Answer:
t
=
26
∘
57
+
k
360
∘
Step-by-step explanation:
tan
t
=
1
2
Calculator and unit circle give 2 solutions for (0, 360) -->
t
=
26
∘
57
, and
t
=
180
+
26.57
=
206
∘
57
General answer:
t
=
26
∘
57
+
k
360
∘
Answer:
Domain is the values that x can take. In terms of real numbers x can take any value and the function will make sense. Since this is linear function the range is not limited too. So the range and domain coincide and are (-infinity;infinity)
Answer:
0.6708 or 67.08%
Step-by-step explanation:
Helen can only make both free throws if she makes the first. The probability that she makes the first free throw is P(C) = 0.78, now given that she has already made the first one, the probability that she makes the second is P(D|C) = 0.86. Therefore, the probability of Helen making both free throws is:
![P(C+D) = P(C) *P(D|C) = 0.78*0.86\\P(C+D) = 0.6708](https://tex.z-dn.net/?f=P%28C%2BD%29%20%3D%20%20P%28C%29%20%2AP%28D%7CC%29%20%3D%200.78%2A0.86%5C%5CP%28C%2BD%29%20%3D%200.6708)
There is a 0.6708 probability that Helen makes both free throws.
Answer:
Keenan's z-score was of 0.61.
Rachel's z-score was of 0.81.
Step-by-step explanation:
Z-score:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Keenan scored 80 points on an exam that had a mean score of 77 points and a standard deviation of 4.9 points.
This means that ![X = 80, \mu = 77, \sigma = 4.9](https://tex.z-dn.net/?f=X%20%3D%2080%2C%20%5Cmu%20%3D%2077%2C%20%5Csigma%20%3D%204.9)
So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{80 - 77}{4.9}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B80%20-%2077%7D%7B4.9%7D)
![Z = 0.61](https://tex.z-dn.net/?f=Z%20%3D%200.61)
Keenan's z-score was of 0.61.
Rachel scored 78 points on an exam that had a mean score of 75 points and a standard deviation of 3.7 points.
This means that
. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{75 - 78}{3.7}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B75%20-%2078%7D%7B3.7%7D)
![Z = 0.81](https://tex.z-dn.net/?f=Z%20%3D%200.81)
Rachel's z-score was of 0.81.