The options are;
1) AB is bisected by CD
2) CD is bisected by AB
3) AE = 1/2 AB
4) EF = 1/2 ED
5) FD= EB
6) CE + EF = FD
Answer:
Options 1, 3 & 6 are correct
Step-by-step explanation:
We are told that Point E is the midpoint of AB. Thus, any line that passes through point E will bisect AB into two equal parts.
The only line passing through point E is line CD.
Thus, we can say that line AB is bisected by pine CD. - - - (1)
Also, since E is midpoint of Line AB, it means that;
AE = EB
Thus, AE = EB = ½AB - - - (2)
Also, we are told that F is the mid-point of CD.
Thus;
CF = FD
Point E lies between C and F.
Thus;
CE + EF = CF
Since CF =FD
Thus;
CE + EF = FD - - - (3)
Answer:
I'm not very sure about this one what icing in probably not going to be right but what I think is that they're not the same because look at the size difference imagine putting an adult's next to a baby that's kind of what it's like the bigger quadrilateral is also the adults while the little one is the baby what do you think about that I probably think that they're not saying
Step-by-step explanation:
sorry I gave it to you in my answer
Answer:
2.
Step-by-step explanation:
1+1=2
We have the frequencies for each of the grades. We can estimate the number of students graded by adding all those frequencies. Let's call N the total number of grades:

We have then a total number of grades of 39.
The corresponding relative frequency for a grade is the ratio of the frequency to the total number of "samples", 39 in this case.
Then, for grade A, the relative frequency (RF) will be:

This will be the fraction of the total grades that are A. Represented as a percentage will be 10.26%, rounded to two decimal places.
Now, to complete the table we do the same for the other frequencies:
For grade B:

For grade C:

For grade D:

For grade F:
Answer:
Step-by-step explanation:
Yes, it's reasonable.
What you are doing is solving the question by rounding. You come up with an answer. Suppose you loose the decimal somewhere and you get 0.36? Is that reasonable? Do you just write the answer in the provided blank and move on. What now?
You get it wrong?!!
But your estimate should be about 9/3 = 3. Now you look at your calculator with great misgivings, because it made a mistake. Did it or did you? Well ultimately you did, but you have to blame something. So the calculator takes the heat.
Who knows? Maybe the decimal doesn't work. It's stuck or something. In any event you should be aware that there's no way the answer could be 0.36 when you estimate it to be 3.