The plot below shows the volume of 11 river water samples collected by an ecologist. All measurements are rounded to the nearest 1/8 gallon.

Answer:

B

Step-by-step explanation:

Period equals pi/2

Answer:

Letter A

Step-by-step explanation:

He is the ticket clerk so it makes sense people would ask about the ticket prices.

Answer:

342 units squared

Step-by-step explanation:

since this is a rectangular prism, all we have to do is multiply the values by one of the other values once and multiply that by 2

Answer:

Many answers including (0,0)

Step-by-step explanation:

Well there is many solutions since it is an inequality, in fact there are infinite possible solutions.

One possible solution is (0,0)

Here's a graph that shows some of the possible solutions:

Answer:it’s 1/4

Step-by-step explanation:

Answer:

82.5

if its wrong sorry

hope this helps you ☺️☺️

Answer:

c) (x+1)²+(y-3)²=25

Step-by-step explanation:

Answer:

-5x=35            divide both sides by 5

-x=7                multiply both sides by -1

x=-7

Answer:

Answer C

Step-by-step explanation:

C.

Answer:

it is the bottom answer or if it was labeled it would be D.

Answer:

42.3

Sorry I haven't been answering your questions I'm just behind on homework.

Answer:

1:1 or 100% i don't know which you need

Step-by-step explanation:

Answer:

9.95 cm

Step-by-step explanation:

Volume of a cylinder = pi * r² * h

Volume = 500 cm³

Radius, r = 8cm / 2 = 4cm

500 = 3.14 * 4² * h

500 = 3.14 * 16 * h

500 = 50.24 * h

h = 500 / 50.24

h = 9.952229

Height, h = 9.95 cm

Answer: 0.09375

Step-by-step explanation: divid 3/8 and 1/4 together and you get 0.09375

Answer:  The range of a set of data is the difference between the highest and lowest values in the set. To find the range, first order the data from least to greatest. Then subtract the smallest value from the largest value in the set.

Answer:

A square unit is a small area of a square used to find a total.

Step-by-step explanation:

Answer:

RS=5.65

Step-by-step explanation:

Please gimme few minutes

6. = 0.0000004834

7. = 0.0000506

8. = 0.00098

9. = 64580000

10. = 1398600000

Answer:

The value of log 8 is 0.9030

Step-by-step explanation:

To find:

Value of log 8

Solution:

Let we take,

x=log8

x=log (2)^3

x=3log2

x=3*0.3010

x=0.9030

Hence, the value of log 8 is 0.90.

Answer:

Im pretty sure you are correct

Step-by-step explanation:

Belive in urself m8

Answer:

no I don't think this is correct

Step-by-step explanation:

(no btw)

Answer:

k = (-1/4)

Step-by-step explanation:

Given that,

The point (-2, k) lies on the line 2x – 4y + 3 = 0.

We need to find the value of k.

Put x = -2 and y = k in the given equation.

2(-2) – 4k + 3 = 0

-4-4k = -3

-4+3 = 4k

-1 = 4k

k = (-1/4)

So, the value of k is equal to (-1/4).

Answer:

45%

Step-by-step explanation:

9/20

Answer:

1, -4, -9, -14, -19, -24, -29, -34, -39, -44, -49, -54 = -318

Step-by-step explanation:

Answer:

The Answer is -318.

Step-by-step explanation:

1 + (-4) + (-9) + (-14) + (-19) + (-24) + (-29) + (-34) + (-39) + (-44) + (-49) + (-54)

Answer:

309

Step-by-step explanation:

To determine the estimated free-flow speed

[tex]FFS = 75.4 -f_{LW}-f_{LC} - 3.22 TRD^{0.84}[/tex]

From the table "Adjustment for lane width," which corresponds to a lane width of 12 feet. As a result, [tex]f_{LW}[/tex] equals 0 mi/h.

Take the meaning from the table "Adjustment for right-shoulder lateral clearance," which corresponds to 6 feet of right shoulder lateral clearance and two lanes in one direction.

The [tex]f_{LC} = 0 mi/h[/tex]

[tex]FFS = 75.4 -0-0-3.22 ( \dfrac{5}{6})^{0.84}[/tex]

[tex]= 72.64 \ mi/h[/tex]

[tex]\simeq 73 \ mi/h[/tex]

Determine the peak-hour factor

[tex]PHF = \dfrac{V}{V_{15}\times 4} \\ \\ =\dfrac{1800}{700\times 4}\\ \\ = 0.6429[/tex]

Now, Find the heavy-vehicle adjustment factor.

[tex]v_P = \dfrac{V}{PHF\times N \times f_{HV}\times f_P} --- (1)[/tex]

Take the value for the 15-minute passenger car equivalent flow rate from the table "LOS requirements for freeway segments" for the FFS and LOS C conditions. The free-flow speed is estimated to be 73 miles per hour

[tex]v_p = 1735+ \dfrac{73-70}{75-70}(1775-1735)[/tex]

[tex]v_p = 1759 \ pc/h/In[/tex]

Think about for familiar users the value of f_p = 1.00

Replace all of the values obtained in (1)

[tex]v_p = \dfrac{V}{PHF \times N\times f_{HV}\times f_p}[/tex]

[tex]1759 = \dfrac{1800}{0.6429\times 2 \times f_{HV}\times 1}[/tex]

[tex]f_{HV} = \dfrac{1800}{0.6429\times 2 }\times \dfrac{1}{1759}[/tex]

= 0.795

Calculate the percentage of trucks and buses in the flow of traffic stream.

[tex]f_{HV} = \dfrac{1}{1+P_{\tau}(E_{\tau}-1) + P_R(E_R-1)}[/tex]

Take the values from the table "Passenger car equivalent for extended highway segments" referring to trucks and buses and rolling terrains for passenger car equivalent for trucks and buses and recreational vehicles. As a result, [tex]E_T[/tex] has a value of 2.5 and [tex]E_R[/tex] has a value of 2.0.

Since [tex]P_R[/tex] is zero and there is no recreational vehicle.

Then;

[tex]0.759 = \dfrac{1}{1+ P_T(2.5 -1) +0}[/tex]

[tex]P_T = 0.1719[/tex]

Finally, to estimate the maximum number of large trucks and buses; we have:

[tex]= V\times P_T = 1800 \times 0.1719[/tex]

Maximum number of large trucks and buses = 309