What is 1/4 less than -3/4

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

ycow [4]

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·v

where P is the given point and v 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)

5 0

2 years ago

Your friend says she used a common

AfilCa [17]

Step-by-step explanation:

An equivalent fraction can be found by either multiplying the numerator and denominator by the same number or dividing the numerator and the denominator by the same number.

7 0

2 years ago

Read 2 more answers

302.406 + 7,710.826 What is the sum?

Alexus [3.1K]

Hi! I believe your answer is 8,013.232. The work is shown below. I hope this helps you! Good luck and have a great day. ❤️✨

4 0

3 years ago

Read 2 more answers

Round 285.39154 to a specified place of the hundreths place

almond37 [142]

Rounding the number to the hundreds place, the answer would be 285.39.

Hope this helped!

Nate

5 0

2 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c) The probability that exactly 1 of the next three vehicles passes is 0.189.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let X = number of vehicles that pass the inspection.

The probability of the random variable X is P (X) = 0.70.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

$=(0.70)^{3}\\=0.343$

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

$=1-0.343\\=0.657$

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

+ P (Only 3rd vehicle passes)

$=(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189$

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

+ P (0 vehicles passes)

$=0.189+(0.30\times0.30\times0.30)\\=0.216$

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let X = all 3 vehicle passes and Y = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

$P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525$

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0

2 years ago