<span>D) 9.0 x 10^10 km
This is more an exercise in handling scientific notation than anything else. Since we have the distance that light travels in 1 second and we want to calculate how far it travels in 5 minutes, we must first calculate how many seconds are in 5 minutes. Simply multiplying 5 by 60 gives us 300 seconds. Now we need to multiply 300 by 3.0x10^8 km. So
300 * 3.0x10^8 = ?
We could first convert 300 into scientific notion, but it's easier to just leave it along and assume that it's 300 x 10^0. So 300 times 3 is 900. And since 0 plus 8 is 8, we have as the answer:
900 x 10^8
But we're not done. The significand has to be greater than or equal to 1 and less than 10. So let's divide 900 by 100 and add 2 to the exponent. So we get
9 x 10^10
Finally, since our data had 2 significant figures, our result should have that as well. So let's add the 2nd digit getting:
9.0 x 10^10
So we know that light travels 9.0x10^10 km in 5 minutes, and that answer matches option "D" from the available choices.</span>
Answer: choice c. GT
===================================
Explanation:
A median is a segment that runs from a vertex point to the midpoint of the opposite side. Segment GT starts at the vertex G and goes to the midpoint T. We know that T is a midpoint since FT and TH are both 3 units long
since FT = TH, we know that FT is half as long as FH; or put another way, FH is twice as long as FT.
---------
Extra info
* Segment HS is an angle bisector: it cuts angle GHF in half.
* If you have two median lines, then you can find the centroid of the triangle.
Both FI and GJ are diameters.
Answer:
Step-by-step explanation:
Consider the options for this question are as follow,
Here, In triangles ABC and PQR,
AB = c, BC = a, AC = b, PQ = r, QR = p and PR = q,
Since,
We know that,
The corresponding sides of similar triangles are in same proportion,
Thus,
A.) For the Junior Varsity Team, mean would be the appropriate measure of center since the data is <span>symmetric or well-proportioned while we should use standard deviation for getting the measure of spread since it also measures the center and how far the values are from the mean.
b.) For the Varsity Team, the median would be the appropriate measure of the center since the data is skewed left and not evenly distributed so median could be used since it does not account for outliers while we use IQR or interquartile range in measuring the spread of data since IQR does not account for the data that is skewed. </span>
