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2 years ago

8

A ball is thrown in the air from a height of 3 m with an initial velocity of 12 m/s. To the nearest tenth of a second, how long

does it take for the ball to land on the ground

1 answer:

wel2 years ago

5 0

Answer: 0.3m/s

Step-by-step explanation:

Height/ velocity

3m/12m per sec

Time= 0.25metres per seconds

To the nearest tenth, 0.3m per seconds

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Kaylis [27]

The correct answer is (20,-50)

7 0

3 years ago

Find the mean median mode and range for the following data 77 60 59 70 89 95

Advocard [28]

Answer:

Mean = 75

Median = 73.5

Mode = 95

Range = 36

Step-by-step explanation:

Given:

77,60,59,70,89,95

Sort:

59,60,70,77,89,95

To find:

Mean
Median
Mode
Range

Mean:

$\displaystyle \large{\dfrac{1}{n}\sum_{i =1}^n x_i = \dfrac{x_1+x_2+x_3+...+x_n}{n}}$

Sum of all data divides by amount.

$\displaystyle \large{\dfrac{59+60+70+77+89+95}{6}=\dfrac{450}{6}}\\\\\displaystyle \large{\therefore mean=75}$

Therefore, mean = 75

Median:

If it’s exact middle then that’s the median. However, if two data or values happen to be in <em>middle</em>:

$\displaystyle \large{\dfrac{x_1+x_2}{2}}$

From 59,60,70,77,89,95, since 70 and 77 are in middle:

$\displaystyle \large{\dfrac{70+77}{2} = \dfrac{147}{2}}\\\displaystyle \large{\therefore median = 73.5}$

Therefore, median = 73.5

Mode:

The highest value or/and the highest amount of data. Mode can have more than one.

From sorted data, there are no repetitive data nor same data. Consider the highest value:

Therefore, mode = 95

Range:

$\displaystyle \large{x_{max}-x_{min}}$ or highest value - lowest value

Thus:

$\displaystyle \large{95-59 = 36}$

Therefore, range = 36

7 0

1 year ago

What’s 6-1/3x=-1 in two step equation???

swat32

Answer:

x=21

Step-by-step explanation:

subtract 6 from both sides get -1/3x=-7

multiply -1/3 or -1/3 as decimal times -7

and get x=21

If brainiest is earned its greatly appreciated. :)

6 0

3 years ago

X + 5 * 3 divided by 4 - 2=

muminat

If it's laptop pres shift nd the key that show ÷ at the same time

6 0

3 years ago

A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the

amm1812

Answer:

$s(t)=(ln|7|+ln|2|)\,ft$

Step-by-step explanation:

Acceleration is second derivative of distance and are related as:

$a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\$

Integrating both sides w.r.to t

$v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\$

Using initial value

$v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}$

We have to calculate the distance covered in time interval [0,5], so:

$\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft$

3 0

3 years ago