In the figure below. MN is the arc of a circle with center L. If the length of arc MN is 6π, what is the area of sector LMN?

Answer:

1.05km/min

Step-by-step explanation:

[tex]Distance = 84 km\\\\Time = 1hr 20 min= 80minuttes\\\\Average \: Speed = \frac{Distance}{Time} \\\\A.S =\frac{84}{80} \\\\A.S = 1.05km/min[/tex]

we, know that

➺ Average Speed = Total Distance/Total Time taken

where,

♛ Substituting these values ♛

◗Average Speed = Distance/Time

◗Average Speed = 84000/4800

◗Average Speed = 840/48

◗Average Speed = 70/4

◗Average Speed = 17.5m/s

hope it helps you

Answer:

116 miles

Step-by-step explanation:

We can solve this by first writing an equation for the cost of the car rental. To begin, the base cost is $17.95, so any further costs must be added to that. Next, the car costs 19 cents (0.19 dollars) for each mile driven, so for each mile, we add 19 cents. This can be written as 0.19 *x if x represents the amount of miles driven. Therefore, we can add the two input costs of the car (the base cost and cost per mile) to get

17.95 + 0.19 * x = total cost.

After that, we want to maximize x/the number of miles with only 40 dollars. We can do this by setting this equal to the total cost, as going over the total cost is impossible and going under would be limiting the amount of miles (this because we are adding money for each mile, so more money means more miles). Therefore, we have

17.95 + 0.19 * x = 40

subtract 17.95 from both sides to isolate the x and its coefficient

22.05 = 0.19 * x

divide both sides by 0.19 to isolate x

22.05/0.19 = x = 116.05

The question asked us to round down, and 116.05 rounded down is 116 for our answer

Answer:

Cooper read the least because

Step-by-step explanation:

1/5 is less than

Answer: Cooper read more

Step-by-step explanation:

Answer:

A = 2025 ft^2

Step-by-step explanation:

The area of a square is given by

A = s^2 where s is the side length

A = 45^2

A = 2025 ft^2

Answer:

Step-by-step explanation:

The area of any square is s^2

s in this case is 45

Area = s^2

Area = 45 * 45

Area = 2025 m^2

Answer:

24 units²

Step-by-step explanation:

A rhombus is divided into 4 right triangles when it's two diagonals intersect at right angles. All the sides are of equal lengths.

Therefore, a simple method to use to find the area of the given rhombus is to calculate the area of one of the right triangles, and multiply by 4.

Area of right triangle = ½*base*height

Height = 3

Base = [tex]\sqrt{5^2 - 3^2} = \sqrt{16} = 4[/tex] (Pythagorean theorem)

Area of right triangle = ½*4*3 = 2*3 = 6 units²

Area of rhombus = 4(6 units²) = 24 units²

Answer:

-6

Step-by-step explanation:

27-3/2x=36

53-3x=72

3x=-18

x=-6

Answer:

Shift the graph of y = x2 right 6 units.

Answer: = a = 0.0206

b = 0.0205.

Step-by-step explanation:

From the question, given that;

Order Accurate         =  328     273    242     142

Order Not Accurate  =  32       54       37       20

Let us make the Total orders given be

T.O = 328+273+242+142+32+54+37+20 = 1128.

a) Let the Prob. that the first order is from restaurant D be

= Number of order from restaurant D / Total number of orders

= 162 / 1128 = 0.1436

Probability of the second order is 0.1436.

This is because, from the question we can tell that the selections are made with replacement, that means the order is the same.

So, the probability of getting 2 orders =

= 0.1436 * 0.1436 = 0.0206

NB: The probability of getting two orders from restaurant B is 0.0206.

This is because choosing the first order does not affect the second order

(independent events).

b) Assuming that the selections are made without replacement , the probability of getting both the orders from restaurant D =

This gives the Total Probability of getting both the orders from restaurant D, without replacement to be =  0.1436*0.1428

= 0.0205.

That is to say choosing the first order affects the second order because of the events are not independent as compared to the first question.

cheers i hope tis helps

Answer:

Step-by-step explanation:

Let us first generate the frequency table from the information given:

Hurricane number(X)       Frequency(f)                f(X)

6                                            9                                   54

8                                            13                                  104

12                                           16                                  192

14                                            12                                168

Total                                     ∑(f) = 50                          ∑f(x) =518

In order to determine the last frequency (the remaining years), we will add the other frequencies and subtract the answer from 50, which is the total frequency (50 years). This is done as follows:

Let the last frequency be f

9 + 3 + 16 + f = 50

38 + f = 50

f = 50 - 38 = 12

Now, calculating mean:

