Answer:
6 cm
Step-by-step explanation:
If you use Tangent-secant product (chapter reference), AB/AC = AD/AB so 4/2 = AD/4. AD = 8, CD = AD - AC = 8 - 2 = 6 cm.
Answer:
The correct option is (b).
Step-by-step explanation:
If X N (µ, σ²), then , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z N (0, 1).
The distribution of these z-variate is known as the standard normal distribution.
The mean and standard deviation of the active minutes of students is:
<em>μ</em> = 60 minutes
<em>σ </em> = 12 minutes
Compute the <em>z</em>-score for the student being active 48 minutes as follows:
Thus, the <em>z</em>-score for the student being active 48 minutes is -1.0.
The correct option is (b).
Answer:
The estimated probability that Ginger will eat a a pizza everyday of the week is;
D. 8/10 = 80%
Step-by-step explanation:
The given parameters are;
The frequency with which Ginger buys launch = Everyday
The percentage of the time the cafeteria has pizza out = 80%
The outcome of 0 and 1 = No pizza available
The outcome of 2, 3, 4, 5, 6, 7, 8, and 9 = Pizza available
Therefore, we have the;
Group number Percentage of time pizza available
1 80%
2 80%
3 80%
4 80%
5 40%
6 100%
7 80%
8 100%
9 80%
10 80%
Therefore, the sum of the percentages outcome the days Ginger eats pizza = 0.8 + 0.8 + 0.8 + 0.8 + 0.4 + 1 + 0.8 + 1 + 0.8 + 0.8 = 8
The number of runs of simulation = 10 runs
The estimated probability that Ginger will eat a a pizza everyday of the week = (The sum of the percentages outcome the days Ginger eats pizza)/(The number of runs of simulation)
∴ The estimated probability that Ginger will eat a a pizza everyday of the week = 8/10
60 / 2 = 30
30 * 5 = 150
he had $150 at first
60, the amount he was left with was 2/5 or 40% of his original amount
By way of example, suppose <em>A</em> = {1, 2, 3} and <em>B</em> = {<em>a</em>, <em>b</em>, <em>c</em>}. Then the Cartesian product of <em>A</em> and <em>B</em> is
<em>A</em> × <em>B</em> = {{1, <em>a</em>}, {1, <em>b</em>}, {1, <em>c</em>}, {2, <em>a</em>}, {2, <em>b</em>}, {2, <em>c</em>}, {3, <em>a</em>}, {3, <em>b</em>}, {3, <em>c</em>}}
That is, each element in <em>A</em> gets a pairing with each element in <em>B</em>, and for each pairing you have <em>n(A)</em> choices for the first element and <em>n(B)</em> choices for the second element.
So if <em>n(A)</em> = <em>p</em> and <em>n(B)</em> = <em>q</em>, then <em>n(A</em> × <em>B)</em> = <em>pq</em>.