The shape of a garden is rectangular at the center and semicircular at the ends. Find the area and perimeter of this garden { length of the rectangle is 20 - (3.5+3.5) meters} The First, correct answer gets BRAINLIEST

Answers

Answer 1

Mensuration:

Mensuration is the branch of mathematics which concerns itself with the measurement of Lengths, areas & volume of different geometrical shapes or figures.

Plane Figure: A figure which lies in a plane is called a plane figure.

For e.g:  a rectangle, square, a rhombus, a parallelogram, a trapezium.

 

Perimeter:

The perimeter of a closed plane figure is the total length of its boundary.

In case of a triangle or a polygon the perimeter is  the sum of the length of its sides.

Unit of perimeter is a centimetre  (cm), metre(m) kilometre(km) e.t.c

Area: The area of the plane figure is the measure of the surface enclose by its boundary.

 

The area of a triangle are a polygon is the measure of the surface enclosed by its sides.

A square centimetre (cm²) is generally taken at the standard unit of an area. We use square metre (m²) also for the units of area.

Circumference of a circle is the perimeter of a circle.

In a circle the radius is half of the diameter.

The approximate value of π( Pi) is= 22/7 

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The Shape Of A Garden Is Rectangular At The Center And Semicircular At The Ends. Find The Area And Perimeter

Related Questions

A certain article indicates that in a sample of 1,000 dog owners, 680 said that they take more pictures of their dog than of their significant others or friends, and 490 said that they are more likely to complain to their dog than to a friend. Suppose that it is reasonable to consider this sample as representative of the population of dog owners.
(a) Construct a 90% confidence interval for the proportion of dog owners who take more pictures of their dog than of their significant others or friends. (Use a table or technology. Round your answers to three decimal places.)
(______),(_________)
(b) Construct a 95% confidence interval for the proportion of dog owners who are more likely to complain to their dog than to a friend. (Use a table or technology. Round your answers to three decimal places.)
(_______),(_______)

Answers

Answer: a) a 90% confidence interval for the proportion of dog owners who take more pictures of their dog than of their significant others or friends is (0.656,0.704).

b) a 95% confidence interval for the proportion of dog owners who are more likely to complain to their dog than to a friend is (0.459,0.521).

Step-by-step explanation:

Confidence interval for a population proportion is given by:-

[tex]\hat{p}\pm z(\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})[/tex]

, where [tex]\hat{p}[/tex] = sample proportion , n= sample size, z= critical z-value.

As per given,

a) n=1000

Sample proportion of dog owners say they take more pictures of their dog than of their significant others or friends =[tex]\hat{p}=\dfrac{680}{1000}=0.68[/tex]

critical value for 90% confidence = 1.645  [By table]

A 90% confidence interval for the proportion of dog owners who take more pictures of their dog than of their significant others or friends.

[tex]0.68\pm (1.645)\sqrt{\dfrac{0.68(1-0.68)}{1000}}\\\\\approx0.68\pm0.024=(0.68-0.024,0.68+0.024)=(0.656,0.704)[/tex]

Hence, a 90% confidence interval for the proportion of dog owners who take more pictures of their dog than of their significant others or friends is (0.656,0.704).

b) Sample proportion of dog owners say they are more likely to complain to their dog than to a friend =[tex]\hat{p}=\dfrac{490}{1000}=0.49[/tex]

critical value for 95% confidence = 1.96  [By table]

A 95% confidence interval for the proportion of dog owners who are more likely to complain to their dog than to a friend:

[tex]0.49\pm (1.96)\sqrt{\dfrac{0.49(1-0.49)}{1000}}\\\\\approx0.49\pm0.031=(0.49-0.031,0.49+0.031)=(0.459,0.521)[/tex]

Hence, a 95% confidence interval for the proportion of dog owners who are more likely to complain to their dog than to a friend is (0.459,0.521).

