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2 years ago

Can someone solve this

Mathematics

1 answer:

marysya [2.9K]2 years ago

4 0

From the Arithmetic Information given, Pr = 0.049375 ≠ -8.050625 ≠ 0.0325 See explanation below.

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What are the step by steps solution to the questions above?</h3>

First, lets us restate the question properly. We have

Pr = 1 - ((6*12+6)/80)² = 1 - ((9*12²)/1600) - ((3*12*6)/1600) - (6²/6400) = 1 - 0.005625 * 12² - 0.001875 * 12 * 6 - 0.00015625 * 6²

Note that there are three equals signs. So lets divide the problem according and solve for the different parts.

Lets 1 - ((6*12+6)/80)² ............A

1 - ((9*12²)/1600) - ((3*12*6)/1600) - (6²/6400) .............B; and

1 - 0.005625 * 12² - 0.001875 * 12 * 6 - 0.00015625 * 6² ........C

Solving for A we have

1 - (78/80)²

= 1 - (0.975)²

= 1 - 0.950625

A = 0.049375

Solving for B we have

1 - ((9*12²)/1600) - ((3*12*6)/1600) - (6²/6400)

= 1- (14,256/1600) - (216/1600) - (36/6400)
= 1 - 8.91 - 0.135 - 0.005625

B = -8.050625

Solving for C we have

1 - 0.005625 * 12² - 0.001875 * 12 * 6 - 0.00015625 * 6²

= 1 - 0.005625 * 144 - 0.001875 * 12 * 6 - 0.00015625 * 144

= 0.0325

In summary we can state that:

A = 0.049375

B = -8.050625

C = 0.0325

Given that there were no abstract quantities, we can state that

Pr = A ≠ B ≠C or

Pr = 0.049375 ≠ -8.050625 ≠ 0.0325

Learn more about equations with equal signs at

brainly.com/question/18924248
#SPJ1

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Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

$\sin{A} = \sin{103\textdegree{}}$ ,
The opposite side of angle A $a = BC = 26$ ,
The angle C is to be found, and
The length of the side opposite to angle C $c = AB = 6$ .

$\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}$ .

$\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}$ .

$\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}$ .

Note that the inverse sine function here $\sin^{-1}()$ is also known as arcsin.

By the law of cosine,

$c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C}$ ,

where

$a$ , $b$ , and $c$ are the lengths of sides of triangle ABC, and
$\cos{C}$ is the cosine of angle C.

For triangle ABC:

$b = 21$ ,
$c = 30$ ,
The length of $a$ (segment BC) is to be found, and
The cosine of angle A is $\cos{123\textdegree}$ .

Therefore, replace C in the equation with A, and the law of cosine will become:

$a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}$ .

$\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}$ .

For triangle ABC:

$a = 14$ ,
$b = 9$ ,
$c = 6$ , and
Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

$b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}$ .

$\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}$ .

$\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree$ .