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2 years ago

If a car travels 825 miles in 15 hours and the car travels at an constant rate how many miles does the car travel in 1 hour?

Mathematics

1 answer:

Doss [256]2 years ago

7 0

Answer:

55 miles

Step-by-step explanation:

Given data

DIstance = 825 miles

Time = 15 hours

Speed= distance/time

Speed= 825/15

Speed= 55 miles per hour

Given that we want to find the distance traveled in 1 hour, we have

Distance= Speed*time

Distance= 55*1

Distance= 55 miles

Hence the car will travel 55 miles in 1 hour

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Leah has $4.35 in her pocket in nickels and dimes. The number of dimes is 2 less than 3 times the number of nickels. How many of

Maksim231197 [3]

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2 years ago

Read 2 more answers

A bus travels on an east-west highway connecting two cities A and B that are 100 miles apart. There are 2 services stations alon

melamori03 [73]

Answer:

51/4

Step-by-step explanation:

To begin with you have to understand what is the distribution of the random variable. If X represents the point where the bus breaks down. That is correct.

X~ Uniform(0,100)

Then the probability mass function is given as follows.

$f(x) = P(X=x) = 1/100 \,\,\,\, \text{if} \,\,\,\, 0 \leq x \leq 100\\f(x) = P(X=x) = 0 \,\,\,\, \text{otherwise}$

Now, imagine that the D represents the distance from the break down point to the nearest station. Think about this, the first service station is 20 meters away from city A, and the second station is located 70 meters away from city A then the mid point between 20 and 70 is (70+20)/2 = 45 then we can represent D as follows

$D(x) =\left\{ \begin{array}{ll} x & \mbox{if } 0\leq x \leq 20 \\ x-20 & \mbox{if } 20\leq x < 45\\ 70-x & \mbox{if } 45 \leq x \leq 70\\ x-70 & \mbox{if } 70 \leq x \leq 100\\ \end{array}\right.$

Now, as we said before X represents the random variable where the bus breaks down, then we form a new random variable $Y = D(X)$ , $Y$ is a random variable as well, remember that there is a theorem that says that

$E[Y] = E[D(X)] = \int\limits_{-\infty}^{\infty} D(x) f(x) \,\, dx$

Where $f(x)$ is the probability mass function of X. Using the information of our problem

$E[Y] = \int\limits_{-\infty}^{\infty} D(x)f(x) dx \\= \frac{1}{100} \bigg[ \int\limits_{0}^{20} x dx +\int\limits_{20}^{45} (x-20) dx +\int\limits_{45}^{70} (70-x) dx +\int\limits_{70}^{100} (x-70) dx \bigg]\\= \frac{51}{4} = 12.75$

3 0

2 years ago

If a simple harmonic motion is described by x = 3 cos(πt), starting from t = 0 s, what is the first occurrence of a value of t f

poizon [28]

Answer:

The first occurence of t for which x = 0 is t = 0.5.

Step-by-step explanation:

The harmonic motion is described by the following equation.

$x(t) = 3\cos{\pi t}$

What is the first occurrence of a value of t for which x = 0?

This is t when $x(t) = 0$ . So

$x(t) = 3\cos{\pi t}$

$3\cos{\pi t} = 0$

$\cos{\pi t} = \frac{0}{3}$

$\cos{\pi t} = 0$

The inverse of the cosine is the arcosine. So we apply the arcosine function to both sides of the equality.

$\arccos{\cos{\pi t}} = \arccos{0}$

$\pi t = \frac{\pi}{2}$

$t = \frac{\pi}{2 \pi}$

$t = \frac{1}{2}$

$t = 0.5$

The first occurence of t for which x = 0 is t = 0.5.

3 0

2 years ago

A score of "3" on a par 5 hole is also known as a(n) _____.

den301095 [7]

The correct answer is "eagle." A score of "3" on a par 5 hole is also known as an eagle. Albatross on the other hand is a score of 3-under par on a hole. A birdie is universally known as a score of 1-under par on a hole. An eagle is 2-under par or 3-under par 5.

Here are the choices.
eagle
birdie
bogey
par

7 0

2 years ago

The population of the town increased from 175000 to 262500 in a decade. What is the average percent increase of population per y

Alenkinab [10]

Answer:

Step-by-step explanation:

As given,

The population of the town increased from 175000 to 262500 in a decade.

⇒ In 10 years , population increased from 175000 to 262500

⇒% increase = $\frac{262500 - 175000}{175000}$ × 100% = $\frac{87500}{175000}$ ×100% = 0.5×100% = 50%