On a coordinate plane, a line is drawn from point J to point K. Point J is at (negative 3, 1) and point K is at (negative 8, 11). What is the y-coordinate of the point that divides the directed line segment from J to K into a ratio of 2:3? y = (StartFraction m Over m + n EndFraction) (y 2 minus y 1) + y 1 –6 –5 5 7

Answers

Answer:   (-5, 5)

Step-by-step explanation:

J = (-3, 1)    K = (-8, 11)          ratio 2 : 3    --> 2 + 3  = 5 segments

x-distance from J to K: -8 - (-3) = -5 units

y-distance from J to K: 11 - 1 = 10 units

Divide those distance into 5 segments:

x = -5/5  = -1 unit per segment

y = 10/5 = 2 units per segment

The partition is 2 segments from J:

x = -3 +2(-1)  = -5

y = 1 + 2(2)   = 5

The partition is located at (-5, 5)

Answer:

5

Step-by-step explanation:

Please help ASAP!!! Thank you so much!!! Just want confirm my answer and it is A) -5. Consider the function f(x)= [tex]x^{3}[/tex] +4[tex]x^{2}[/tex]-25x-50. (x^3+4x^2-25x-50). If there is a remainder of 50 when the function is divided by (x-a) what is the value of a? A) -5 B) -2 C)-3 D)

Answers

Answer:

Correct is A

Step-by-step explanation:

Correct is A , notice that when you perform synthetic division you get that the reminder is 50, please refer to the image I attach to see it.

i will give brainliest and 50 points pls help ASP

pls can u show how to work this out thx !!! : )​
its a simultaneous equation

Answers

Answer:

x = 3

y = -2

Step-by-step explanation:

Given:

8x - 3y = 30 ..................(1)

3x + y = 7 .......................(2)

Eliminate y by adding (1)+3*(2)

8x-3y + 3*( 3x+y) = 30 + 3*7

8x + 9x -3y + 3y = 51

17x = 51

x = 3  .....................(3)

substitute (3) in (2)

3(3) + y = 7

y = 7-9 = -2

y = -2 ....................(4)

Answer:

Step-by-step explanation:

[tex]8x-3y=30\\y=7-3x\\\\8x-3(7-3x)=30\\8x-21+9x=30\\17x=51\\x=3\\\\y=7-3(3)\\y=7-9\\y=-2[/tex]

Write the equation of the line in slope intercept form that passes through the points (4,-2) and (2,-1)

Answers

Answer:

y + 2 = (-1/2)(x - 4)

Step-by-step explanation:

Let's move from (2, -1) to (4, -2) and measure the changes in x and y.  x increases by 2 units from 2 to 4, and y decreases by 1 unit from -1 to -2.  Thus, the slope of the line connecting the two points is m = rise / run =

-1

--- = (-1/2).

2

Using the point-slope formula, we get:

y + 2 = (-1/2)(x - 4)

stephano walks 2/5 mile in 1/4 hour. What is stephano's speed in miles?

Answers

Answer: See below

Explanation:

V = d/t
V = 2/5 divide by 1/4
V = 2/5 x 4/1
V = 8/5 mi/hr

What are some key words used to note addition operations?

Answers

Answer:

The correct answer is

For addition, Caulleen used the words total, sum, altogether, and increase. But we could also have used the words combine, plus, more than, or even just the word "and". For subtraction, Caulleen used the words, fewer than, decrease, take away, and subtract. We also could have used less than, minus, and difference.

Step-by-step explanation:

hope this helps u!!!

Question 5(Multiple Choice Worth 4 points) (05.05)Based on the graph, what is the initial value of the linear relationship?

Answers

Answer:  A) -2

Step-by-step explanation:

The y-intercept (where the graph crosses the y-axis) is the "initial value".

Looking at the given graph, it crosses the y-axis when y = -2

                                                                                                   

HELP number 12 pls i do nor have long more​

Answers

Answer:

Dian has $250 originally.

Step-by-step explanation:

Let the total money Dian has originally = $S

Dian gave [tex]\frac{2}{5}[/tex] of her total money to Justin,

Money given to Justin = [tex]\frac{2}{5}(\text{S})[/tex]

Money left with Dian = S - [tex]\frac{2}{5}(\text{S})[/tex]

                                   = [tex]\frac{\text{5S-2S}}{5}[/tex]

                                   = [tex]\frac{3S}{5}[/tex]

Since Dian has $150 left then the equation will be,

[tex]\frac{3S}{5}=150[/tex]

S = [tex]\frac{150\times 5}{3}[/tex]

S = $250

Therefore, Dian has $250 originally.

A copy machine makes 153 copies in 4 minutes 15 seconds how many copies does it make per minute

Answers

Answer:

36 copies.

Step-by-step explanation:

4mins 15 seconds is the same as 4+1/4 minsutes. Since 1 minute is less than 4+1/4minutes and 4+1/4 minutes produces 153 copies. 1 minute will produce less.

(1÷4+1/4)×153

= 36 copies.

