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Reil [10]
2 years ago
11

Sara found that the older person is, fewer cartoons that person watches.

Mathematics
1 answer:
Misha Larkins [42]2 years ago
4 0

Answer:

B

Step-by-step explanation:

The scatter graph shows a decreasing correlation.

Its not A because it's a increasing correlation

its not C because it's neutral

its not D because it's a no correlation graph

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Consider the absolute value inequality: |x - 6| > 20. Will the graph of the solutions to the inequality result in an "and" or
Dmitrij [34]

Step-by-step explanation:

Case 1 : |x| > a, => x > a or x < -a

Case 2 : |x| < a, => -a < x < a

Since this question follows Case 1, we will have an "or" inequality.

3 0
3 years ago
A large pizza at Palanzio’s Pizzeria costs $6.80 plus $0.90 for each topping.
Nookie1986 [14]

Answer:2

Step-by-step explanation:

.9x+6.8

.65x+7.3

Find x by graphing it

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(WORTH 95 POINTS) PLEASE HELP I CANNOT FIGURE IT OUT AND ITS SUPPOSED TO BE TURNED IN ALREADY!
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Hope this helped :3

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5 0
3 years ago
Not really getting this can someone plz explain.
Komok [63]

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7 0
3 years ago
A cold drink is poured out at 52°F. After 2 minutes of sitting in a 72°F room, its temperature has risen to 55°F. Find an equati
I am Lyosha [343]

Answer:

The model for the temperature of the drink can be written as

T=72-20e^{-0.08t}

Step-by-step explanation:

For a cold drink in a hotter room, we can say that the rate of change of temperature of the drink is proportional to the difference of temperature between the drink and the room.

We can model that in this way

\frac{dT}{dt}=k*(T_r-T)

If we rearrange and integrate

\int\frac{dT}{(T-Tr)} =-k*\int dt\\\\ln(T-T_r)=-kt+C1\\\\T-T_r=Ce^{-kt}\\\\T=T_r+Ce^{-kt}

We know that at time 0, the temperature of the drink was 52°F. Then we have:

T=T_r+Ce^{-kt}\\\\52=72+Ce^0=72+C\\\\C=-20

We also know that at t=2, T=55°F

T=T_r+Ce^{-kt}\\\\55=72-20e^{-k*2}\\\\e^{-k*2}=(72-55)/20=0.85\\\\-2k=ln(0.85)=-0.1625\\\\k=0.08

The model for the temperature of the drink can be written as

T=72-20e^{-0.08t}

7 0
3 years ago
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