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3 years ago

How would you characterize the relationships between the hours spent on homework and the test scores?

Mathematics

1 answer:

Elena-2011 [213]3 years ago

5 0

Answer:

There is a positive correlation because the more hours spent on homework the better the test score generally it.

Step-by-step explanation:

Since the graph has a positive slope and the data is close together, it looks to have positive correlation.

There is a positive correlation because the more hours spent on homework the better the test score generally it.

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What is two thirds or 2/3 of the whole 120

Zepler [3.9K]

It would be 80 because 2 times 120 is 240. 240 divided by 3 is 80

3 0

3 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago

the utility bill for the Millers home was it in April with $132 42% of the bill was for gas and the rest for electricity how muc

Anit [1.1K]

In the given question, there are several data of immense importance. based on these data's the answer to the question can be easily deduced. it is already given that the total utility bill for the month of April in Millers house is $132 and 42% of the bill was paid for gas. remaining amount is paid as electricity charges.
Now
Amount of money paid for gas = 132 * (42/100) dollars
      = 5544/100 dollars
   = 55.44 dollars
Then
The amount of money paid for electricity = (132 - 55.44) dollars
   = 76.56 dollars
So the Millers paid 55.44 dollars for gas and 76.56 dollars for electricity in the month of April.

6 0

3 years ago

Read 2 more answers

Can someone help with my homework please

EleoNora [17]

B is the answer to this problem

8 0

3 years ago

The sandwiches at the bakery come in four sizes:

kaheart [24]

13 kids sandwiches X 4 inches = 52 inches
22 small sandwiches X 6 inches = 132 inches
17 medium sandwiches X 8 inches = 136 inches
17 large sandwiches X 12 inches = 204 inches

52+132+136+204= 524 inches = 43.66 feet = 43.7 feet

7 0

3 years ago