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jekas [21]
3 years ago
10

Suppose Celine wants to choose a box that maximizes the amount of cereal it can hold. Volume of Box 1: 3x5 Volume of Box 2: 4x5

– x4 If Celine decides the width of the cereal boxes will be greater than 1, which box will hold more cereal? Explain.
Mathematics
2 answers:
tekilochka [14]3 years ago
5 0

Answer:

If a box has a greater volume, it can hold more cereal.

Box 2 can hold more cereal. Although the value of x is unknown, it represents a length, so it must be greater than zero. For any x > 1, the volume of Box 2 is greater than the volume of Box 1.

Step-by-step explanation:

finlep [7]3 years ago
3 0

The <em><u>correct answer</u></em> is:

Box 2

Explanation:

In these expressions, x represents the width of the box.  We are asked about widths greater than 1; let's us 1.1 as an example:

Box 1:

3(1.1⁵) = 4.83153

Box 2:

4(1.1⁵)-(1.1⁴) = 6.44204 - 1.4641 = 4.97794

With a difference as small as a tenth, the volume of box 2 is larger.  The larger the difference in width, the larger the difference in volume, and the volume of box 2 will be larger every time.

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How can I solve this?<br>Please use the simplest format​
IgorLugansk [536]

Answer:

option D is true.

Step-by-step explanation:

The right-angled triangle is shown.

From the right-angled triangle,

The angle Ф = 60°

We know that the trigonometric ratio

tan Ф = opposite / adjacent

  • opposite = 4
  • adjacent = n

Thus,

tan 60 = 4 / n

√3 = 4/n

n = 4/√3

Thus,

n = 4/√3

  = (4 × √3) / (√3 × √3)

  = 4√3 / 3

Thus,

n  = 4√3 / 3

Using Pythagorean theorem

m = √n²+4²

m=\sqrt{\left(4\cdot \frac{\sqrt{3}}{3}\right)^2+4^2}

m=\sqrt{\frac{4^2}{3}+4^2}

m=\sqrt{\frac{64}{3}}

m=\frac{8}{\sqrt{3}}

m=\frac{8\sqrt{3}}{3}

Thus,

  • m=\frac{8\sqrt{3}}{3}
  • n  = 4√3 / 3

Therefore, option D is true.

3 0
2 years ago
Match this system of linear equations to the correct description of its solution set.
Misha Larkins [42]

Answer:

Infinite number of solutions.

Step-by-step explanation:

Are the equations written correctly?  They are identical.  Since they overlap, every point is a solution for as long as the lines reach.  Since no one is stopping them, the answers are infinite.

8 0
2 years ago
The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =
stich3 [128]

I'm assuming \alpha is the shape parameter and \beta is the scale parameter. Then the PDF is

f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}

a. The expectation is

E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx

To compute this integral, recall the definition of the Gamma function,

\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt

For this particular integral, first integrate by parts, taking

u=x\implies\mathrm du=\mathrm dx

\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}

E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x

E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx

Substitute x=3y^{1/2}, so that \mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy:

E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy

\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}

The variance is

\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2

The second moment is

E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx

Integrate by parts, taking

u=x^2\implies\mathrm du=2x\,\mathrm dx

\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}

E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx

E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx

Substitute x=3y^{1/2} again to get

E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9

Then the variance is

\mathrm{Var}[X]=9-E[X]^2

\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}

b. The probability that X\le3 is

P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx

which can be handled with the same substitution used in part (a). We get

\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}

c. Same procedure as in (b). We have

P(1\le X\le3)=P(X\le3)-P(X\le1)

and

P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}

Then

\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}

7 0
2 years ago
The coordinates of the vertices of the triangle are (–8, 8), (–8, –4), and . Consider QR the base of the triangle. The measure o
Mademuasel [1]

The coordinates of the vertices of the triangle are
(–8, 8), (–8, –4), and<span> (10, –4)</span>.

Consider QR the base of the triangle. The measure of the base is b = 18 units, and the measure of the height is h = <span>12</span> units.

The area of triangle PQR is<span>108</span> square units.

4 0
3 years ago
Read 2 more answers
A. Suppose you have just inherited a sum of money. Let’s say you choose to invest the money. How much did you inherit? Choose an
leva [86]

Amount =$48003.20

Step-by-step explanation:

Here apply the compound interest formula;

A=P(1+\frac{r}{n} )^{nt}

where;

P=Principal amount invested = $37500

r=rate of interest as a decimal, 2.5% =0.025

n=number of compounding per year=1

t=time period the amount in invested=10

In our case, the amount after investing will be;

A=37,500(1+0.025)^{10} \\\\A=37,500(1.025)^{10} \\\\A=48003.20

Interest earned after the period= $48003.20-$37500 =$10503.20

Learn More

Compound Interest formula application: brainly.com/question/7014337

Keywords: inherit, sum, money, interest

#LearnwithBrainly

3 0
3 years ago
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