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3 years ago

In a bag there are 40 pieces of candy: 7

Mathematics

2 answers:

Sonja [21]3 years ago

8 0

The answer is 13/40

Slav-nsk [51]3 years ago

3 0

Answer:

37.5

Step-by-step explanation:

because if you calculte percent = to 37.5 that is why it sis the correct answer

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Consider the function V=g(x), where g(x) =x(6-2x)(8-2x), with x being the length of a cutout in cm and V being the volume of an

Andrej [43]

Answer:

The maximum volume of the open box is 24.26 cm³

Step-by-step explanation:

The volume of the box is given as $V=g(x)$ , where $g(x)=x(6-2x)(8-2x)$ and $0\le x\le3$ .

Expand the function to obtain:

$g(x)=4x^3-28x^2+48x$

Differentiate wrt x to obtain:

$g'(x)=12x^2-56x+48$

To find the point where the maximum value occurs, we solve

$g'(x)=0$

$\implies 12x^2-56x+48=0$

$\implies x=1.13,x=3.54$

Discard x=3.54 because it is not within the given domain.

Apply the second derivative test to confirm the maximum critical point.

$g''(x)=24x-56$ , $g''(1.13)=24(1.13)-56=-28.88\:$

This means the maximum volume occurs at $x=1.13$ .

Substitute $x=1.13$ into $g(x)=x(6-2x)(8-2x)$ to get the maximum volume.

$g(1.13)=1.13(6-2\times1.13)(8-2\times1.13)=24.26$

The maximum volume of the open box is 24.26 cm³

See attachment for graph.

6 0

3 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

vredina [299]

Answer:

2.35%

Step-by-step explanation:

Mean number of months (M) = 39 months

Standard deviation (S) = 10 months

According to the 68-95-99.7 rule, 95% of the data is comprised within two standard deviations of the mean (39-20 to 39+20 months), while 99.7% of the data is comprised within two standard deviations of the mean (39-30 to 39+30 months).

Therefore, the percentage of cars still in service from 59 to 69 months is:

$P_{59\ to\ 69}=\frac{P_{9\ to\ 69}-P_{19\ to\ 59}}{2} \\P_{59\ to\ 69}=\frac{99.7-95}{2}\\P_{59\ to\ 69}=2.35\%$

The approximate percentage of cars that remain in service between 59 and 69 months is 2.35%.

8 0

2 years ago

How do three planes intersect at one point?

natima [27]

Imagine a corner of a cube.
It is made of three planes interesting at a single point
The image should help.

7 0

3 years ago

Suppose f (x) = x^2. Find the graph of f(x+2)

NISA [10]

$\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)={{ A}}({{ B}}x+{{ C}})+{{ D}}
\\\\
~~~~y={{ A}}({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)={{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\\\
f(x)={{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\\\
f(x)={{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\\\\
--------------------$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
~~~~~~\textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
~~~~~~if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
~~~~~~if\ {{ D}}\textit{ is negative, downwards}\\\\
~~~~~~if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

with that template in mind,

$\bf f(x)=x^2\qquad \quad f(x+2)=(x+2)^2\implies f(x+2)=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+2})^2\stackrel{D}{+0}$

C = 2 B = 1 C/B = 2/1 or +2, horizontal left shift of 2 units

f(x) shifted left by 2 units is f(x+2).

8 0

2 years ago

Consider triangle ABC where AB=X+5, BC=X-2, area is 30cm squared. Find X by solving X^2+3X-70=0, and find the perimeter.

klasskru [66]

Answer:

What the hell AC is not given so you can't find the Perimeter

Step-by-step explanation:

x^2+3x-70=0

(x-7)(x+10)=0

x-7=0

x+10=0

x=-10

AB=x+5

AB=12

AB=-5

BC=x-2

BC=5

BC=-12

Area of ABC = 30cm2

Perimeter of ABC = AB+BC+AC

8 0

2 years ago