[tex]\bar {X} = \frac{\sum f(x)}{\sum(f)} \\\\\bar {X} = \frac{518}{50} \\\\\bar {X} = 10.36[/tex]

Therefore mean number of hurricanes = 10.4 (to one decimal place)

Answer:

too complex:<

Step-by-step explanation:

120cm+120cm=240cm (2 squirrels + 2 air spaces)

90cm+90cm=180cm (2 rats + 2 air spaces)

240cm-180cm=(2 squirrels + 2 air spaces) - (2 rats + 2 air spaces)

                       =2 squirrels + 2 air spaces - 2 rats - 2 air spaces

                       =2 squirrels - 2 rats

                       =60cm

1 squirrel - 1 rat = 60cm divided by 2

                         = 30cm

120cm + 90cm = squirrel + air space + rat + air space

                        = 210cm

I've no idea!! This qn is too challenging!!

But i hope the above workings might help you in a way or another:>

The table is 105cm tall.

Two or more expressions with an Equal sign is called as Equation.

Let the table is x

Squirrel is y

Rat is z

From 1st diagram

x+y-z=120...(1)

From 2nd diagram

x+z-y=90...(2)

Add 1 and 2

x+y-z+x+z-y=120+90

2x=210

Divide both sides by 2

x=105

Hence, the table is 105cm tall.

To learn more on Equation:

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#SPJ2

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Answer:

147

Step-by-step explanation:

The number is three times the other number

so 7 and 21

cause 7 x 3 = 21

finally check 21 x 7 = 147

Answer:

A; The first choice.

Step-by-step explanation:

We have the equation [tex]x^4+6x^2+5=0[/tex] and we want to solve using u-substitution.

When solving by u-substitution, we essentially want to turn our equation into quadratic form.

So, let [tex]u=x^2[/tex]. We can rewrite our equation as:

[tex](x^2)^2+6(x^2)+5=0[/tex]

Substitute:

[tex]u^2+6u+5=0[/tex]

Solve. We can factor:

[tex](u+5)(u+1)=0[/tex]

Zero Product Property:

[tex]u+5=0\text{ and } u+1=0[/tex]

Solve for each case:

[tex]u=-5\text{ and } u=-1[/tex]

Substitute back u:

[tex]x^2=-5\text{ and } x^2=-1[/tex]

Take the square root of both sides for each case. Since we are taking an even root, we need plus-minus. Thus:

[tex]x=\pm\sqrt{-5}\text{ and } x=\pm\sqrt{-1}[/tex]

Simplify:

[tex]x=\pm i\sqrt{5}\text{ and } x=\pm i[/tex]

Our answer is A.

hypotenuse goes to the line across from the right angle. adjacent is the bottom one. lastly opposite is the left one.

Step-by-step explanation:

Hi, there!!

According to the question, we should find the hypotenuse, adjacent, and opposite to the refrence angle A ,right.

so, let's simply work with it,

hypotenuse (h)= AC {side opposite to the 90° is always a hypotenuse}.

opposite (p)= BC { as the side opposite to the refrence angle is always perpendicular or opposite}

adjacent (b)= AB { as remaining side is always base or adjacent}

Hope it helps....

Answer:

19 purses

Step-by-step explanation:

Set up an inequality where x represents the number of purses:

127 [tex]\geq[/tex] 7x

Solve for x by dividing each side by 7:

18.14 [tex]\geq[/tex] x

Round up to 19 because purses have to be whole

So, 19 purses have to be sold.

Answer: Choice A

Explanation: A segment (or line segment) is whenever we have two endpoints connected with a straight line. It does not go on forever in either direction.

Answer:

Step-by-step explanation:

B is a ray. One of its ends is stationary.

C is an arc which means it bends

D is a line.

That means that A is the answer. Both its ends are stationary.

Let

ATQ

[tex]\\ \sf \longmapsto Area=Length\times Width[/tex]

[tex]\\ \sf \longmapsto x(x+3)=28[/tex]

[tex]\\ \sf \longmapsto x^2+3x=28[/tex]

[tex]\\ \sf \longmapsto x^2+3x-28=0[/tex]

[tex]\\ \sf \longmapsto x^2+7x-4x-28=0[/tex]

[tex]\\ \sf \longmapsto x(x+7)-4(x+7)=0[/tex]

[tex]\\ \sf \longmapsto (x-4)(x+7)=0[/tex]

[tex]\\ \sf \longmapsto x=4\:or\:x=-7[/tex]

[tex]\\ \sf \longmapsto Width=4units[/tex]

[tex]\\ \sf \longmapsto Length=4+3=7units[/tex]

Answer:

Answer is below

Step-by-step explanation:

The width of the CI is directly proportional to critical value. When t* is greater than z value, the t value would then cause the margin of error to be larger and this will in turn cause the width of the confidence interval to be larger.

