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3241004551 [841]

3 years ago

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The results of a series of surveys revealed a population with a mean of 4.73 and a standard deviation of 0.865. If each survey h

as a sample size of 200, which value falls within the interval where 95% of the sample means occur? A. 4.88 B. 4.63 C. 4.91 D. 4.55

2 answers:

nika2105 [10]3 years ago

7 0

Answer: for plato users

= 4.63

Step-by-step explanation:

creativ13 [48]3 years ago

6 0

Answer:

4.63

Please give me brainliest, I really need it.

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HELP ASAP PLZ 39 POINTS<br> Find x, y, u, v

Ad libitum [116K]

Answer:

x = 9; y = 12; u = 24; v = 32

Step-by-step explanation:

The corresponding sides of similar triangles are in the same ratio to each other.

3/5 = (3 + x)/20 Multiply each side by 20

20×3/5 = 3 + x

12 = 3 + x Subtract 3 from each side

x = 9

=====

4/5 = (4 + y)/20 Multiply each side by 20

20×4/5 = 4 + y

16 = 4 + y Subtract 4 from each side

y = 12

=====

3/5 = (3 + x + u)/60 Multiply each side by 60

60×3/5 = 3 + x + u

36 = 3 + x + u Insert the value of x

36 = 3 + 9 + u

36 = 12 + u Subtract 12 from each side

u = 24

=====

4/5 = (4 + y + v)/60 Multiply each side by 60

60×4/5 = 4 + y + v

48 = 4 + y + v Insert the value of y

48 = 4 + 12 + v

48 = 16 + v Subtract 16 from each side

v = 32

x = 9; y = 12; u = 24; v = 32

5 0

3 years ago

The publisher of a recently released nonfiction book expects that over the first 20 months after its release, the monthly profit

wolverine [178]

Answer:

(a) $\frac{dP}{dt}=\frac{4800-1600t-240t^2}{(t^2+20)^2}$

(b)P'(5)=-($4.54) Thousand

(c)P'(11)=-($2.10) Thousand

(d)The fifth Month

Step-by-step explanation:

Given the monthly profit model:

$P(t)=\frac{240t-40t^2}{t^2+20}$

(a)We want to derive a model that gives the Marginal Profit, P' of the book.

We differentiate

$P(t)=\frac{240t-40t^2}{t^2+20}$ using quotient rule.

$\frac{dP}{dt}=\frac{(t^2+20)(240-80t)-(240t-40t^2)(2t)}{(t^2+20)^2}$

Simplifying

$\frac{dP}{dt}=\frac{4800-1600t-240t^2}{(t^2+20)^2}$

We have derived a model for the marginal profit.

(b) After 5 months, at t=5

Marginal Profit=P'(5)

$\frac{dP}{dt}=\frac{4800-1600t-240t^2}{(t^2+20)^2}$

$P^{'}(5)=\frac{4800-1600(5)-240(5)^2}{(5^2+20)^2}$

=-($4.54) Thousand of dollars

(c)Marginal Profit 11 Months after book release

$P^{'}(11)=\frac{4800-1600(11)-240(11)^2}{(11^2+20)^2}$

=-($2.10) Thousand of dollars

(d) Since the marginal profit at t=5 is negative, after the 5th Month, the profit starts to experience a steady decrease.

6 0

3 years ago

Ari has 10 boards that are each 7.8 ft long. He plans to cut as many pieces as he can from the boards to make some siding for th

Georgia [21]

Answer:

The total length of the board is 78 ft.

The number of pieces of wood for porch cut from the boards be 34 .

Total length of left 10 pieces that were left from each board after he cut the 2.25 ft long boards be 10.5 ft .

Step-by-step explanation:

Case (a)

As given

Ari has 10 boards that are each 7.8 ft long.

Total length of board = Total number of board × Length of each board

Putting the values in the above

Total length of board = 10 × 7.8

= 78 ft

Therefore the total length of the board is 78 ft.

Case (b)

As given

Each piece of wood for the porch has to be 2.25 ft long.

Total length of the 10 board = 78 ft

Let us assume that the number of piece of wood for porch cut from board be x.

