How is the product of 4 and 2 shown using integer tiles?

Samples of a cast aluminum part are classified on the basis of surface finish (in microinches) and edge finish. The results of 1

liubo4ka [24]

Answer:

a) P (A)= 88/102= 0.8627

(b) P (B)= 89/102= 0.8725

c) P (A`) = 14/102 = 0.1372

(d) P (A∩B) = 84/102 =0.8235

(e) P(AUB)= 93/102 = 0.9117

(f)P (A`UB) = 98/102= 0.96078

Step-by-step explanation:

edge

finish excellent good Total

<u>(A )surface finish excellent 84 4 88</u>

good ( B) ⇵ 5 9 14

Total 89 13 102

a) Upper P left-parenthesis Upper A right-parenthesis equals

P (A)= 88/102= 0.8627

All the elements of set A = 84+4= 88

(b) Upper P left-parenthesis Upper B right-parenthesis equals

P (B)= 89/102= 0.8725

All the elements of set B = 84+5= 89

c) Upper P left-parenthesis Upper A prime right-parenthesis equal

P (A`) = 14/102 = 0.1372

All the elements of Universal set U which are not elements of set A = 102- 88= 14

(d) Upper P left-parenthesis Upper A intersection Upper B right-parenthesis equals

P (A∩B) = 84/102 =0.8235

Only those elements of set A and set B which are common

(e) Upper P left-parenthesis Upper A union Upper B right-parenthesis equals

P(AUB)= 93/102 = 0.9117

Totalling elements of set A and B= 88+5= 93

(f) Upper P left-parenthesis Upper A prime union Upper B right-parenthesis equals

P (A`UB) = 98/102= 0.96078

All the elements of Universal set U which are not elements of set A and the elements of Set B = 5+9+ 84= 98

6 0

3 years ago

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to <em>0.002652</em> for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0

3 years ago

If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .

The area (in square inches) of the circle that the hexagon did not cover would be:

$\begin{aligned}&\text{Area of circle} - \text{Area of hexagon} \\ &= 100\pi - 150\sqrt{3}\end{aligned}$ .

3 0

3 years ago

-4/5 to the power of 2 times -3/50

Crazy boy [7]

=(-4/5)^2 * -3/50
square -4/5 first; remember -4/5^2 is the same as -4/5 * -4/5

=(-4/5 * -4/5) * -3/50
multiply -4 numerators; multiply 5 denominators

=(-4 * -4)/(5 * 5) * -3/50

=16/25 * -3/50
multiply numerators 16 & -3; multiply denominators 25 & 50

=(16 * -3)/(25 * 50)
= -48/1250

simplify by 2
= -24/625 (or -0.0384)

ANSWER: -24/625 (or -0.0384)

Hope this helps! :)

6 0

3 years ago

Read 2 more answers

8.4×0.5 showing work

Katen [24]

Can u add a picture?

8 0

3 years ago