36

Change the time into seconds so they’ll be in the same unit, (4*60) + 15= 255. Do cross multiplication,
153. 255
? 60
(153*60) / 255= 36

round 235,674 to the nearest thousand​

Answers

Answer:

236,000

Step-by-step explanation:

When rounding to thousands, if the hundred is 500 or over, you round up. If it is less, round down.

Answer:

Hey! The answer is 236,000.... steps will be below!

Step-by-step explanation:

Since there is 5,674 it is more closer to 6,000 than 5,000

So we change the number to 236,000

ANSWER TO YOUR QUESTION: 236,000

Hope this helps! :)

⭐️Have a wonderful day!⭐️

Select the correct answer. Sarah wants to print copies of her artwork. At the local print shop, it costs her $1 to make 5 copies and $5 to make 25 copies. How much would it cost Sarah to make 100 copies? A. $15 B. $20 C. $25 D. $30

Answers

$1 = 5copies means

$5 = 25 copies obviously

then

$x = 100 copies

100 / 5 = $x

so she needs

$20

The cost of printing 100 copies of artwork is $20.

What is a unitary method?

A unitary method is a mathematical way of obtaining the value of a single unit and then deriving any no. of given units by multiplying it with the single unit.

Suppose 2 pens cost $10, So the cost of 1 pen is (10/2) = $5.

From this unitary cost of pens, we can determine the cost of any no. of pens by multiplying the unit cost by the no. of pens.

Given, The cost of printing 5 artworks is $1.

∴ The cost of printing 100 copies is $(100/5),

= $20.

learn more about the unitary method here :

https://brainly.com/question/22056199

#SPJ2

a beetle sets out on a journey. on the first day it crawls 1m in a straight line. on the second day it makes a right-angled turn (in either direction) and crawls 2m in a straight line. on the third day it makes a right-angled turn (in either direction) and crawls 3m in a straight line. this continues each day with the beetle making a right-angled turn ( in either direction and crawling 1m further than it did the day before. what is the least number of days before the beetle could find itself stopped at its starting point?

Answers

Answer:

  7

Step-by-step explanation:

The signed sum of sequential odd numbers must be zero, as must the signed sum of sequential even numbers.

The minimum number of sequential even numbers that have a sum of 0 is 3: 2+4-6 = 0.

The minimum number of sequential odd numbers with a sum of zero is 4: 1-3-5+7=0.

Since we start with an odd number, we can get these sets of numbers in 7 days. The attached diagram shows one possible route.

The radius of a sphere is 11 cm.what is the surface area of the sphere? Round to the nearest tenth.

Answers

Answer:

The answer is

1520.5 cm²

Step-by-step explanation:

Surface area of a sphere is given by

S = 4πr²

where r is the radius

From the question r = 11cm

So the surface area of the sphere is

4π(11)²

= 4(121)π

= 484π

= 1520.5308

Which is

1520.5 cm² to the nearest tenth

Hope this helps you

Write the equation of the line in point-slope form that passes through (1, -4) and has a slope of 1/4.

Answers

Answer:

y + 4 =(1/4)(x - 1)

Step-by-step explanation:

The point slope equation is y - k = m(x - h), where (h, k) is the given point and m is the given slope.

In this particular case we have y + 4 =(1/4)(x - 1)

A magazine article states that the mean weight of one-year-old boys is the same as that of one-year-old girls. Does the confidence interval contradict this statement? The confidence interval this statement

Answers

Answer:

Yes, the confidence interval contradict this statement.

Step-by-step explanation:

The complete question is attached below.

The data provided is:

[tex]n_{1}=318\\n_{2}=297\\\bar x_{1}=25\\\bar x_{2}=24.1\\s_{1}=3.6\\s_{2}=3.8[/tex]

Since the population standard deviations are not provided, we will use the t-confidence interval,

[tex]CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\cdot s_{p}\cdot\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}[/tex]

Compute the pooled standard deviation as follows:

[tex]s_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}=\sqrt{\frac{(318-1)(3.6)^{2}+(297-1)(3.8)^{2}}{318+297-2}}=2.9723[/tex]

The critical value is:

[tex]t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (318+297-2)}=t_{0.025, 613}=1.962[/tex]

*Use a t-table.

The 95% confidence interval is:

[tex]CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\cdot s_{p}\cdot\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}[/tex]

     [tex]=(25-24.1)\pm 1.962\times 2.9723\times \sqrt{\frac{1}{318}+\frac{1}{297}}\\\\=0.90\pm 0.471\\\\=(0.429, 1.371)\\\\\approx (0.43, 1.37)[/tex]

The 95% confidence interval for the difference between the mean weights is (0.43, 1.37).

