The answer is deductive reasoning. Deductive reasoning begins from a premise or general statement accepted as true and logically progresses to conclusions that are definitively valid. This method employs a top-down approach to link premises to conclusions.
Result:
20.19°
Detailed explanation:
Refer to the attached diagram related to the question. We need to determine <CAB
Utilizing the sine rule on triangle ABC:
cross-multiply
Thus, the angle <CAB measures 20.19°
Answer:
y 
y = StartFraction 3 + 6 StartRoot 2 EndRoot Over 4 EndFraction y = StartFraction 3 menos 6 StartRoot 2 EndRoot Over 4 EndFraction
Explicación paso a paso:
La ecuación cuadrática que tenemos es (4y - 3)² = 72
Debemos encontrar el valor de y.
Ahora, 4y - 3 = ± 6√2
⇒ 4y = 3 ± 6√2
⇒ y
Por lo tanto, las soluciones son y = StartFraction 3 + 6 StartRoot 2 EndRoot Over 4 EndFraction y y = StartFraction 3 menos 6 StartRoot 2 EndRoot Over 4 EndFraction (Respuesta)
1.4×5=7
0.8×10=8
1.4×10=14
1×15=15
15+14+8+7=44
44÷4=11
LQ of 44=11
LQ=10 minutes
11×3=33
UQ= 29 minutes
The Range is 19 minutes
Detailed breakdown:
Commence with the individual boxes. For determining the number of students in each category, calculate Frequency density × The difference in the category. (if it's 5-15, the difference is 10)
This results in the counts of students in each range.
Next, determine the LQ of 44, which is 11.
Then locate the 11th student's score; in this instance, it resides in the 5-15 range. 7 students have already surpassed it, with 8 in the 5-15 range. Hence, the 11th lies within the bounds of 5-15, making the middle 10.
Repeat this process for the UQ.
The interquartile range is calculated as UQ-LQ, yielding 29-10=19 minutes.
I hope this helps, though I'm not entirely sure if my explanation is coherent and I'm unclear on the terminology I've used for these categories.
Answer:
89
Step-by-step explanation:
Given that
2 O, 2 T and 1 M
Now based on this, the following arrangements exist
1
Three arrangements i.e. {M,T,O}
2
XX or XY
XX in 2C1 = two arrangements i.e. {OO or TT}
XY in 3C2 × 2! = six arrangements
3
XXY or XYZ
XXY in 2C1 × 2C1 × 3! ÷ 2! = twelve arrangements
XYZ in 3C3 × 3! = six arrangements
4
XXYY or XXYZ
XXYY = 4! ÷ (2! × 2!) = six arrangements
XXYZ in 2C1 × 4! ÷ 2! = twenty four arrangements
5
= 5! ÷ (2! ×2!)
= 120 ÷ 4
= 30
Thus, the overall total is
= 3 + (2 +6)+ (12 +6) + (6 +24) + 30
= 89
