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frosja888 [35]
3 years ago
11

Find parametric equations through point P=(2,2,7) in the direction of the vector v = (44, 14, -20)?

Mathematics
1 answer:
ycow [4]3 years ago
5 0

9514 1404 393

Answer:

  (x, y, z) = (2+44t, 2+14t, 7-20t)

Step-by-step explanation:

One way to write parametric equations for line L is ...

  L = P + t·<em>v</em>

where P is the given point and <em>v</em> is the given direction vector. Using that form, we have ...

  (x, y, z) = (2+44t, 2+14t, 7-20t)

__

If you like, you can remove a common factor of 2 from the coefficients of t.

  (x, y, z) = (2+22t, 2+7t, 7-10t)

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James and Lin got married and got new jobs. Lin earns ​$3100 more per year than James. Together they earn ​$81,960. How much doe
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Step-by-step explanation:

We can turn this into an equation.

Let’s denote the amount of money James makes as X.

Then Lin makes X + 3100.

Together they make 81960.

So combine their earning expressions.

X + X + 3100 = 81960

Solve for X

X = 39430

So James makes $39430 since his earnings are demoted by x.

Lin then makes 3100 + 39430, which is $42530.

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3 years ago
Find two consecutive integers whose sum is 19
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The answer is 9 and 10.
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The daily sales S (in thousands of dollars) that are attributed to an advertising campaign are given by S = 11 + 3 t + 3 − 18 (t
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The parking rate at the hospital a parking garage are 50 cents for the 1st hour and 25 cents for each additional hour if guan part in the hospital parking garage for 8 hours how much will the total of charge for parking be

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Gizmos come in a carton. A carton holds 6 packages. Each package holds 10 small boxes. Each small box holds 12 gizmos. How many
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720

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8 0
2 years ago
You are to play the following game of cards. Cards are worth their face value, Jacks, Queens and Kings are also worth 10 and Ace
ryzh [129]

Answer:

$82.875

Step-by-step explanation:

From the given information:

Assume F is used to denote the two cards;

If there are four aces among 52 playing cards, the chance of selecting the first ace is =  \dfrac{4}{52}

After selecting the first ace, we have only 3 aces remaining and a total of 51 playing cards. Thus, the chance of selecting the second ace will be \dfrac{3}{51}

By applying product rule, we can determine the chance of selecting two aces without replacement as follows:

i.e.

P(F=22)=\dfrac{4}{52}\times \dfrac{3}{51}

= \dfrac{1}{221}

The probability of getting one ace, one face is:

P(F=21) =( \dfrac{4}{52}\times \dfrac{12}{51})+( \dfrac{12}{52}\times \dfrac{4}{51})

P(F=21) = \dfrac{4}{221}\times\dfrac{4}{221}

P(F=21) = \dfrac{8}{221}

Since there are 4 aces, 4 nine, and 12 faces in a card deck

The probability of getting one ace, one nine, or two faces now will be:

P(F=20) = (\dfrac{4}{52} \times \dfrac{4}{51})+ (\dfrac{4}{52} \times \dfrac{4}{51}) + (\dfrac{12}{52} \times \dfrac{11}{51})

P(F=20) = (\dfrac{53}{663}  )

Now, the probability  of at least 20 now is:

\text{P(F at least 20)} = \dfrac{1}{221}+\dfrac{8}{221}+\dfrac{53}{663}

\text{P(F at least 20)} = \dfrac{80}{663}

If H represents the amount of prize of the expected winnings:

Then;

(H - 10) (\dfrac{80}{663}) + (-10)(\dfrac{663-80}{663}) = 0

\dfrac{80(H-10)}{663}-\dfrac{5830}{663}=0

\dfrac{80(H-10)}{663}=\dfrac{5830}{663}

80H - 800 = 5830

80H = 5830 +800

80H = 6630

H = 6630/80

H = $82.875

The prize should be $82.875 to make a winning positive.

4 0
3 years ago
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