To test the magazine's claim the hypothesis can be defined as follows:

H₀: There is no difference between the mean weight of 1-year old boys and girls, i.e. [tex]\mu_{1}-\mu_{2}=0[/tex].

Hₐ: There is a significant difference between the mean weight of 1-year old boys and girls, i.e. [tex]\mu_{1}-\mu_{2}\neq 0[/tex].

Decision rule:

If the confidence interval does not consists of the null value, i.e. 0, the null hypothesis will be rejected.

The 95% confidence interval for the difference between the mean weights does not consists the value 0.

Thus, the null hypothesis will be rejected.

Conclusion:

There is a significant difference between the mean weight of 1-year old boys and 1-year old girls.

How would 2X - 2Y - 6 = 0 be
written in slope-intercept form?

A. 2X - 2Y = 6
B. -2Y = 2X = 6
C. Y=X - 3
D. Y = -2X - 3
E. Y = x + 3

Answers

Answer:

D. Y= -2x - 3

Step-by-step explanation:

The reason the answer is D is because it is the reverse form of standard. D shows that the form of slope-intercept for is y=mx + b.

Answer:

C. Y=X - 3

Step-by-step explanation:

[tex]2X - 2Y - 6 = 0\\\mathbf{y=mx+b}\:\mathrm{is\:the\:slope\:intercept\:form\:of\:a\:line\:where}\: \mathbf{m}\:\mathrm{is\:the\:slope\:and}\:\mathbf{b}\:\mathrm{is\:the}\:\mathbf{y}\:\mathrm{intercept}\\\mathrm{For\:a\:line\:in\:the\:form\:of\:}\mathbf{y=mx+b}\mathrm{,\:the\:slope\:is}\:\mathbf{m}\:\mathrm{and}\:\mathbf{y}\:\mathrm{intercept\:is}\:\mathbf{b}\\\\Y=X-3[/tex]

What is the volume of air that the beach ball will hold?

Answers

Answer:

In order to solve this problem, we need to know the volume of a sphere.

The volume of a sphere is (4/3)pir^3

In this case our r value, or radius, is 30cm.

If we plug in 30cm to the equation like so:

(4/3)pi(30)^3

The answer is 113097 cm^3 of air

Hope you understand





























































































Item 25
The linear function m=45−7.5b represents the amount m (in dollars) of money that you have after buying b books. Select all of the values that are in the domain of the function.

0

1

2

3

4

5

6

7

8

9

10





















































































Item 25
The linear function m=45−7.5b represents the amount m (in dollars) of money that you have after buying b books. Select all of the values that are in the domain of the function.

0

1

2

3

4

5

6

7

8

9

10

Answers

Answer:

5

Step-by-step explanation:

From which sphere of earth did this food did this food originate

Answers

Answer:

I'm not entirely sure what you are asking, could you comment on this answer the full question so I can edit this question to provide you an answer?

Answer: biosphere

Step-by-step explanation:

I am not sure what picture you are looking at but if it is 3 barbeque chicken legs in one image than this is your answer. The reason being that chickens can only be found on land and the land is considered part of the biosphere because bio = life

f(x)=−7x^2 −x and g(x)=9x^2 −4x , find (f−g)(x) and (f−g)(1)

Answers

Answer:

f(x) = -7x² - x

g(x) = 9x² - 4x

To find ( f - g)(x) subtract g(x) from f(x)

That's

( f - g)(x) = - 7x² - x - ( 9x² - 4x)

= -7x² - x - 9x² + 4x

Group like terms

( f - g)(x) = - 7x² - 9x² - x + 4x

( f - g)(x) = - 16x² + 3x

To find (f - g)(1) substitute 1 into ( f - g)(x)

That's

(f - g)(1) = - 16(1) + 3(1)

= -16 + 3

= - 13

Hope this helps you

Write a system of equations in x and y describing the situation. Do not solve the system.
Keiko has a total of $5500, which she has invested in two accounts. The larger account is $700 greater than the smaller account. (Let x be the amount of money in the
larger account and y be the amount of money in the smaller account.)
Choose a system of equations describing the situation,
X+ y + 5500 = 0
x+y=5500
X= y + 700
X-y+ 700 = 0
O
2x = 5500
2x + 2y = 5500
x-y=700
x- y = 700​

Answers

X=y+700
X is larger acct
Y is smaller acct
Larger acct equals amount in smaller account plus $700.