Than the equation becomes

2.25 × x = 78

$x = \frac{78}{2.25}$

x = 34.7 (Approx)

Thus

x = 34

Therefore the number of pieces of wood for porch cut from the boards be 34 .

Case (c)

As given

As the length of the one board be 7.8 ft and the length of the each piece of wood for porch is 2.25 ft .

Let us assume that the number of piece of wood for porch cut from one board be y .

Than the equation becomes

y × 2.25 = 7.8

$y = \frac{7.8}{2.25}$

y = 3.5 (Approx)

y = 3

Thus the number of piece of wood for porch cut from one board be 3 .

Than

Length of piece left after cutting 2.25 ft piece of wood from each board = Length of each boards - Number of pieces of wood × Length of each pieces of wood.

Length of piece left after cutting 2.25 ft piece of wood from each board = 7.8 - 3 × 2.25

= 7.8 - 6.75

= 1.05 ft

Thus

Total length of left 10 pieces = 10 × Length of piece left after cutting 2.25 ft piece of wood from each board

= 10 × 1.05

= 10.5 ft

Therefore total length of left 10 pieces that were left from each board after he cut the 2.25 ft long boards be 10.5 ft .

5 0

3 years ago

A ship sails 250km due North qnd then 150km on a bearing of 075°.1)How far North is the ship now? 2)How far East is the ship now

olga_2 [115]

Answer:

1) 288.8 km due North

2) 144.9 km due East

3) 323.1 km

4) 207°

Step-by-step explanation:

<u>Bearing</u>: The angle (in degrees) measured clockwise from north.

<u>Trigonometric ratios</u>

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

<u>Cosine rule</u>

$c^2=a^2+b^2-2ab \cos C$

where a, b and c are the sides and C is the angle opposite side c

-----------------------------------------------------------------------------------------------

Draw a diagram using the given information (see attached).

Create a right triangle (blue on attached diagram).

This right triangle can be used to calculate the additional vertical and horizontal distance the ship sailed after sailing north for 250 km.

<u>Question 1</u>

To find how far North the ship is now, find the measure of the short leg of the right triangle (labelled y on the attached diagram):

$\implies \sf \cos(75^{\circ})=\dfrac{y}{150}$

$\implies \sf y=150\cos(75^{\circ})$

$\implies \sf y=38.92285677$

Then add it to the first portion of the journey:

⇒ 250 + 38.92285677... = 288.8 km

Therefore, the ship is now 288.8 km due North.

<u>Question 2</u>

To find how far East the ship is now, find the measure of the long leg of the right triangle (labelled x on the attached diagram):

$\implies \sf \sin(75^{\circ})=\dfrac{x}{150}$

$\implies \sf x=150\sin(75^{\circ})$

$\implies \sf x=144.8888739$

Therefore, the ship is now 144.9 km due East.

<u>Question 3</u>

To find how far the ship is from its starting point (labelled in red as d on the attached diagram), use the cosine rule:

$\sf \implies d^2=250^2+150^2-2(250)(150) \cos (180-75)$

$\implies \sf d=\sqrt{250^2+150^2-2(250)(150) \cos (180-75)}$

$\implies \sf d=323.1275729$

Therefore, the ship is 323.1 km from its starting point.

<u>Question 4</u>

To find the bearing that the ship is now from its original position, find the angle labelled green on the attached diagram.

Use the answers from part 1 and 2 to find the angle that needs to be added to 180°:

$\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{Total\:Eastern\:distance}{Total\:Northern\:distance}\right)$

$\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{150\sin(75^{\circ})}{250+150\cos(75^{\circ})}\right)$

$\implies \sf Bearing=180^{\circ}+26.64077...^{\circ}$

$\implies \sf Bearing=207^{\circ}$

Therefore, as bearings are usually given as a three-figure bearings, the bearing of the ship from its original position is 207°

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2 years ago

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Which shows how the distributive property can be used to evaluate 7 times 8 and four-fifths?

ICE Princess25 [194]

Answer:

2 + 2 = fish

Step-by-step explanation:

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