A shopping center keeps track of the number of customers in each store at lunchtime. The data shows the number of customers in t
he 15 different stores in the shopping center last Sunday.
4, 18, 20, 17, 16, 23, 19, 14, 8, 8, 6, 12, 10, 14, 18
Create a histogram of this data.
To create a histogram, hover over each number of customers range on the x-axis. Then click and drag up to plot the data.
4, 18, 20, 17, 16, 23, 19, 14, 8, 8, 6, 12, 10, 14, 18
Create a histogram of this data.
To create a histogram, hover over each number of customers range on the x-axis. Then click and drag up to plot the data.
1 answer:
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Answer:
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Explanation:
This should help you figure out why this is the answer:
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A) There are a number of ways to compute the determinant of a 3x3 matrix. Since k is on the bottom row, it is convenient to compute the cofactors of the numbers on the bottom row. Then the determinant is ...
1×(2×-1 -3×1) -k×(3×-1 -2×1) +2×(3×3 -2×2) = 5 -5k
bi) Π₁ can be written using r = (x, y, z).
Π₁ ⇒ 3x +2y +z = 4
bii) The cross product of the coefficients of λ and μ will give the normal to the plane. The dot-product of that with the constant vector will give the desired constant.
Π₂ ⇒ ((1, 0, 2)×(1, -1, -1))•(x, y, z) = ((1, 0, 2)×(1, -1, -1))•(1, 2, 3)
Π₂ ⇒ 2x +3y -z = 5
c) If the three planes form a sheath, the ranks of their coefficient matrix and that of the augmented matrix must be 2. That is, the determinant must be zero. The value of k that makes the determinant zero is found in part (a) to be -1.
A common approach to determining the rank of a matrix is to reduce it to row echelon form. Then the number of independent rows becomes obvious. (It is the number of non-zero rows.) This form for k=-1 is shown in the picture.
1×(2×-1 -3×1) -k×(3×-1 -2×1) +2×(3×3 -2×2) = 5 -5k
bi) Π₁ can be written using r = (x, y, z).
Π₁ ⇒ 3x +2y +z = 4
bii) The cross product of the coefficients of λ and μ will give the normal to the plane. The dot-product of that with the constant vector will give the desired constant.
Π₂ ⇒ ((1, 0, 2)×(1, -1, -1))•(x, y, z) = ((1, 0, 2)×(1, -1, -1))•(1, 2, 3)
Π₂ ⇒ 2x +3y -z = 5
c) If the three planes form a sheath, the ranks of their coefficient matrix and that of the augmented matrix must be 2. That is, the determinant must be zero. The value of k that makes the determinant zero is found in part (a) to be -1.
A common approach to determining the rank of a matrix is to reduce it to row echelon form. Then the number of independent rows becomes obvious. (It is the number of non-zero rows.) This form for k=-1 is shown in the picture.
Answer:
1. 7 outcomes.
2. 64 outcomes.
3. D and H
Step-by-step explanation:
1. The experimenter will record the total number of heads after flipping the coin 6 times, therefore, the possible outcomes are: the experimenter got 0 heads, the experimenter got 1 head,..., the experimenter got 6 heads. Therefore there are 7 possible outcomes: (0, 1, 2, 3, 4, 5, 6).
2. If the experimenter records the result after each flip (H or T), for the first flip she has 2 results (heads or tails), for the second one she has 2 results, same for the third, fourth, fifth and sixth flip.
Therefore the possible outcomes are: .
3. Now the experimenter will end up the experiment earlier if she gets two tails or two heads in a row.
A.TTHTT is not possible because in this case she would've ended up the experiment after the second flip since she got two tails.
B.TTH is not possible either for the same reason.
C. HT is not possible because she still needs to get another T or two H in a row.
D. This is a possible outcome because in this one she flipped the coin 6 times and she didn't get two heads or two tails in a row so she didn't finish it earlier.
E. Is not possible because she should've stopped after the second flip.
F. Is not possible because it has 7 letters, therefore, 7 flips.
G. Is not possible because she should've stopped once she got heads in the fourth and fifth flip.
H. Is possible because she got heads in the second and third flip.