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3 years ago

Urgent. Explain in words, how you would plot (0, 4) on a coordinate plane.

Mathematics

2 answers:

Naddika [18.5K]3 years ago

8 0

0 means u do not move right or left at all and the 4 means u move up 4 units.

MArishka [77]3 years ago

7 0

Answer:

in the ordered pair, (0,4), the zero represents the x number, meaning that you would not move in any direction on the x axis( it stays on the origin). the 4 means you will move 4 units up the y axis. in the end you will have one point that in 4 units up from the origin.

Step-by-step explanation:

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The perimeter of a scalene triangle is 14.5 cm. The longest side is twice that of the shortest side. Which equation can be used

e-lub [12.9K]

The answer is:
a = c ÷ 2
b = P - a - c

The perimeter is the sum of all sides of a polygon. The perimeter of the triangle is:
P = a + b + c
where
P - perimeter of the triangle,
a, b, c - sides of the triangle.

It is given:
P = 14.5 cm
c = 6.2 cm

If, the longest side (c) is twice that of the shortest side (a), then
c = 2a
Therefore, the equation that can be used to find the length of the shortest side is a = c ÷ 2.

⇒ a = c ÷ 2 = 6.2 cm ÷ 2
⇒ a = 3.1 cm

By knowing two sides (c and a) and perimeter (P), we can calculate the length of the third side (b).
If P = a + b + c, then the equation that can be used to find the length of the side b is b = P - a - c.

⇒ b = P - a - c = 14.5 cm - 6.2 cm - 3.1 cm
⇒ b = 5.2 cm

6 0

2 years ago

Suiting at 6 a.m., cars, buses, and motorcycles arrive at a highway loll booth according to independent Poisson processes. Cars

dem82 [27]

Answer:

Step-by-step explanation:

From the information given:

the rate of the cars = $\dfrac{1}{5} \ car / min = 0.2 \ car /min$

the rate of the buses = $\dfrac{1}{10} \ bus / min = 0.1 \ bus /min$

the rate of motorcycle = $\dfrac{1}{30} \ motorcycle / min = 0.0333 \ motorcycle /min$

The probability of any event at a given time t can be expressed as:

$P(event \ (x) \ in \ time \ (t)\ min) = \dfrac{e^{-rate \times t}\times (rate \times t)^x}{x!}$

∴

(a)

$P(2 \ car \ in \ 20 \ min) = \dfrac{e^{-0.20\times 20}\times (0.2 \times 20)^2}{2!}$

$P(2 \ car \ in \ 20 \ min) =0.1465$

$P ( 1 \ motorcycle \ in \ 20 \ min) = \dfrac{e^{-0.0333\times 20}\times (0.0333 \times 20)^1}{1!}$

$P ( 1 \ motorcycle \ in \ 20 \ min) = 0.3422$

$P ( 0 \ buses \ in \ 20 \ min) = \dfrac{e^{-0.1\times 20}\times (0.1 \times 20)^0}{0!}$

$P ( 0 \ buses \ in \ 20 \ min) = 0.1353$

Thus;

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.1465 × 0.3422 × 0.1353

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.0068

(b)

the rate of the total vehicles = 0.2 + 0.1 + 0.0333 = 0.3333

the rate of vehicles with exact change = rate of total vehicles × P(exact change)

$= 0.3333 \times \dfrac{1}{4}$

= 0.0833

∴

$P(zero \ exact \ change \ in \ 10 minutes) = \dfrac{e^{-0.0833\times 10}\times (0.0833 \times 10)^0}{0!}$

P(zero exact change in 10 minutes) = 0.4347

The probability of the 7th motorcycle after the arrival of the third motorcycle is:

$P( 4 \ motorcyles \ in \ 45 \ minutes) =\dfrac{e^{-0.0333\times 45}\times (0.0333 \times 45)^4}{4!}$

$P( 4 \ motorcyles \ in \ 45 \ minutes) =0.0469$

Thus; the probability of the 7th motorcycle after the arrival of the third one is = 0.0469

P(at least one other vehicle arrives between 3rd and 4th car arrival)

= 1 - P(no other vehicle arrives between 3rd and 4th car arrival)

The 3rd car arrives at 15 minutes

The 4th car arrives at 20 minutes

The interval between the two = 5 minutes

For Bus:

P(no other vehicle other vehicle arrives within 5 minutes is)

= $\dfrac{6}{12} = 0.5$

For motorcycle:

$= \dfrac{2 }{12} = \dfrac{1 }{6}$

∴

The required probability = $1 - \Bigg ( \dfrac{e^{-0.5 \times 0.5^0}}{0!} \times \dfrac{e^{-1/6}\times (1/6)^0}{0!} \Bigg)$

= 1- 0.5134

= 0.4866

6 0

2 years ago

I am very stuck with this

Bingel [31]

Answer:

see explanation

Step-by-step explanation:

To multiply the vector by a scalar, multiply each of the elements by the scalar.

To add 3 vectors add the corresponding elements of each vector

2a + 3b + 4c

= 2 $\left[\begin{array}{ccc}2\\3\\\end{array}\right]$ + 3 $\left[\begin{array}{ccc}-4\\1\\\end{array}\right]$ + 4 $\left[\begin{array}{ccc}3\\2\\\end{array}\right]$

= $\left[\begin{array}{ccc}4\\6\\\end{array}\right]$ + $\left[\begin{array}{ccc}-12\\3\\\end{array}\right]$ + $\left[\begin{array}{ccc}12\\8\\\end{array}\right]$