Answer:
B) The heights of the bear should equal the class frequency.
Step-by-step explanation:
In drawing a histogram, the heights of the bear should equal the class frequency. as in histogram height of bar represent the frequency density and it´s area represent the frequency of class interval. Frequency distribution are represented by mean of rectangle. Earlier in the bar graph, width of bar does not represent any information, however, histogram´s bar width represent class interval.
That’s a good one Because what is the domain of the function Y equals 2X
Hi there!
Your question:
8.5 + 49(1-2.5k)=24.5
My answer:
8.5 + 4(1-2.5k) = 24.5 >>>>>>> first, you have to distribute
8.5 + 4 - 10k = 24.5 >>>>>>>> now, we are going to combine like terms.
12.5 - 10k = 24.5 >>>>>> now, we are going to subtract 12.5 on both sides
-10k = 12 >>>>>>>>> finally, divide -10 on both sides
k = -1.2
Hope this helps you!
Answer:
probably C.
Step-by-step explanation:
Answer:
<h2>
110.41m</h2>
Step-by-step explanation:
To get the value f x, we need to use SOH, CAH, TOA identity on ΔAEB but first we need to know any of the sides EB or AB.
From ΔBED, sum of the the angle in the triangle is 180° i.e ∠BED+∠EDB+∠EBD = 180°
65+55+∠EBD = 180
∠EBD = 180-120
∠EBD = 60
Also ∠DBC = 90- ∠EBD
∠DBC = 90- 60
∠DBC = 30°
Applying sine rule on ΔBCD;
52/sin∠DBC = DB/sin115
52/sin30 = DB/sin115
DB = 52sin115/sin30
DB = 52sin115/0.5
DB = 104sin115
DB = 114.75 m
Also applying sine rule on ΔBED to get side BE;
BE/sin55 = 114.75/sin65
BEsin65 = 114.75sin55
BE = 114.75sin55/sin65
BE = 93.998/0.906
BE = 103.75m
Finally, applying the trigonometry identity SOH on ΔABE
sinΔEAB = opp/hyp
sin70° = BE/AE
Since AE = x;
sin70° = 103.75/x
x = 103.75/sin70°
x = 110.41m
Hence, the value of x missing is approximately 110.41m
