sets are manufactured and sold for 180$ each

C) When each set is taxed at ​$55​, the maximum profit is 99125$ when 1850 sets are manufactured and sold for 207.5$ each.

A)p(x)=300−(x/20​),

revenue R(x)=p*x

revenue R(x)=300x -(x2/20)

for maximum revenue dR/dx =0 ,

=>300-(2x/20)=0

=>x/10=300

=>x=3000

maximum revenue = R(3000)=300*3000 -(30002/20)

maximum revenue = R(3000)=450000$

B) profit =revenue -cost

profit P(x)=300x -(x2/20)-72000-60x

profit P(x)=240x -(x2/20)-72000

for maximum cost dP/dx =0

240 -(2x/20)=0

x=240*10

x=2400

p(2400)=300−(2400/20​)=180

profit P(2400)=240*2400 -(24002/20)-72000 =216000

The maximum profit is 216000$ when 2400 sets are manufactured and sold for 180$ each

c)

profit =revenue -cost -tax

profit P(x)=300x -(x2/20)-72000-60x-55x

profit P(x)=185x -(x2/20)-72000

for maximum cost dP/dx =0

185-(2x/20)=0

x=185*10

x=1850

p(1850)=300−(1850/20​)=207.5

profit P(1850)=185*1850 -(18502/20)-72000

profit P(1850)=99125$

When each set is taxed at ​$55​, the maximum profit is 99125$ when 1850 sets are manufactured and sold for 207.5$ each.

To know more about maximum profit check the below link:

brainly.com/question/4166660

#SPJ4

5 0
1 year ago
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