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3 years ago

13

You bought a car that had factory installed tires marked as "P185 75/R14". You bought and installed tires on your car that are m

arked "P205 75/R14". What is the circumference of the P185 tire?
A) 24.93 inches
B) 27.93 inches
C) 78.31 inches
D) 81.99 inches

1 answer:

Slav-nsk [51]3 years ago

4 0

D is the right answer I think

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6, 3, 4, 7, 3, 7, 8, 5, 7 i. Find the median ii. Find the mean iii. Find the modal mark

4vir4ik [10]

Answer:

i. 6

ii. $5\frac{5}{9}$

iii. 7

Step-by-step explanation:

First organize the data from least to greatest. 3,3,4,5,6,7,7,7,8

To find the median, remove the extremes from the data over and over.

3,4,5,6,7,7,7

4,5,6,7,7

5,6,7

6

To find the mean, add all of the numbers and divide by 9

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50/9= $5\frac{5}{9}$

To find the modal mark, simply find the number present most in the data set: 7(occurs 3 times)

Hope it helps <3

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3 years ago

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

7 0

3 years ago