The expression sin 80 cos 70 + cos 80 sin 70 is equivalent to

Five applicants are to be scheduled for job interviews at five given times. On an application form, each applicant was asked to

dlinn [17]

Answer:

0.01024

Step-by-step explanation:

Assume marking was done at random : Hence, each of the 5 time slots have equal Chamves of being marked ;

Number of time slots, n = 5

Required to mark, number of preferred timeslot x = 2

P(x) = 2 /5 = 0.4

Probability of 0.4 that an interviewee gets one of his preferred timeslot.

Probability that each of the 5 interviewees gets one of their preferred time slots :

Using the multiplication rule of independence :

0.4 * 0.4 * 0.4 * 0.4 * 0.4 = 0.4^5 = 0.01024

0.01024 * 100% = 1.024%

6 0

3 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

7 0

3 years ago

I ONLY GOT 1 MIN!!!! PER QUESTION HELP

belka [17]

Answer:

your answer is

<h2>a traepoizd </h2>

Step-by-step explanation:

reason ----

1 parallelogram 

it is opposite side quadrilateral
opposite angle

2 rhombus 

four congruence side in rhombus

3 traepoizd 

one pair of congruent sides that were not parallel. It also had one pair of parallel lines.

4 kite 

two pair of quadrilateral adjacent

<h2>hope it's helpful thanks please mark me as brainlist please </h2>

4 0

2 years ago

Read 2 more answers

What is the mixed number for 8/2

deff fn [24]

Simply 8. 8 fits into 2 4 times.

5 0

2 years ago

Read 2 more answers

i love how this website was created for people to get help with work and ask questions about it but instead people ask stupid qu

Luden [163]

Answer:

mk, eee

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers