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3 years ago

Need help!!Do the values in the table represent a proportional relationship?

Mathematics

2 answers:

ikadub [295]3 years ago

7 0

Yes y equals 2.5 times x

irakobra [83]3 years ago

5 0

Y = 2.5x

answer

1st : are
2nd : proportional

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Two kinds of crated cargo, A and B, are to be shipped by truck. Each crate of cargo A is 50 cubic feet in volume and weighs 200

BlackZzzverrR [31]

Answer:

The correct answer is option b)

50a+10b<1000; 200a+360b < 7200; a> 0; b>0

Step-by-step explanation:

We are given that Two kind of crated cargo namely A and B to be shipped by truck.

Cargo A:

Volume of each crate of cargo A = 50 cubic ft

Weight of each crate of cargo A = 200 pounds

Let number of crates of cargo A to be shipped = a

Total volume of 'a' crates of cargo A = 50a cubic ft

Total weight of 'a' crates of cargo A = 200a pounds

Cargo B:

Volume of each crate of cargo B = 10 cubic ft

Weight of each crate of cargo B = 360 pounds

Let number of crates of cargo B to be shipped = b

Total volume of 'b' crates of cargo B = 10b cubic ft

Total weight of 'b' crates of cargo B = 360b

Total volume allowed in the truck is 1000 cubic ft

Total volume of 'a' crates of Cargo A and Total volume of 'b' crates of Cargo B = 50a+10b cubic ft (This sum should be less than volume of truck so that it can fit in the truck)

So, the inequality becomes

$50a+10b$ ....... (1)

Total weight allowed (load limit) in the truck is 7200 pounds

Total weight of 'a' crates of Cargo A and Total weight of 'b' crates of Cargo B = 200a+360b cubic ft (This sum should be less than volume of truck so that it can fit in the truck)

So, the inequality becomes

$200a+360b$ ....... (1)

And number of crates of cargo A and B are always a positive number.

So, a > 0 and b > 0.

So, the correct answer is option b.

b. 50a+10b<1000; 200a+360b < 7200; a> 0; b>0

3 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

All angles below are right angles. What is the area of the figure?

disa [49]

The correct answer is 36

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2 years ago

If you buy a ticket, what is the Probability that your ticket will have the winning numbers?

dedylja [7]

It truly depends on the ticket you got.

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2 years ago