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3 years ago

I NEED HELP:

Mathematics

2 answers:

Dmitry_Shevchenko [17]3 years ago

6 0

I believe it's b.

Hope this helped!

Tju [1.3M]3 years ago

4 0

Answer:

<em>B.</em><em> Mean: 339.20 (Median: 318) (Mode: 445.20) (Range: 243.80)</em>

Step-by-step explanation:

A store sells models of cameras for $190.00, $420.00, $270.00, $300.00, and $420.00.

As a 6% tax will be levied on this as sales tax, the new prices will be, $201.40, $445.20, $286.20, $318, $445.20

Arranging them in ascending order,

201.40, 286.20, 318, 445.20, 445.20

Mean of the sample will be,

$\overline{x}=\dfrac{201.40+286.20+318+445.20+445.20}{5}=\dfrac{1696}{5}=339.20$

Median is the middle value of the sample. So median will be 318.

The highest number of occurrences is the mode of the set. So mode is 445.20

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21. Who is closer to Cameron? Explain.

pickupchik [31]

Problem 21

<h3>Answer: Jamie is closer</h3>

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

Explanation:

A = Arthur's location = (20,35)
J = Jamie's location = (45,20)
C = Cameron's location = (65,40)

To find out who's closer to Cameron, we need to compute the segment lengths AC and JC. Then we pick the smaller of the two lengths.

We use the distance formula to find each length

Let's find the length of AC.

$A = (x_1,y_1) = (20,35)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from A to C} = \text{length of segment AC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-65)^2 + (35-40)^2}\\\\d = \sqrt{(-45)^2 + (-5)^2}\\\\d = \sqrt{2025 + 25}\\\\d = \sqrt{2050}\\\\d \approx 45.2769257\\\\$

The distance from Arthur to Cameron is roughly 45.2769257 units.

Let's repeat this process to find the length of segment JC

$J = (x_1,y_1) = (45,20)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from J to C} = \text{length of segment JC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(45-65)^2 + (20-40)^2}\\\\d = \sqrt{(-20)^2 + (-20)^2}\\\\d = \sqrt{400 + 400}\\\\d = \sqrt{800}\\\\d \approx 28.2842712\\\\$

Going from Jamie to Cameron is roughly 28.2842712 units

We see that segment JC is shorter than AC. Therefore, Jamie is closer to Cameron.

=================================================

Problem 22

<h3>Answer: Arthur is closest to the ball</h3>

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

Explanation:

We have these key locations:

A = Arthur's location = (20,35)
J = Jamie's location = (45,20)
C = Cameron's location = (65,40)
B = location of the ball = (35,60)

We'll do the same thing as we did in the previous problem. This time we need to compute the following lengths:

These segments represent the distances from a given player to the ball. Like before, the goal is to pick the smallest of these segments to find out who is the closest to the ball.

The steps are lengthy and more or less the same compared to the previous problem (just with different numbers of course). I'll show the steps on how to get the length of segment AB. I'll skip the other set of steps because there's only so much room allowed.

$A = (x_1,y_1) = (20,35)\\\\B = (x_2,y_2) = (35,60)\\\\d = \text{Distance from A to B} = \text{length of segment AB}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-35)^2 + (35-60)^2}\\\\d = \sqrt{(-15)^2 + (-25)^2}\\\\d = \sqrt{225 + 625}\\\\d = \sqrt{850}\\\\d \approx 29.1547595\\\\$

Segment AB is roughly 29.1547595 units.

If you repeated these steps, then you should get these other two approximate segment lengths:

JB = 41.2310563

CB = 36.0555128

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

So in summary, we have these approximate segment lengths

AB = 29.1547595
JB = 41.2310563
CB = 36.0555128

Segment AB is the smallest of the trio, which therefore means Arthur is closest to the ball.

3 0

3 years ago

for her phone service, ann pays a monthly fee offor her phone service and pays a monthly fee of $18 and she pays an additional $

trapecia [35]

$92,28-$18=$74.28
$74.28<span>÷$0.06= 1238minutes</span>

6 0

3 years ago

Please answer sorry if it's not clear

Nitella [24]

Move the figure 10 units to the right and then rotate it 90º

4 0

3 years ago

A rocket leaves the surface of Earth at time t=0 and travels straight up from the surface. The height, in feet, of the rocket ab

erastovalidia [21]

Answer:

Remember that:

Speed = distance/time.

Then we can calculate the average speed in any segment,

Let's make a model where the average speed at t = t0 can be calculated as:

AS(t0) = (y(b) - y(a))/(b - a)

Where b is the next value of t0, and a is the previous value of t0. This is because t0 is the middle point in this segment.

Then:

if t0 = 100s

AS(100s) = (400ft - 0ft)/(200s - 0s) = 2ft/s

if t0 = 200s

AS(200s) = (1360ft - 50ft)/(300s - 100s) = 6.55 ft/s

if t0 = 300s

AS(300s) = (3200ft - 400ft)/(400s - 200s) = 14ft/s

if t0 = 400s

AS(400s) = (6250s - 1360s)/(500s - 300s) = 24.45 ft/s

So for the given options, t = 400s is the one where the velocity seems to be the biggest.

And this has a lot of sense, because while the distance between the values of time is constant (is always 100 seconds) we can see that the difference between consecutive values of y(t) is increasing.

Then we can conclude that the rocket is accelerating upwards, then as larger is the value of t, bigger will be the average velocity at that point.

3 0

3 years ago

Which expression correctly uses commutative properties to rewrite the expression above?

Anuta_ua [19.1K]

Theirs no picture it’s a black screen

8 0

3 years ago