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Artist 52 [7]
3 years ago
13

Mr. Baker's class has three pet mice. The students observe the mice every day for two weeks. Every hour, they write down what th

e mice are doing. At the end of the two weeks, they look at the data. They see that the mice are sleeping most of the time. What does this most likely tell the students about mice?
Mathematics
2 answers:
ASHA 777 [7]3 years ago
6 0
The mice are nocturnal, meaning they are awake at night and sleep during the day at most times
gizmo_the_mogwai [7]3 years ago
5 0
They are dead due to the law of dead
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To the nearest​ dollar, the average tuition at a public four−year college was $3088 in 1997 and $3292 in 2000. ​Find, to the nea
Delicious77 [7]

the fees in 1997 was $3088

fees in 2000 was $3292

average rate of increase= (3292-3088)/(2000-1997) = 204/3=68

so average rate of increase is $68 per year

4 0
3 years ago
Show steps using Pythagorean Theorem please
neonofarm [45]
Same I don’t get this
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2 years ago
2x to the third power - 4y to the second power-10
SCORPION-xisa [38]
2 (x^3 - 2y^2 -5)

hope this helps

you are trying to factor as much out as possible from each

in this case, it is 2
5 0
3 years ago
Question 9 (1 point)
Luden [163]

Answer:

7

Step-by-step explanation:

16x-80+18=4x+22

16x-62=4x+22

12x-62=22

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8 0
3 years ago
A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min
sesenic [268]
A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})
\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}
A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4

We're given that A(0)=30. Multiply both sides by the integrating factor e^{t/100}, then

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4
\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4
e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C
A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}

Given that A(0)=30, we have

30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02

so the amount of salt in the tank at time t is

A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}
3 0
3 years ago
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