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algol [13]
3 years ago
12

Assume a standard deck of 52 cards is being used. Find the possible way 2 black cards can be chose

Mathematics
1 answer:
GREYUIT [131]3 years ago
7 0

We know there are 4 suites the deck can be divided into, and two of the 4 are black. So you divide 2/4, and get 1/2. So the probability that a black card will be chosen is 1/2.

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Evaluate expression 20+3c+17d when c=8 and d=2.
Tems11 [23]
Just plug in the numbers for the variables.


20+3(8)+17(2)=
3 0
4 years ago
Can you Solve this plzzz nerd ur help<br><br>6÷2(1+2)=​
Mkey [24]

Answer:

9

Step-by-step explanation:

(do 1+2 first) then (6 divided by 2) then 3x3 and you will get 9

3 0
3 years ago
Cot(pi/2)=<br><br>1<br>-1<br>undefined<br>0
GenaCL600 [577]

Answer:

the answer is zeroo

Step-by-step explanation:


4 0
4 years ago
Read 2 more answers
Suppose u = (1,2,3), v = (2, -3, -4), and w = (1,0,-2). (a) Find u-2v + 3w. (b) Write the vector (3,4,5) as a linear combination
marishachu [46]
<h2>Answer with explanation:</h2>

We are given vectors u,v and w as follows:

      u=(1,2,3)\,\ v=(2,-3,-4),\ and\ w=(1,0,-2)

(a)

Now we are asked to find the value of:

  u-2v+3w

The value is calculated as follows:

u-2v+3w=(1,2,3)-2(2,-3,-4)+3(1,0,-2)\\\\i.e.\\\\u-2v+3w=(1,2,3)+(-2\times 2,-2\times -3,-2\times -4)+(3\times 1,3\times 0,3\times -2)\\\\i.e.\\\\u-2v+3w=(1,2,3)+(-4,6,8)+(3,0,-6)\\\\i.e.\\\\u-2v+3w=(1-4+3,2+6+0,3+8-6)\\\\i.e.\\\\u-2v+3w=(0,8,5)

                              Hence,

                 u-2v+3w=(0,8,5)

(b)

We are asked to represent (3,4,5) as a linear combination of u, v, and w.

Let:

(3,4,5)=au+bv+cw

(3,4,5)=a(1,2,3)+b(2,-3,-4)+c(1,0,-2)

i.e.

(3,4,5)=(a,2a,3a)+(2b,-3b,-4b)+(c,0,-2c)\\\\i.e.\\\\(3,4,5)=(a+2b+c,2a-3b,3a-4b-2c)\\\\i.e.\\\\a+2b+c=3----------(1)

2a-3b=4-------------(2)

3a-4b-2c=5------------(3)

on multiplying equation(1) by 2 an then adding equation (1) to equation (3)

5a=11\\\\i.e.\\\\a=\dfrac{11}{5}

on putting the value of a in equation (2) we get:

2\times \dfrac{11}{5}-3b=4\\\\i.e.\\\\\dfrac{22}{5}-3b=4\\\\i.e.\\\\3b=\dfrac{22}{5}-4\\\\i.e.\\\\3b=\dfrac{2}{5}\\\\i.e.\\\\b=\dfrac{2}{15}

Now on putting the value of a and b in equation (1)

\dfrac{11}{5}+2\times \dfrac{2}{15}+c=3\\\\i.e.\\\\\dfrac{11}{5}+\dfrac{4}{15}+c=3\\\\i.e.\\\\c=3-\dfrac{11}{5}-\dfrac{4}{15}\\\\i.e.\\\\c=\dfrac{3\times 15-11\times 3-4}{15}\\\\i.e.\\\\c=\dfrac{45-33-4}{15}\\\\i.e.\\\\c=\dfrac{8}{15}

Hence, we have:

(3,4,5)=\dfrac{11}{5}(1,2,3)+\dfrac{2}{15}(2,-3,-4)+\dfrac{8}{15}(1,0,-2)\\\\i.e.\\\\(3,4,5)=\dfrac{11}{5}u+\dfrac{2}{15}v+\dfrac{8}{15}w

5 0
4 years ago
Identify the reflection of the figure with vertices E(8,4), F(-16,-8), and G(24,-16) across the y-axis.
Dafna1 [17]

Answer:

\boxed{\text{E$'$(-8, 4), F$'$(16,-8), G$'$(-24,-16)}}

Step-by-step explanation:

When you reflect a point (x, y) in the y-axis, the y-coordinate remains the same, but the x-coordinate gets the opposite sign. Thus,

E (8,4) ⟶ E' (-8,4)

F (-16,-8) ⟶ F' (16,-8)

G (24,-16) ⟶ G' (-24,-16)

\text{The reflected figure has coordinates } \boxed{\textbf{E$'$(-8, 4), F$'$(16,-8), G$'$(-24,-16)}}

The figure EFG and its reflection E'F'G' are shown in the diagram below.

6 0
3 years ago
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