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loris [4]
3 years ago
8

A bag contains 2 white marbles, 4 blue marbles, and 6 red marbles. if a marble is drawn from the bag, what is the probability th

at it is not white?
Mathematics
1 answer:
Lerok [7]3 years ago
8 0
5/6. 10/12 reduce is 5/6
10 are not white 12 is the total
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N is prime given that it has 2 digits
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Minimum 11, Maximum 97
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Students at Hampton Middle School sold T-shirts as a school fundraiser. Sylvie asked 12 random seventh-grade students how many T
AveGali [126]

Answer:

b. 4.1 shirts

Step-by-step explanation:

Given data:

number of terms = 12

Terms given are 3, 4, 8, 5, 2, 5, 0, 5, 3, 4, 3, 7

Mean = (sum of terms)/ (number of terms)

Mean = (3 +4+ 8+ 5+2+5+0+ 5+ 3+ 4+3+ 7)/12

Mean = 49/12

Mean = 4.083

Mean = 4.1 (<em>to the nearest tenth)</em>

7 0
3 years ago
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A meuseum had x visitors on friday. On Saturday it had 150 more visitors than it did on friday. The total number of visitors on
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Answer:

75 people.

Step-by-step explanation:

If Friday is x, then Saturday is 150 + x. The number of people on Sunday then is (Friday + Saturday)/2, thus (x + (150 + x))/2. This then simplifies to (x + 150 + x)/2 ---> (2x + 150)/2 ---> 2x/2 + 150/2 ---> x + 75. So, Friday is x and Sunday is x + 75, to find out how many more visitors the museum have on Sunday than Friday, you subtract the two numbers. (x + 75) - x = 75.

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3 years ago
Help anyone ??? Please
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Angle F is equal to angle G so 5x + 18 = 7x-12 so you would solve for x which would equal 15
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All vectors are in Rn. Check the true statements below:
Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that ||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}. Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then ||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||.

C) Consider S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2. This set is orthogonal because (0,2)\cdot(2,0)=0(2)+2(0)=0, but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in \mathbb{R}^n. Then the columns of A form an orthonormal set. We have that A^{-1}=A^t. To see this, note than the component b_{ij} of the product A^t A is the dot product of the i-th row of A^t and the jth row of A. But the i-th row of A^t is equal to the i-th column of A. If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then A^t A=I    

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set \{u_1,u_2\cdots u_p\} and suppose that there are coefficients a_i such that a_1u_1+a_2u_2\cdots a_nu_n=0. For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then a_i||u_i||=0 then a_i=0.  

5 0
3 years ago
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