Greater t*df than z* gives us a bigger margin of error. This would in turn give bigger width of confidence interval. t distribution has greater width confidence interval compared to z distribution.

The width of confidence interval is a function of the margin of error. If the critical value of t(t*) is slightly larger than the critical value of z(z*), then the width of the confidence interval will be larger.

The margin of error is the product of the critical value and the standard error. Therefore, given the same standard error value, the value of the margin of error will increases based on the value of the critical value.

Since, t* is slightly larger than z*, then the confidence interval, t will be wider.

Learn more : https://brainly.com/question/18405415

Answer:

Step-by-step explanation:

From the given information;

Let consider p to be the number of large buses and q to be the number of small buses.

The company has 18 large buses which can carry up to 30 students and 19 small buses which can carry up to 15 students.

The linear inequality represent the students with respect to the total students s:

30p + 15q   ≥  450   -----(1)

They are only 20 drivers available on the day of the field trip, therefore only 20 buses can be used. So,

p  +   q    ≤  20      ----- (2)

Also , There  are only 19 small buses and 18 large buses:

∴

0 ≤  p ≤ 18

0 ≤  q ≤ 19

The total cost of operating one large bus is $225 a day, and the total cost of operating one small bus is $100 per day

In order to minimize the cost of transporting all 450 students, let z be the minimal cost. i.e

z = 225p + 100q

From equation 1 and 2 ; if we plot them graphically on the graph, the following points of intersection were obtained as shown in the sketch below, in which the shaded region lies the answer.

The point of intersection between equation (1) and (2) is (p,q) = (10,10)

From the critical point in the sketch of our graph attached below, the following values of z can be determined.

Point(s)              Value for Z

(15,0)                 15 ×  225 = 3375

(18,0)                 18 ×  225 =  4050

(18,2)                 (18 × 225 )+ (2 × 100 ) = 4250

(10,10)                (10 × 225) + (10 × 100) = 3250

Thus ; the minimum cost of transporting all 450 students = $3250

Answer:

6(x-3)

Step-by-step explanation:

the common number for 6 and 18 is 6 so if you extract that from the expression then it turns to 6(x-3) which cannot be factored further

Answer:

Option B:  6(x - 3)

Step-by-step explanation:

Answer:

See below

Step-by-step explanation:

we want to prove that A.M, G.M. and H.M between any two unequal positive numbers satisfy the following relations.

well, to do so let the two unequal positive numbers be [tex]\text{$x_1$ and $x_2$}[/tex] where:

the AM,GM and HM of [tex]x_1[/tex] and[tex] x_2[/tex] is given by the following table:

[tex]\begin{array}{ |c |c|c | } \hline AM& GM& HM\\ \hline \dfrac{x_{1} + x_{2}}{2} & \sqrt{x_{1} x_{2}} & \dfrac{2}{ \frac{1}{x_{1} } + \frac{1}{x_{2}} } \\ \hline\end{array}[/tex]

[tex] \displaystyle \rm AM \times HM = \frac{x_{1} + x_{2}}{2} \times \frac{2}{ \frac{1}{x_{1} } + \frac{1}{x_{2}} } [/tex]

simplify addition:

[tex] \displaystyle \frac{x_{1} + x_{2}}{2} \times \frac{2}{ \dfrac{x_{1} + x_{2}}{x_{1} x_{2}} } [/tex]

reduce fraction:

[tex] \displaystyle x_{1} + x_{2} \times \frac{1}{ \dfrac{x_{1} + x_{2}}{x_{1} x_{2}} } [/tex]

simplify complex fraction:

[tex] \displaystyle x_{1} + x_{2} \times \frac{x_{1} x_{2}}{x_{1} + x_{2}} [/tex]

reduce fraction:

[tex] \displaystyle x_{1} x_{2}[/tex]

rewrite:

[tex] \displaystyle (\sqrt{x_{1} x_{2}} {)}^{2} [/tex]

[tex] \displaystyle AM \times HM = (GM{)}^{2} [/tex]

hence, PROVEN

[tex] \displaystyle x_{1} > x_{2}[/tex]

square root both sides:

[tex] \displaystyle \sqrt{x_{1} }> \sqrt{ x_{2}}[/tex]

isolate right hand side expression to left hand side and change its sign:

[tex]\displaystyle\sqrt{x_{1} } - \sqrt{ x_{2}} > 0[/tex]

square both sides:

[tex]\displaystyle(\sqrt{x_{1} } - \sqrt{ x_{2}} {)}^{2} > 0[/tex]

expand using (a-b)²=a²-2ab+b²:

