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ICE Princess25 [194]
3 years ago
10

Write the number 63 in four ways

Mathematics
2 answers:
Sever21 [200]3 years ago
7 0
60+3=63
30+33=63
126/2=63
3x21=63
Scorpion4ik [409]3 years ago
6 0
60+3=63
 10x6+3=63
 7x9=63
 20x3+3=63
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Helpppppppp please i dont understand ​
RUDIKE [14]

Answer:

(a) 64 days

(b) 2 phones calls

Step-by-step explanation:

8 0
3 years ago
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Find an equation of the plane orthogonal to the line
jolli1 [7]

The given line is orthogonal to the plane you want to find, so the tangent vector of this line can be used as the normal vector for the plane.

The tangent vector for the line is

d/d<em>t</em> (⟨0, 9, 6⟩ + ⟨7, -7, -6⟩<em>t </em>) = ⟨7, -7, -6⟩

Then the plane that passes through the origin with this as its normal vector has equation

⟨<em>x</em>, <em>y</em>, <em>z</em>⟩ • ⟨7, -7, -6⟩ = 0

We want the plane to pass through the point (9, 6, 0), so we just translate every vector pointing to the plane itself by adding ⟨9, 6, 0⟩,

(⟨<em>x</em>, <em>y</em>, <em>z</em>⟩ - ⟨9, 6, 0⟩) • ⟨7, -7, -6⟩ = 0

Simplifying this expression and writing it standard form gives

⟨<em>x</em> - 9, <em>y</em> - 6, <em>z</em>⟩ • ⟨7, -7, -6⟩ = 0

7 (<em>x</em> - 9) - 7 (<em>y</em> - 6) - 6<em>z</em> = 0

7<em>x</em> - 63 - 7<em>y</em> + 42 - 6<em>z</em> = 0

7<em>x</em> - 7<em>y</em> - 6<em>z</em> = 21

so that

<em>a</em> = 7, <em>b</em> = -7, <em>c</em> = -6, and <em>d</em> = 21

4 0
3 years ago
How do you write a word problem using 5621÷23
zepelin [54]
Megan has $5,621 to give to charity. She wanrs to give the money to 23 different charities. Each charity will get the same amount. How much money will each charity get? How much will be left over? Answer: 22r9
4 0
3 years ago
A number is selected, at random, from the set {1,2,3,4,5,6,7,8}.
Olegator [25]

Applying the formula, you have:

A = the number is prime

B = the number is odd

I assume that with "random" you imply that all numbers can be chosen with the same probability. So, we have

P(A) = \dfrac{4}{8} = \dfrac{1}{2}

because 4 out of 8 numbers are prime: 2, 3, 5 and 7.

Similarly, we have

P(B) = \dfrac{4}{8} = \dfrac{1}{2}

because 4 out of 8 numbers are odd: 1, 3, 5 and 7.

Finally,

P(A \land B) = \dfrac{3}{8}

because 3 out of 8 numbers are prime and odd: 3, 5 and 7.

So, applying the formula, we have

P(\text{prime } | \text{ odd}) = \dfrac{P(\text{prime and odd})}{P(\text{odd})} = \dfrac{\frac{3}{8}}{\frac{1}{2}} = \dfrac{3}{8}\cdot 2 = \dfrac{3}{4}

Note:

I think that it is important to have a clear understanding of what's happening from a conceptual point of you: conditional probability simply changes the space you're working with: you are not asking "what is the probability that a random number, taken from 1 to 8, is prime?"

Rather, you are adding a bit of information, because you are asking "what is the probability that a random number, taken from 1 to 8, is prime, knowing that it's odd?"

So, we're not working anymore with the space {1,2,3,4,5,6,7,8}, but rather with {1,3,5,7} (we already know that our number is odd).

Out of these 4 odd numbers, 3 are primes. This is why the probability of picking a prime number among the odd numbers in {1,2,3,4,5,6,7,8} is 3/4: they are literally 3 out of 4.

5 0
3 years ago
A deli offers a lunch special that comes with a soup, a sandwich., and a desert. The soup choices are tomato or onion; the sandw
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8 0
3 years ago
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