Answer:
50.020 and 50.02
Step-by-step explanation:
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80=6th graders
40=5th graders
80+40=120
120/22= 60/11
120/25=24/5
120/30= 4
There are 30 students in each class because he teaches 4 class and 120/30 will give you 4.
21-7=14 is a relate subtraction problem
For each coin, there are <u>2 outcomes</u>, heads (H) or tails (T), so for the <u>3 coins</u>, the number of outcomes are:
2 x 2 x 2 = <u>8 outcomes:</u>
HHH / HHT / HTH / THH / TTT / TTH / THT / HTT
Out of all these outcomes, there are <u>3 ways</u>, we get exactly 2 Heads:
HHT / HTH / THH
So, the probability of exactly 2 coins landing on Heads is 3/8, or 0.375
Like XZ divides the cord YV into two congruent parts (YW=5.27 cm=WV), this segment XZ must be perpendicular to the segment YV, then the angle XWY in triangle XWY is a right angle (90°) and the triangle XWY is a right angle.
We can apply the trigonometric ratios in triangle XWY:
Hypotenure: XY
sin 44°=(Opposite leg to 44°)/(hypothenuse)
sin 44°=YW/XY
sin 44°=(5.27 cm)/XY
Solving for XY. Cross multiplication:
sin44° XY=5.27 cm
Dividing both sides of the equation by sin 44°:
sin 44° XY / sin 44° = (5.27 cm)/sin 44°
XY=(5.27/sin 44°) cm
XY=(5.27/0.694658370) cm
XY=7.586462929 cm
This value XY is the radius of the circle, then:
XZ=XY→XZ=7.586462969 cm
tan 44°=(Opposite leg to 44°) / (Adjacent leg to 44°)
tan 44°=YW/XW
tan 44°=(5.27 cm)/XW
Solving for XW. Cross multiplication:
tan 44° XW=5.27 cm
Dividing both sides of the equation by tan 44°:
tan 44° XW / tan 44°=(5.27 cm)/tan 44°
XW=(5.27/tan 44°) cm
XW=(5.27/0.965688775) cm
XW=5.457244753 cm
WZ=XZ-XW
WZ=7.586462969 cm-5.457244753 cm
WZ=2.129218216 cm
Rounded to 2 decimal places:
WZ=2.13 cm
Answer: The <span>measurement is closest to the measure of segment WZ is
2.13 cm</span>