Lacey's mom makes her a birthday cake in the shape of an "L" . Lacey loves frosting, so her mom covers the entire outside of the cake in frosting, even the bottom of the cake. How much space does Lacey's mom cover in frosting? cm2\text{cm}^2cm2start text, c, m, end text, squared

Answers

Answer:

1360cm²

Step-by-step explanation:

Since the shape of the cake is in L shape, we can divide the cake in to rectangles..

The amount of space covered by the frosting = The sum of the areas of the sides that we can find in this L shaped cake diagram.

The sides of this cake, are shaped like a rectangle.

Hence, Area of a Rectangle = Length × Width

a) Side 1 = Rectangle on the left

Area of a Rectangle = Length × Breadth

Length = 30cm

Breadth =10cm

Area = 30 × 10 = 300cm²

Since we have another side with this measurement/ dimensions also,

Side 2 = 300cm²

Side 3 = The front face of the cube by the right

Area of a Rectangle = Length × Breadth

Length = 22cm - 10cm = 12cm

Breadth =10cm

Area = 12 × 10 = 120cm²

Likewise, we have the another side with the same dimensions as well

Hence, Side 4 = 120cm²

Side 5

30 × 5 = 150cm²

Side 6

10 × 5 = 50cm²

Side 7

20 × 5 = 100cm²

Side 8

22cm × 5 cm = 110cm²

Side 9

10cm × 5cm = 50cm²

Side 10

12cm × 5cm = 60cm²

The amount of space covered by the frosting = Area of Sides( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)

= (300 + 300 + 120 + 120 + 150 + 50 + 100 + 110 + 50 + 60) cm²

= 1360cm²

for f(x)=1/x-5 and g(x)=x^2+2 find the expression for g(x) and substitute the value of g(x) into the function in place of x to find the value of f(g(x))

Answers

Answer:

f(x)  = (1/x) - 5

g(x) = x^2 + 2

=> f[g(x)] = [1/(x^2 +2)] - 5

Use identities to find values of the sine and cosine functions of the function for the angle measure
a. theta, given that cos2theta=28/53 and 0theta < theta < 90degrees
b. 2theta, given sin theta= - sqrt 7 over 5 and cos theta > 0
c. 2x, given tan x=2 and cos x<0

Answers

Answer:

Step-by-step explanation:

a) Given cos2theta=28/53 and 0degrees< theta < 90degrees

From cos2theta=28/53

[tex]2\theta = cos^{-1}\frac{28}{53}[/tex]

[tex]2\theta = cos^{-1}0.5283\\ \\2\theta = 58.12\\\\Dividing\ both \ sides\ by \ 2\\\\\frac{2\theta}{2} = \frac{58.12}{2}\\ \\\theta = 29.06^0[/tex]

b) Given

[tex]sin\theta = \frac{-\sqrt{7} }{5} \\\\\theta = sin^{-1} \frac{-\sqrt{7} }{5}\\\\\\\theta = sin^{-1} \frac{-2.6458}{5}\\\\\theta = sin^{-1} -0.5292\\\\\theta = -31.95^0[/tex]

If cos theta [tex]\gneq[/tex] 0, this means we need to look for the quadrant where sin is negative and cos is positive. That will be the fourth quadrant. In the fourth quadrant, theta = 360 - 31.95° = 328.05°

2theta = 2 * 328.05

2theta = 656.1°

c) Given tan x=2 and cos x<0, lets find the angle of x first.

If tan x = 2

x = tan^-1 2

x = 63.4°

Sine cos is less than 0, then we need to find the angle of x where tan is positive and cos is negative. That will be the third quadrant. In the third quadrant, ew value of x = 180+63.4

x = 243.4°

Since we are to find 2x,

2x = 2(243.4)

2x = 486.8°

If Joe drives 50 mph for 0.5 hours and then 60 mph for 1.5 hours, then how far did he drive?

Answers

Answer:

115 mi

Step-by-step explanation:

speed = distance/time

distance = speed * time

0.5 hours at 50 mph

distance = 50 mph * 0.5 h = 25 mi

1.5 hours at 60 mph

distance = 60 mph * 1.5 h = 90 mi

total distance = 25 mi + 90 mi = 115 mi

What is the missing segment?

Answers

Answer:

78.

