Can someone help me with number 2

iogann1982 [59]

Answer:

the median marks the middle or the mid-point of the data and is often shown by the line that divides the box into two parts.

Step-by-step explanation:

hope this helps:)

6 0

3 years ago

Read 2 more answers

Simplify 16k/7k a) k+23 b) 9k c) 23k d) 23k^2

alexira [117]

Answer:

16/7

Step-by-step explanation:

16k / 7k

Divide the numbers

16/7

Divide the variables

k/k = 1

16/7

3 0

3 years ago

Given f(x)=4x-6 and g(x)=8x+4 findthe value of x for which f(x)=g(x)

lana [24]

The value of "x" is $\frac{-5}{2}$

Solution:

Given f(x) = 4x - 6

g(x) = 8x + 4

To find: value of x when f(x) = g(x)

We can equate f(x) and g(x) as per given statement and solve for x

f(x) = g(x)

Substituting the values of f(x) and g(x) in above expression,

4x - 6 = 8x + 4

4x - 8x = 4 + 6

-4x = 10

On solving,

$x = \frac{10}{-4}\\\\x = \frac{-5}{2}$

Thus value of "x" is $\frac{-5}{2}$

4 0

3 years ago

The acceleration of an object (in m/s^2) is given by the function a(t) = 9 sin(t). The initial velocity of the object is v(0) =

pentagon [3]

a) Acceleration is the derivative of velocity. By the fundamental theorem of calculus,

$v(t)=v(0)+\displaystyle\int_0^ta(u)\,\mathrm du$

so that

$v(t)=\left(-11\frac{\rm m}{\rm s}\right)+\int_0^t9\sin u\,\mathrm du$

$\boxed{v(t)=-\left(2+9\cos t)\right)\frac{\rm m}{\rm s}}$

b) We get the displacement by integrating the velocity function like above. Assume the object starts at the origin, so that its initial position is $s(0)=0\,\mathrm m$ . Then its displacement over the time interval [0, 3] is

$s(0)+\displaystyle\int_0^3v(t)\,\mathrm dt=-\int_0^3(2+9\cos t)\,\mathrm dt=\boxed{-6-9\sin3}$

c) The total distance traveled is the integral of the absolute value of the velocity function:

$s(0)+\displaystyle\int_0^3|v(t)|\,\mathrm dt$

$v(t)$ for $0\le t$ and $v(t)\ge0$ for $\cos^{-1}\left(-\frac29\right)\le t\le3$ , so we split the integral into two as

$\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}-v(t)\,\mathrm dt+\int_{\cos^{-1}\left(-\frac29\right)}^3v(t)\,\mathrm dt$

$=\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}(2+9\cos t)\,\mathrm dt-\int_{\cos^{-1}\left(-\frac29\right)}^3(2+9\cos t)\,\mathrm dt$

$\displaystyle=\boxed{2\sqrt{77}-6+4\cos^{-1}\left(-\frac29\right)-9\sin3}$

4 0

3 years ago

Mr. Jackson lengthened the fence along the back of his yard. Before he added 130 feet, the fence was 300 feet long. He divided t

hram777 [196]

Answer:

The answer is A, 86 sections.

Step-by-step explanation:

Step one : Add the 130 feet on to the 300 feet to find the new length of the fence.

130 + 300 = 430

Step two : find how many 5 foot sections he has now, divide 430 by 5.

430/5 = 86

Mr. Jackson can divide the new fence into 86 5-foot sections.

The answer is A, 86 sections.

8 0

3 years ago