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3 years ago

Assume that the probability of a driver getting into an accident is 8.6%, the

Mathematics

1 answer:

allsm [11]3 years ago

3 0

- The answer is option B. The insurance company must request a premium of $ 1095.16

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ABCD is a trapezoid with AD = 5 ft, DC = 7 ft, and AB = 15 ft. Find the area of the trapezoid. 30 ft2 33 ft2 44 ft2 55 ft2

Nataly_w [17]

Answer: 33 ft².

Step-by-step explanation:

1. As you can see in the figure attached, the trapezoid is formed by a rectangle and two equal right triangles. The base of each triangle is 4 feet and you know the lenght of the hypotenuse AD, therefore, you can calculate the height by applying the Pythagorean Theorem:

$h=\sqrt{(5ft)^{2}-(4ft)^{2}}=3ft$

2. Now, you can calculate the area with the following formula:

$A=(\frac{B+b}{2})h$

Where B is the longer base, b is the shorter base and h is the height.

Then:

$A=(\frac{15ft+7ft}{2})(3ft)=33ft^{2}$

8 0

4 years ago

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What statement best describes the solution of the system of the equations shown 2x-y=1

Aliun [14]

I’m pretty sure it’s B because you can put in any number and it could still have a difference of 1

8 0

3 years ago

(5x^5) - (80x^3) factoring

elena55 [62]

Answer:

Factoring the term $(5x^5) - (80x^3)$ we get $\mathbf{5x^3(x-4)(x+4)}$

Step-by-step explanation:

We need to factor the term $(5x^5) - (80x^3)$

First we can see that $5x^3$ is common in both terms

So, taking $5x^3$ common:

$(5x^5) - (80x^3)\\=5x^3(x^2-16)$

We can write $x^2-16$ as $(x)^2-(4)^2$

$=5x^3((x)^2-(4)^2)$

Now we can solve $(x)^2-(4)^2$ using the formula: $a^2-b^2=(a+b)(a-b)$

We can write:

$=5x^3((x)^2-(4)^2)\\=5x^3(x-4)(x+4)$

So, factoring the term $(5x^5) - (80x^3)$ we get $\mathbf{5x^3(x-4)(x+4)}$

4 0

3 years ago

All boxes with a square base, an open top, and a volume of 60 ft cubed have a surface area given by S(x)equalsx squared plus

Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval $(0,\infty)$ is $S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

The dimensions of the box with minimum surface area are: the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

Step-by-step explanation:

We are given the surface area of a box $S(x)=x^2+\frac{240}{x}$ where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval $(0,\infty)$ and the dimensions of the box with minimum surface area.

1. To find the absolute minimum you must find the derivative of the surface area ( $S'(x)$ ) and find the critical points of the derivative ( $S'(x)=0$ ).

$\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}$

Next,

$2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120$

There is a undefined solution $x=0$ and a real solution $x=2\sqrt[3]{15}$ . These point divide the number line into two intervals $(0,2\sqrt[3]{15})$ and $(2\sqrt[3]{15}, \infty)$

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

$\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}$

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before $x=2\sqrt[3]{15}$ , increases after it, and is defined at $x=2\sqrt[3]{15}$ . So f(x) has a relative minimum point at $x=2\sqrt[3]{15}$ .

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for $x=2\sqrt[3]{15}$ .

$\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}$

and for $x=2\sqrt[3]{15}$ we get:

$2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0$

Therefore S(x) has a minimum at $x=2\sqrt[3]{15}$ which is:

$S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

2. To find the third dimension of the box with minimum surface area:

We know that the volume is 60 $ft^3$ and the volume of a box with a square base is $V=x^2h$ , we solve for h

$h=\frac{V}{x^2}$

Substituting V = 60 $ft^3$ and $x=2\sqrt[3]{15}$

$h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft$

The dimension are the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

6 0

3 years ago

Craig started the summer with 10 freckles. The number of freckles, F, has doubled each week, W, since summer began. Which equati

mixer [17]

Answer:

The answer to your question is letter D

Step-by-step explanation:

Letter A. It is not possible because it is necessary to considers the variable (number of weeks)and this option does not consider it.

Letter B. It is not possible besides it consider the number of weeks the results do not correspond with the conditions given.

Example: for week 1, the number of freckles is 10 + 2¹ = 10 + 2 = 12, and the number of freckles must be 20.

Letter C. It is incorrect because the numbers of freckles do not correspond with the conditions.

Example: week 2 f = 2(10) ² = 2(100) = 200, it is higher than the expected.

Letter D. This is the correct answer because it satisfied the conditions given.

4 0

3 years ago

Read 2 more answers