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Kay [80]
3 years ago
10

What t represent for the equation a(t)=210-15t

Mathematics
1 answer:
vampirchik [111]3 years ago
4 0
T represents x (the input) for the equation.
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Anoki bought 360 millimeters of fabric. How many centimeters of fabric does Anoki have? It’s either 360, 36.0,3.60, or 0.360
Stels [109]
Hey!


First, we need to know how many millimeters are in a centimeter.

<em>10 mm = 1 cm</em>

Okay, now we can write an equation.

<em>360 mm ÷ 10</em>

To solve this equation, we simply divide.

<em>360 mm ÷ 10 = 36 cm</em>

<em>So, if </em><span><em>Anoki bought 360 millimeters of fabric, that means he has </em> 36.0 centimeters  <em>of fabric.</em>

Hope this helps!


- Lindsey Frazier ♥</span>
8 0
3 years ago
F(x)= 1/2x^2-4x+3 in standard form
Finger [1]

Answer:

Already in standard form

Step-by-step explanation:

A quadratic equations standard form is ax^2 + bx + c. This equation is already in standard form.

ax^2 + bx + c

1/2 x^2 - 4x + 3

5 0
3 years ago
2, 3, 5, 9, 17...
BARSIC [14]

Answer:

Step-by-step explanation:

i'm sure the answer is C

8 0
3 years ago
Read 2 more answers
Customers arrive at a service facility according to a Poisson process of rate λ customers/hour. Let X(t) be the number of custom
mash [69]

Answer:

Step-by-step explanation:

Given that:

X(t) = be the number of customers that have arrived up to time t.

W_1,W_2... = the successive arrival times of the customers.

(a)

Then; we can Determine the conditional mean E[W1|X(t)=2] as follows;

E(W_!|X(t)=2) = \int\limits^t_0 {X} ( \dfrac{d}{dx}P(X(s) \geq 1 |X(t) =2))

= 1- P (X(s) \leq 0|X(t) = 2) \\ \\ = 1 - \dfrac{P(X(s) \leq 0 , X(t) =2) }{P(X(t) =2)}

=  1 - \dfrac{P(X(s) \leq 0 , 1 \leq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}

=  1 - \dfrac{P(X(s) \leq 0 ,P((3 \eq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}

Now P(X(s) \leq 0) = P(X(s) = 0)

(b)  We can Determine the conditional mean E[W3|X(t)=5] as follows;

E(W_1|X(t) =2 ) = \int\limits^t_0 X (\dfrac{d}{dx}P(X(s) \geq 3 |X(t) =5 )) \\ \\  = 1- P (X(s) \leq 2 | X (t) = 5 )  \\ \\ = 1 - \dfrac{P (X(s) \leq 2, X(t) = 5 }{P(X(t) = 5)} \\ \\ = 1 - \dfrac{P (X(s) \LEQ 2, 3 (t) - X(s) \leq 5 )}{P(X(t) = 2)}

Now; P (X(s) \leq 2 ) = P(X(s) = 0 ) + P(X(s) = 1) + P(X(s) = 2)

(c) Determine the conditional probability density function for W2, given that X(t)=5.

So ; the conditional probability density function of W_2 given that  X(t)=5 is:

f_{W_2|X(t)=5}}= (W_2|X(t) = 5) \\ \\ =\dfrac{d}{ds}P(W_2 \leq s | X(t) =5 )  \\ \\  = \dfrac{d}{ds}P(X(s) \geq 2 | X(t) = 5)

7 0
3 years ago
Highest to lowest -0.6,0.38, -0.09, 0.7
algol13

Hey there!

Guide:

• NEGATIVES are BELOW zero

• POSITIVES are ABOVE zero

• DECIMALS are BELOW zero

Highest to lowest: –0.6, 0.38, –0.09, 0.7

0.7 ⬅️ HIGHEST

0.38 ⬅️ 2nd HIGHEST

–0.09 3rd HIGHEST ⬅️

–0.6 ⬅️ LOWEST

Answer: 0.7 , 0.38, –0.09, and –0.6 ☑️

Good luck on your assignment and enjoy your day!

~LoveYourselfFirst:)

3 0
3 years ago
Read 2 more answers
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