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3 years ago

A tour bus travels 2160 miles in 36 hours. What is the average distance the bus travels in one hour?

Mathematics

2 answers:

puteri [66]3 years ago

8 0

2160/36 is equal to 60

Butoxors [25]3 years ago

6 0

60 would be the answer

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What is 5x5x3x5x3x2 simplified using exponents

8090 [49]

Answer: 5 to the 3rd times 3 to the second times 2

Step-by-step explanation:

8 0

3 years ago

Which answer is the approximate standard deviation of the data set?

Anna35 [415]

Answer:

3.4

Step-by-step explanation:

Standard deviation of a population is defined as:

σ² = ∑(xᵢ − μ)² / n

The standard deviation of a sample is defined as:

s² = ∑(xᵢ − x)² / (n - 1)

It's not clear which one we have, so let's calculate both.

First, we must find the mean.

μ = (5+12+15+10+12+6+8+8) / 8

μ = 9.5

Now we find the squares of the differences:

(5-9.5)² + (12-9.5)² + (15-9.5)² + (10-9.5)² + (12-9.5)² + (6-9.5)² + (8-9.5)² + (8-9.5)²

= 80

Divide by n:

σ² = 80 / 8

σ² = 10

And take the square root:

σ = √10

σ ≈ 3.2

That's not one of the answers, so let's try the standard deviation of a sample instead of a population.

Instead of dividing by n, we'll divide by n-1:

s² = 80 / 7

And take the square root:

s = √(80/7)

s ≈ 3.4

So that must be it.

6 0

3 years ago

There are 2 pecks in one kenning there are 2 kennings in 1 bushel there are 8 bushels in1 quarter there are 5 quarters in 1 load

Delicious77 [7]

Answer:

yes 1 2 3

Step-by-step explanation:

7 0

2 years ago

A plane is an undefined term because it

igor_vitrenko [27]

No on can explain it but it has a definition.

8 0

3 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

2 years ago