All categories

3 years ago

If a car is traveling at a rate of 75 miles per hour, how far will the car travel in 1 hour??

Mathematics

1 answer:

puteri [66]3 years ago

5 0

Answer:

75 miles

Step-by-step explanation:

Recall that distance = rate times time.

Here,

distance = 75 miles/hour times 1 hour = 75 miles

You might be interested in

Find the values of the 30th and 90th percentiles of the data. Please show your work.

MaRussiya [10]

Answer:

30th percentile:109

90th percentile: 188

Step-by-step explanation:

The given data set is:

129, 113, 200, 100, 105, 132, 100, 176, 146, 152

We arrange the data set in ascending order to obtain;

Array= $100,100,105,113,129,132,146,176,200$

The 30th percentile is located at

$(\frac{30}{100}\times 10 )}$ -th position, where n=10 is the number of items in the data set

This implies that the 30th percentile is at the 3rd position.

Since 3 is an integer, the 30th percentile is the average of the 3rd and 4th occurrence in the array;

This implies that the 30th percentile = $\frac{105+113}{2}=\frac{218}{2}=109$

The 90th percentile is located at

$(\frac{90}{100}\times 10 )}$ -th position, where n=10 is the number of items in the data set

This implies that the 90th percentile is at the 9th position.

Since 9 is an integer, the 90th percentile is the average of the 9th and 10th occurrence in the array;

This implies that the 90th percentile = $\frac{176+200}{2}=\frac{376}{2}=188$

6 0

3 years ago

Doug hits a baseball straight towards a 15 ft high fence that is 400 ft from home plate. The ball is hit 2.5ft above the ground

Scorpion4ik [409]

Answer:

$125.4\ \text{m/s}$

Step-by-step explanation:

u = Initial velocity of baseball

$\theta$ = Angle of hit = $30^{\circ}$

x = Displacement in x direction = 400 ft

y = Displacement in y direction = 15 ft

$y_0$ = Height of hit = 2.5 ft

$a_y$ = g = Acceleration due to gravity = $32.2\ \text{ft/s}^2$

t = Time taken

Displacement in x direction

$x=u_xt\\\Rightarrow x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{400}{u\cos30^{\circ}}\\\Rightarrow t=\dfrac{400}{u\dfrac{\sqrt{3}}{2}}\\\Rightarrow t=\dfrac{800}{u\sqrt{3}}$

Displacement in y direction

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=y_0+u\sin\theta t+\dfrac{1}{2}a_yt^2\\\Rightarrow 15=2.5+u\sin30^{\circ}(\dfrac{800}{u\sqrt{3}})+\dfrac{1}{2}\times -32.2\times (\dfrac{800}{u\sqrt{3}})^2\\\Rightarrow \dfrac{400}{\sqrt{3}}-\dfrac{10304000}{3u^2}-12.5=0\\\Rightarrow 218.44=\dfrac{10304000}{3u^2}\\\Rightarrow u=\sqrt{\dfrac{10304000}{3\times218.44}}\\\Rightarrow u=125.4\ \text{m/s}$