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PtichkaEL [24]
2 years ago
6

How many seconds are in 35 min

Mathematics
2 answers:
kotegsom [21]2 years ago
4 0

Answer: 2,100 seconds

Step-by-step explanation:

There are 60 seconds in 1 minute.

Multiply 60 by 35, and you get 2,100.

So, there are 2,100 seconds in 35 minutes.

Hope this helps you out!

GenaCL600 [577]2 years ago
3 0

Answer:

2100

Step-by-step explanation:

Key: there are 60 secondes in a minute

60x35= 2100

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Use the formula C = pi * d
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3 years ago
Please help me with this
bija089 [108]

Answer: 20 inches

Step-by-step explanation:  You will use the equation 48 x 28 x N = 26,880. To find the solution divide 48 and 28 from 26,880 to get N. To check your work just multiply what you got from N by 48 and 28.

3 0
3 years ago
An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect
Serggg [28]

Answer:

k = 1

P(x > 3y) = \frac{2}{3}

Step-by-step explanation:

Given

f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y  \leq x }  & { \text 0, { elsewhere. } } \end{array} \right.

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1

Where

{ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y  \leq x }

Substitute values for the interval of x and y respectively

So, we have:

\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1

Isolate k

k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1

Integrate y, leave x:

k \int\limits^2_{0} y {dx} \, [0,x/2]= 1

Substitute 0 and x/2 for y

k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1

k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1

Integrate x

k * \frac{x^2}{2*2} [0,2]= 1

k * \frac{x^2}{4} [0,2]= 1

Substitute 0 and 2 for x

k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1

k *[ \frac{4}{4} - \frac{0}{4} ]= 1

k *[ 1-0 ]= 1

k *[ 1]= 1

k = 1

Solving (b): P(x > 3y)

We have:

f(x,y) = k

Where k = 1

f(x,y) = 1

To find P(x > 3y), we use:

\int\limits^a_b \int\limits^a_b {f(x,y)}

So, we have:

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0  dxdy

Integrate x leave y

P(x > 3y) = \int\limits^2_0  x [0,y/3]dy

Substitute 0 and y/3 for x

P(x > 3y) = \int\limits^2_0  [y/3 - 0]dy

P(x > 3y) = \int\limits^2_0  y/3\ dy

Integrate

P(x > 3y) = \frac{y^2}{2*3} [0,2]

P(x > 3y) = \frac{y^2}{6} [0,2]\\

Substitute 0 and 2 for y

P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}

P(x > 3y) = \frac{4}{6} -\frac{0}{6}

P(x > 3y) = \frac{4}{6}

P(x > 3y) = \frac{2}{3}

8 0
2 years ago
Please help. Thanks. Work is due tomorrow. Thanks
eimsori [14]
2 hole pizzas equal 16 because 1 pizza has 8 slices so make 1/4 a decimal which is 4 4/16 equals 4 so for 4 friends 1/4 is left and do the same to the 21/3 which equals 7 so 7 +4 = half the amount to fead the friends
7 0
3 years ago
Ap calc multiple choice question; attached below<br> please explain how, please!
anygoal [31]

Answer:

C. \frac{f(b)-f(1)}{b-1}=20

General Formulas and Concepts:

<u>Calculus</u>

  • Mean Value Theorem (MVT) - If f is continuous on interval [a, b], then there is a c∈[a, b] such that  f'(c)=\frac{f(b)-f(a)}{b-a}
  • MVT is also Average Value

Step-by-step explanation:

<u>Step 1: Define</u>

f(x)=e^{2x}

f'(c) = 20

Interval [1, b]

<u>Step 2: Check/Identify</u>

Function [1, b] is continuous.

Derivative [1, b] is continuous.

∴ There exists a c∈[1, b] such that f'(c)=\frac{f(b)-f(a)}{b-a}

<u>Step 3: Mean Value Theorem</u>

  1. Substitute:                    20=\frac{ f(b)-f(1)}{b-1}
  2. Rewrite:                        \frac{ f(b)-f(1)}{b-1}=20

And we have our final answer!

5 0
2 years ago
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