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ycow [4]
3 years ago
8

Among the licensed drivers in the same age group, what is the probability that

Mathematics
2 answers:
GarryVolchara [31]3 years ago
3 0

Answer:

"12.5%?"

Step-by-step explanation:

before you use my answer use other peoples answers. I'm not good at som questions, sorry.

Dmitry_Shevchenko [17]3 years ago
3 0

Answer:

10.5%

Step-by-step explanation:

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What is the solution to 2x^+8x=x^-16
malfutka [58]

Answer:

If you are solving for x then the answer is x= -4

Step-by-step explanation:

8 0
3 years ago
PLS HELP ME ASAP FOR 1 and 2 (SHOW WORK!! + Brainliest!) LOTS OF POINTS!! (Kinda)
liubo4ka [24]
For number one, the answer will be C because the ratio between the side and the perimeter of the Pentagon will be 1/5 or 2/10 or 3/15. For number 2, we know that the Pentagon had 5 sides, and the side of it is 8, so the perimeter of this Pentagon will be 8+8+8+8+8 or 40 inches, which is C. Hope it help!
8 0
3 years ago
I really need some help
motikmotik
It is -12,10 and your welcome for me helping you im only trying to get the point s
8 0
2 years ago
Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify
irina [24]

Answer:

x-coordinates of relative extrema = \frac{-1}{2}

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}

Differentiate with respect to x

f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}

f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}

Differentiate f'(x) with respect to x

f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1

At x = \frac{-1}{2},

f''\left ( \frac{-1}{2} \right )=\frac{4\left ( -1+4\left ( \frac{-1}{2} \right ) \right )}{3\left ( \frac{-1}{2} \right )^{\frac{5}{3}}}>0

We know that if f''(a)>0 then x = a is a point of minima.

So, x=\frac{-1}{2} is a point of minima.

For inflexion points:

Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.

So, x-coordinates of the inflexion points are 0, 1

7 0
3 years ago
because the angles in a rectangle are 90 degrees it is not a parallelogram true or false? please explain why
ra1l [238]
<span>A parallelogram is a 4-sided shape where opposites sides are parallel. A rectangle is a special case of a parallelogram. All rectangles are parallelograms.

But a rectangle is a shape where opposites sides are parallel *and* all the corners are 90 degree angles. So you can't say that all parallelograms would be rectangles. Some parallelograms would be rectangles, but not all.

Rectangles are a subset of the shapes called parallelograms. But parallelograms are *not* a subset of the shapes called rectangles.

It's similar to saying all cars are vehicles. But you can't say that all vehicles are cars. </span>
6 0
3 years ago
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