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2 years ago

6

How many seconds are in 35 min

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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The circumference of a circle

Dmitry_Shevchenko [17]

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.

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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>

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

And we have our final answer!

5 0

2 years ago