Step-by-step explanation:

12 / (48-12) = 26 / x

12/36 = 26/x

1/3 = 26/x

x = 26*3 = 78.

How many solutions does the following equation have? 14(z+3)=14z+21

Answers

Answer:

No solutions

Step-by-step explanation:

14(z + 3) = 14z + 21

Expand brackets.

14z + 42 = 14z + 21

Subtract 14z on both sides.

42 = 21

There are no solutions.

Answer:

No solution

Step-by-step explanation:

First, We have to simplify the right side.

Distribute 14, 14z+42.

Now the equation stands as 14z+42=14z+21

Subtract 14z from both sides,

this makes it 42=21.

We know when the solution is #=#, our answer is no solution.

Of the following points, name all that lie on the same horizontal line? (1,-2) (-1,2) (-2,1) (2,-1)

Answers

Hey there!

If a line is horizontal, the entire line is at the same elevation. On a line, the elevation is the y-value. So, all of the points with the same y-value are on the same horizontal line.

We see that all of these points have different y-values so none of them would be on the same horizontal line.

I hope that this helps! Have a wonderful day!

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.68 and standard deviation 0.92.
A. If a random sample of 25 specimens is selected, what is theprobability that the sample average sediment density is at most3.00? Between 2.68 and 3.00
B. How large a sample size would be required to ensure thatthe first probability in part (a) is at least .99 ?

Answers

Answer:

The sample size must be 45 large enough that would ensure that the first probability in part (a) is at least 0.99.

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.68 and standard deviation 0.92.

Let [tex]\bar X[/tex] = sample  average sediment density

The z-score probability distribution for the sample mean is given by;

                             Z  =  [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = population mean = 2.68

           [tex]\sigma[/tex] = population standard deviation = 0.92

           n = sample of specimens = 25

(a) The probability that the sample average sediment density is at most 3.00 is given by = P([tex]\bar X[/tex] [tex]\leq[/tex] 3.00)

   

   P([tex]\bar X[/tex] [tex]\leq[/tex] 3.00) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{3.00-2.68}{\frac{0.92}{\sqrt{25} } }[/tex] ) = P(Z [tex]\leq[/tex] 1.74) = 0.9591

The above probability is calculated by looking at the value of x = 1.74 in the z table which has an area of 0.9591.

Also, the probability that the sample average sediment density is between 2.68 and 3.00 is given by = P(2.68 < [tex]\bar X[/tex] < 3.00)

P(2.68 < [tex]\bar X[/tex] < 3.00) = P([tex]\bar X[/tex] < 3.00) - P([tex]\bar X[/tex] [tex]\leq[/tex] 2.68)

 

   P([tex]\bar X[/tex] < 3.00) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{3.00-2.68}{\frac{0.92}{\sqrt{25} } }[/tex] ) = P(Z < 1.74) = 0.9591

   P([tex]\bar X[/tex] [tex]\leq[/tex] 2.68) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{2.68-2.68}{\frac{0.92}{\sqrt{25} } }[/tex] ) = P(Z [tex]\leq[/tex] 0) = 0.50

The above probability is calculated by looking at the value of x = 1.74 and x = 0 in the z table which has an area of 0.9591 and 0.50.

Therefore, P(2.68 < [tex]\bar X[/tex] < 3.00) = 0.9591 - 0.50 = 0.4591.

(b) Now, we have to find a sample size that would ensure that the first probability in part (a) is at least 0.99, that is;

      P([tex]\bar X[/tex] [tex]\leq[/tex] 3.00) [tex]\geq[/tex] 0.99

      P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{3.00-2.68}{\frac{0.92}{\sqrt{n} } }[/tex] ) [tex]\geq[/tex] 0.99

      P(Z [tex]\leq[/tex] [tex]\frac{3.00-2.68}{\frac{0.92}{\sqrt{n} } }[/tex] ) [tex]\geq[/tex] 0.99

Now, in the z table; the critical value of x which has an area of at least 0.99 is given by 2.3263, that is;

                 [tex]\frac{3.00-2.68}{\frac{0.92}{\sqrt{n} } }=2.3263[/tex]

                 [tex]\sqrt{n} } }=\frac{ 2.3263\times 0.92}{0.32}[/tex]

                  [tex]\sqrt{n} } }=6.69[/tex]

                    n = 44.76 ≈ 45          {By squaring both sides}

Hence, the sample size must be 45 large enough that would ensure that the first probability in part (a) is at least 0.99.

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