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3 years ago

She decide to use graph A to predict the number of bottles of water that would be sold when the temperature is 100

Mathematics

1 answer:

UNO [17]3 years ago

7 0

Answer:

Graph A has a stronger correlation

Step-by-step explanation:

The data in Graph A has a strong correlation coefficient.

so we would be able predict more accurately.

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Task 1:

laiz [17]

Answer:

The toy's maximum height is $10.12\ \text{feet}$ .

Step-by-step explanation:

The equation that models the toy's height in feet from the ground at t seconds after he threw it is given by :

$h = -2t^2 + 7t + 4$ .....(1)

It is required to find the toy's maximum height. For maxima or minima, put

$\dfrac{dh}{dt}=0$

Plugging the value of t in above equation,

$\dfrac{d(-2t^2 + 7t + 4)}{dt}=0\\\\-4t+7=0\\\\t=\dfrac{7}{4}\\\\t=1.75\ s$

Now put t = 1.75 s in equation (1) such that,

$h = -2(1.75)^2 + 7(1.75) + 4\\\\h=10.12\ \text{feet}$

So, the toy's maximum height is $10.12\ \text{feet}$ .

4 0

3 years ago

What is 3/4 to the power of 2

natima [27]

Answer:

9/16

Step-by-step explanation:

5 0

3 years ago

R minus the sum of 4 and p equals

UkoKoshka [18]

Answer:

r - (4 + p)

Step-by-step explanation:

{that's how you write it if that's what you are asking}

4 0

3 years ago

6. A tank contains 100-gallon of pure water. At time t = 0, a solution containing 2 lb of salt per gallon flows into the tank at

SOVA2 [1]

Answer:

The answer is below

Step-by-step explanation:

Let Q represent the amount of salt in the tank at time t.

$\frac{dQ}{dt}= flow\ in - flow\ out\\ \\flow\ in=3\ gal/min*2\ lb/gal=6\ lb/min\\\\net\ gain\ in\ tank\ volume=3-4=-1, henceflow\ out= \frac{4Q}{100-t} \\\\\frac{dQ}{dt}= 6-\frac{4Q}{100-t} \\\\\frac{dQ}{dt}+ \frac{4Q}{100-t}=6\\\\The \ integrating\ factor\ is:\\\\IF=e^{\int\limits {\frac{4}{100-t} } \, dt }=e^{-4\int\limits {\frac{-1}{100-t}}=e^{-4ln(100-t)}=(100-t)^{-4}}\\\\Multiplying\ through \ by\ IF: \\\\(100-t)^{-4}\frac{dQ}{dt}+ (100-t)^{-4}\frac{4Q}{100-t}=6(100-t)^{-4}\\\\$

$Integrating:\\\\A(100-t)^{-4}=-2(100-t)^{-3}+c\\\\A=-2(100-t)+\frac{c}{(100-t)^{-4}} \\\\at, t=0,A=0\\\\0=-2(100-0)+\frac{c}{(100-0)^{-4}}\\\\c=0.02\\\\A=-2(100-t)+\frac{0.02}{(100-t)^{-4}}$

7 0

3 years ago

52 > 43 + w

zvonat [6]

Hopes this helps

Answer: w < 9 yes

Please mark as brainlist

4 0

3 years ago