[tex]\displaystyle x_{1} -2\sqrt{x_{1} }\sqrt{ x_{2}} + x_{2} > 0[/tex]

move -2√x_1√x_2 to right hand side and change its sign:

[tex]\displaystyle x_{1} + x_{2} > 2 \sqrt{x_{1} } \sqrt{ x_{2}}[/tex]

divide both sides by 2:

[tex]\displaystyle \frac{x_{1} + x_{2}}{2} > \sqrt{x_{1} x_{2}}[/tex]

[tex]\displaystyle \boxed{ AM>GM}[/tex]

again,

[tex]\displaystyle \bigg( \frac{1}{\sqrt{x_{1} }} - \frac{1}{\sqrt{ x_{2}}} { \bigg)}^{2} > 0[/tex]

expand:

[tex]\displaystyle \frac{1}{x_{1}} - \frac{2}{\sqrt{x_{1} x_{2}} } + \frac{1}{x_{2} }> 0[/tex]

move the middle expression to right hand side and change its sign:

[tex]\displaystyle \frac{1}{x_{1}} + \frac{1}{x_{2} }> \frac{2}{\sqrt{x_{1} x_{2}} }[/tex]

[tex]\displaystyle \frac{\frac{1}{x_{1}} + \frac{1}{x_{2} }}{2}> \frac{1}{\sqrt{x_{1} x_{2}} }[/tex]

[tex]\displaystyle \rm \frac{1}{ HM} > \frac{1}{GM} [/tex]

cross multiplication:

[tex]\displaystyle \rm \boxed{ GM >HM}[/tex]

hence,

[tex]\displaystyle \rm A.M>G.M>H.M[/tex]

PROVEN

Answer:

20000 kg

Step-by-step explanation:

Recall that 1 kg = 2.2 lb approximately.  Then:

22 tons        1 kg        2000 lb

------------ * ------------ * --------------  =  20000 kg

      1           2.2 lb          1 ton

Answer:

yes, the volume = 16 ft^3

Answer: 3/8

Step-by-step explanation:

A=3

B=5

3/3+5=3/8

Answer:

A. 1000/1 = x/2500

Step-by-step explanation:

1000 g = 1 kg

Using ratios

1000 g         x g

------------ = --------------

1 kg             25 00 kg

Answer:

A. 1000/1 = x/2500

Step-by-step explanation:

KHD - DCM

Kilo - Hecta - Deca - Base - Deci - Centi - Mili

*When moving from left to right, each spot you pass multiplis your number by 10.

2500 kilo → x grams (base)

Proportions:

grams/kilos = x/2500

*For every 1 kilo, there is 1000 grams.

1000/1 = x/2500

Full Question:

Tony rounded each of the numbers 1,143 and 1,149 to the nearest hundred. Which choice correctly compares the rounded numbers?

[tex]1,000 = 1,000[/tex]

[tex]1,140< 1,150[/tex]

[tex]1,100 = 1.100[/tex]

[tex]1,140>1,150[/tex]

Answer:

[tex]1,100 = 1,100[/tex]

Step-by-step explanation:

Given

1,143 and 1,149

Required

Which of the option is correct

We start by approximating both numbers to nearest digit

1,143; when approximated to nearest hundred is 1,100

1,149; when approximated to nearest hundred is also 1,100

Hence;

1,143 ≅ 1,100

1,149 ≅ 1,100

Comparing both results, we have that

[tex]1,100 = 1,100[/tex]

From the list of given options, option C is correct;

Answer:

1.)  [tex]\frac{1}{9^4}*9^3[/tex]

2.) [tex]\frac{1}{w^7}[/tex]

3.)

Step-by-step explanation:

When you have a negative exponent, rewrite:

[tex]x^{-a}=\frac{1}{x^a}[/tex]

Rewrite using this to change all negative exponents.

Answer:

Multiple Answers

Step-by-step explanation:

Note: When multiplying numbers with exponents, you add the exponents. When dividing numbers with exponents, you subtract exponents.When you have a negative exponent, flip the fraction and write it as a positive exponent.

1) -4 + 3= -1  

So we have  (9^-4) + (9^3)= (1/(9^1)

2) (1/w)^7

3) cannot read problem, but just apply the rules I wrote under "Note"

4) 14/y

5) cannot read problem,but just apply the rules I wrote under "Note"

6) 20d^4 n^? --Cannot read n exponents--.

7) cannot read problem

8) Cannot read problem

9) 90/z^4---only if exponents are 5,-3,and-6

10) 1/(9^5)

11) 54b^4

12) Cannot read problem

13) 16d^8c^8 ---if exponents are 5,3,6,2--

14) s^8

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