Answer:
I would say it would be D
Step-by-step explanation:
Answer:the answer is d hope this helped
Step-by-step explanation:
The new median of the data set will be 616.
To multiply means to add a number to itself a particular number of times. Multiplication can be viewed as a process of repeated addition.
Given that;
The data set is,
⇒ 14, 20, 18, 58, 71, 36
Now,
Since, The data set is,
⇒ 14, 20, 18, 58, 71, 36
Multiply by 22 in each number , we get;
⇒ 14 × 22, 20 × 22, 18 × 22, 58 × 22, 71 × 22, 36 × 22
⇒ 308, 440, 396, 1276, 1562, 792
Arrange the number is ascending order we get;
⇒ 308, 396, 440, 792, 1276, 1562
So, The median of data set = (440 + 792) / 2
= 1232/2
= 616
Thus, The new median of data set = 616
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Answer:
If each of the numbers in the following data set were multiplied by 31, what would be the median of the data set?
28, 58, 20, 14, 18, 71, 36
A. 558
B. 1116
C. 434
D. 868
answer
D.868
–329,472x5y8
–41,184x8y5
41,184x8y5
Answer:
It's A
Step-by-step explanation:
On Edg
Answer:
q = 0.105uC
Step-by-step explanation:
We can determine the force on one ball by assuming two balls are stationary, finding the E field at the lower right vertex and calculate q from that.
Considering the horizontal and vertical components.
First find the directions of the fields at the lower right vertex. From the lower left vertex the field will be at 0° and from the top vertex, the field will be at -60° or 300° because + charge fields point radially outward in all directions. The distances from both charges are the same since this is an equilateral triangle. The fields have the same magnitude:
E=kq/r²
Where r = 20cm
= 20/100
= 0.2m
K = 9.0×10^9
9.0×10^9 × q /0.2²
9.0×10^9/0.04
2.25×10^11 q
These are vector fields of course
Sum the horizontal components
Ecos0 + Ecos300 = E+0.5E
= 1.5E
Sum the vertical components
Esin0 + Esin300 = -E√3/2
Resultant = √3E at -30° or 330°
So the force on q at the lower right corner is q√3×E
The balls have two forces, horizontal = √3×E×q
and vertical = mg, therefore if θ is the angle the string makes with the vertical tanθ = q√3E/mg
mg×tanθ = q√3E.
..1
Then θ will be...
Since the hypotenuse = 80cm
80cm/100
= 0.8m
The distance from the centroid to the lower right vertex is 0.1/cos30 =
0.1/0.866
= 0.1155m
Hence,
0.8×sinθ = 0.1155
Sinθ = 0.1155/0.8
Sin θ = 0.144375
θ = arch sin 0.144375
θ = 8.3°
From equation 1
mg×tanθ = q√3E
g = 9.8m/s^2
m = 3.0g = 0.003kg
0.003×9.8×tan(8.3)
0.00428 = q√3E
0.00428 = q×1.7320×E
Where E=kq/r²
Where r = 0.2m
0.0428 = kq^2/r² × 1.7320
K = 9.0×10^9
0.0428/1.7320 = 9.0×10^9 × q² / 0.2²
0.02471×0.04 = 9.0×10^9 × q²
0.0009884 = 9.0×10^9 × q²
0.0009884/9.0×10^9 = q²
q² = 109822.223
q = √109822.223
q = 0.105uC
Answer:
Hey! the answer is e
Step-by-step explanation:
If you were to draw two lines intercepting each other, one point would be the same, right?
Here there is line AE, and line DE. The repeated point is E, which means it's the intersection.
I can't add visuals at the moment, but imagine like an x, a being on one end, d on the other, and e in the middle.
Hope this helps!