I don’t know how to do this one.
2 answers:
Part (a)
The unit circle is centered at (0,0). This point is the origin.
The radius of the unit circle is 1
The term "unit" means "one", which describes the radius length.
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Part (b)
The equation of the unit circle is x^2+y^2 = 1
The general equation of any circle is (x-h)^2+(y-k)^2 = r^2
In this case, the unit circle has center (h,k) = (0,0) and radius r = 1.
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Part (c)
The following four points are on the unit circle
- (1,0)
- (0,1)
- (-1,0)
- (0,-1)
As shown below in the diagram. The idea is to plug each given coordinate into the equation from part (b), and solve for the missing variable.
For example, if we know x = 1, then...
x^2+y^2 = 1
1^2+y^2 = 1
1+y^2 = 1
y^2 = 1-1
y^2 = 0
y = sqrt(0)
y = 0
Meaning x = 1 leads to y = 0. So (x,y) = (1,0) is one point on the circle. The other parts are handled in a similar fashion.
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Answer:
(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.
(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.
(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.
Step-by-step explanation:
We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.
Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.
Let X = <u><em>back-to-college family spending on electronics</em></u>
SO, X ~ Normal()
The z score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = population mean family spending = $237
= standard deviation = $54
(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)
P(X < $150) = P( <
) = P(Z < -1.61) = 1 - P(Z
1.61)
= 1 - 0.9463 = <u>0.0537</u>
The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.
(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)
P(X > $390) = P( >
) = P(Z > 2.83) = 1 - P(Z
2.83)
= 1 - 0.9977 = <u>0.0023</u>
The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.
(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)
P($120 < X < $175) = P(X < $175) - P(X $120)
P(X < $175) = P( <
) = P(Z < -1.15) = 1 - P(Z
1.15)
= 1 - 0.8749 = 0.1251
P(X < $120) = P( <
) = P(Z < -2.17) = 1 - P(Z
2.17)
= 1 - 0.9850 = 0.015
The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.
Therefore, P($120 < X < $175) = 0.1251 - 0.015 = <u>0.1101</u>