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ozzi
2 years ago
12

You weigh six packages and find the weights to be 28, 22, 52, 25, 49, and 46 ounces. If you include a package that weighs 142 ou

nces, which will increase more,
the median or the mean?
A. The median and mean are affected the same amount.
OB. The median and the mean will stay the same.
C. The median increases more.
D. The mean increases more.

Mathematics
2 answers:
Viefleur [7K]2 years ago
5 0

Answer:

chk photo

Step-by-step explanation:

erastovalidia [21]2 years ago
4 0
The mean increases more
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Which statement completes the proof that the sum of the angles of triangle PRQ is 180°? Statement Justification
vichka [17]
M∠1 + m∠RPQ + m∠5 = 180° Definition of a Straight Angle

proof :  mes1= PRQ (alternate interior)

RPQ = 3 Opposite angle (vertex)
5 =Q  (alternate interior)
PRQ + RPQ + RQP = 180
7 0
3 years ago
Read 2 more answers
Find the slope of the line passing through the points (-6,2), (0,-6).
Serhud [2]

C. -4/3

slope =  \frac{(y _{2} - y _{1}) }{(x _{2} - x _{1} )}  \\  =  \frac{( - 6 - 2)}{(0 -( -  6)}  \\  =  \frac{ - 8}{ 6}  \\  =   - \frac{4}{3}

3 0
3 years ago
The points ​(​15, 18​) and ​(​35,42) form a proportional relationship. Find the slope of the line through the points. Then use t
Ann [662]

Answer:

6/5

Step-by-step explanation:

(42-18) / (35 - 15)  = 24 / 20  = 6 / 5

8 0
3 years ago
5. Earl runs 75 meters in 30 seconds. How many meters does Earl<br> run per second?
Rudik [331]

Answer:

He runs 2.5 meters per second

Step-by-step explanation:

all you do is divide 75/30 and the answer is 2.5.

8 0
3 years ago
A football player is running the length of a 100 yard long football field. For the first part of the field, he runs at a rate of
notka56 [123]

Answer:

\frac{2}{5}th part of the field is covered in the second sprint.

Step-by-step explanation:

A football player is running the length of a 100 yard long football field.

Let player sprints x yards with the speed = 2 yards per second.

So time taken to cover x yards player will take time = \frac{x}{2} seconds

Now rest distance (100 - x) yards when covered with the speed of 4 yards per second, so time taken to cover this distance = \frac{Distance}{Speed}

= \frac{100-x}{4} seconds

Now total time taken by the player can be represented by the equation

\frac{x}{2}+\frac{100-x}{4}=40

Now we can solve this equation for the value of x.

\frac{2x+100-x}{4}=40

x + 100 = 40×4

x + 100 = 160

x = 160 - 100 = 60 yards

And length of the second part will be = 100 - 60 = 40 yards

Now the fraction of the field covered by the player in second sprint will be

= \frac{40}{100}

= \frac{2}{5} or 40%

Therefore, \frac{2}{5}th part of the field was covered in second sprint.

5 0
3 years ago
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