You have a bowl with 5 orange, 6 blue, 3 green, 4 red and 7 yellow candies. What is the probability that you will

Mathematics

1 answer:

shutvik [7]3 years ago

8 0

Answer:

с ) 20%

Step-by-step explanation:

There are 5 orange candies, and 25 total candies to choose from. If you only want to pick the orange candies, then the probability would be 5/25, or 1/5.

1/5, in percentage, is 20%.

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Can someone help me solve this differentiation/tangent problem?

nirvana33 [79]

A)

$\bf g'(x)=\stackrel{product~rule}{2x\cdot f(x)+x^2\cdot f'(x)}\quad 
\begin{cases}
x=5\\
f(5)=5\\
f'(5)=5
\end{cases}
\\\\\\
g'(5)=2(5)\cdot f(5)+(5)^2\cdot f'(5)\implies g'(5)=50+500\\\\\\ g'(5)=550\\\\
-------------------------------\\\\
g(5)=(5)^2f(5)\implies g(5)=125
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}
\begin{cases}
x=5\\
y=125\\
\stackrel{m}{g'(5)}=550
\end{cases}\implies y-125=550(x-5)
\\\\\\
y-125=550x-2750\implies y=550x+400$

b)

$\bf h'(x)=\stackrel{quotient~rule}{\cfrac{f'(x)(x-6)~~-~~f(x)\cdot 1}{(x-6)^2}}\quad 
\begin{cases}
x=5\\
f(5)=5\\
f'(5)=5
\end{cases}
\\\\\\
h'(5)=\cfrac{f'(5)(5-6)~~-~~f(5)\cdot 1}{(5-6)^2}
\\\\\\
h'(5)=\cfrac{5(-1)-5}{(-1)^2}\implies h'(5)=-10\\\\
-------------------------------\\\\$

$\bf h(5)=\cfrac{f(5)}{5-6}\implies h(5)=\cfrac{5}{-1}\implies h(5)=-5
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\quad 
\begin{cases}
x=5\\
y=-5\\
\stackrel{m}{-10}
\end{cases}\implies y-(-5)=-10(x-5)
\\\\\\
y+5=-10x+50\implies y=-10x+45$

3 0

4 years ago

Show your work to evaluate g(f(x)) if f(x)=x+2/2 and g(x)=1x−1

uranmaximum [27]

Answer:

x/2.

Step-by-step explanation:

We replace the x in g(x) by f(x):

g(f(x)) = 1((x + 2/2) - 1

= (x + 2)/2 - 1

= (x + 2 - 2)/2

= x/2.