Answer:
12
Step-by-step explanation:
The price of a 7 -minute phone call is $1.75. What is the price of a 14 -minute phone call?
Answer:
$3.50
Step-by-step explanation:
14 is double 7
So the price should be double
1.75 x 2 = 3.5
(Assuming there's no base fee and the charge is purely for minutes)
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In an all boys school, the heights of the student body are normally distributed with a
mean of 69 inches and a standard deviation of 2.5 inches. Using the empirical rule,
what percentage of the boys are between 61.5 and 76.5 inches tall?
Answer:
99.7%
Step-by-step explanation:
99.7% of boys fall between 61.5 and 76.5
The percentage of the boys that are between 61.5 and 76.5 inches tall is 99.73%
How to determine the percentage between the range?The given parameters are:
Mean = 69Standard deviation = 2.5Start by calculating the z score for x = 61.5 and 76.5 using:
[tex]z = \frac{x - \mu}{\sigma}[/tex]
So, we have:
[tex]z_1 = \frac{61.5 - 69}{2.5} = -3[/tex]
[tex]z_2 = \frac{76.5 - 69}{2.5} = 3[/tex]
The percentage is then represented as:
Percentage = P(-3 < x < 3)
Using the z table of probabilities, we have:
Percentage = 0.9973
Express as percentage
Percentage = 99.73%
Hence, 99.73% of the boys are between 61.5 and 76.5 inches tall
Read more about normal distribution at:
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3r + 6y - 2z = -6 2x + y + 4z = 19 -5x - 2y+8z = 62
solve each system of equations
This is finding exact values of sin theta/2 and tan theta/2. I’m really confused and now don’t have a clue on how to do this, please help
First,
tan(θ) = sin(θ) / cos(θ)
and given that 90° < θ < 180°, meaning θ lies in the second quadrant, we know that cos(θ) < 0. (We also then know the sign of sin(θ), but that won't be important.)
Dividing each part of the inequality by 2 tells us that 45° < θ/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(θ/2) > 0 and sin(θ/2) > 0.
Now recall the half-angle identities,
cos²(θ/2) = (1 + cos(θ)) / 2
sin²(θ/2) = (1 - cos(θ)) / 2
and taking the positive square roots, we have
cos(θ/2) = √[(1 + cos(θ)) / 2]
sin(θ/2) = √[(1 - cos(θ)) / 2]
Then
tan(θ/2) = sin(θ/2) / cos(θ/2) = √[(1 - cos(θ)) / (1 + cos(θ))]
Notice how we don't need sin(θ) ?
Now, recall the Pythagorean identity:
cos²(θ) + sin²(θ) = 1
Dividing both sides by cos²(θ) gives
1 + tan²(θ) = 1/cos²(θ)
We know cos(θ) is negative, so solve for cos²(θ) and take the negative square root.
cos²(θ) = 1/(1 + tan²(θ))
cos(θ) = - 1/√[1 + tan²(θ)]
Plug in tan(θ) = - 12/5 and solve for cos(θ) :
cos(θ) = - 1/√[1 + (-12/5)²] = - 5/13
Finally, solve for sin(θ/2) and tan(θ/2) :
sin(θ/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)
tan(θ/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2