Y= the number riders over time
x= hours into Tara's shift
y= 33x + 132
Answer:
Step-by-step explanation:
hello :
√(5/4)= √5/ √4 =√5/2 = (1/2)√5
3+√5/4−√5 = 3+(1/2)√5 so : a=3 and b= 1/2
<span>1220
Subtracting the lower boundary of 1492 grams from the mean of 3234 gives you 1742 grams below the mean. Dividing 1742 by the standard deviation of 871 gives you 2 standard deviations below the curve. Now doing the same with the upper limit of 4976 grams also gives you 2 standard deviations above the mean (4976-3234)/871 = 2
So you now look for what percentage of the population lies within 2 standard deviations of the mean. Standard lookup tables will indicate that 95.4499736% of the population will be within 2Ď of the mean. So multiply 1278 by 0.954499736 giving 1219.851. Then round to the nearest whole number and you have an estimated 1220 babies that weigh between 1492 grams and 4976 grams.</span>
The discount sales price would be $218
Answer: see proof below
<u>Step-by-step explanation:</u>
Use the Double Angle Identity: sin 2Ф = 2sinФ · cosФ
Use the Sum/Difference Identities:
sin(α + β) = sinα · cosβ + cosα · sinβ
cos(α - β) = cosα · cosβ + sinα · sinβ
Use the Unit circle to evaluate: sin45 = cos45 = √2/2
Use the Double Angle Identities: sin2Ф = 2sinФ · cosФ
Use the Pythagorean Identity: cos²Ф + sin²Ф = 1
<u />
<u>Proof LHS → RHS</u>
LHS: 2sin(45 + 2A) · cos(45 - 2A)
Sum/Difference: 2 (sin45·cos2A + cos45·sin2A) (cos45·cos2A + sin45·sin2A)
Unit Circle: 2[(√2/2)cos2A + (√2/2)sin2A][(√2/2)cos2A +(√2/2)·sin2A)]
Expand: 2[(1/2)cos²2A + cos2A·sin2A + (1/2)sin²2A]
Distribute: cos²2A + 2cos2A·sin2A + sin²2A
Pythagorean Identity: 1 + 2cos2A·sin2A
Double Angle: 1 + sin4A
LHS = RHS: 1 + sin4A = 1 + sin4A 
