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3 years ago

The ratio 25:45 in simplest form is

Mathematics

2 answers:

Inessa05 [86]3 years ago

7 0

Answer:

Simplest form 5 : 9.

Step-by-step explanation:

Given : ratio 25:45

To find : Convert it into simplest form.

Solution : We have given 25:45

On dividing both by 5

25 ÷5 : 45 ÷ 5

5 : 9

There is no common factor between them so, we can to divide further.

Therefore, Simplest form 5 : 9.

stepladder [879]3 years ago

5 0

25/45=(5∗5)/(5∗9)=5/9

5/9 is the answer.

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Twice the sum of two numbers is 24. if one number is 2 times larger than the other, what is the smaller of the two numbers?

Vikki [24]

2(x+y)=24
2x=y

Use substitution. If y=2x, than we can substitute 2x for it in the other equation, because x and y equal the same things in both equations.

So using that we get 2(x+2x)=24

SIMPLIFY
2(3x)=24

DISTRIBUTE
6x=24.

DIVIDE: x=4.

If we substitute x into the original equations we get x=4 and y=8. Those are your two numbers.

Have a nice day! :)

4 0

3 years ago

Perform the indicated operation. (-7) • (-5)<br> -12 <br> 35<br> 12<br> -35

Flauer [41]

Hi,

Equation:

$- 7 \times (- 5)$

Multiplying two negatives equals a positive.

$( - ) \times ( - ) = ( + )$

Now equation looks like:

$7 \times 5 = 35$

Hope this helps.
r3t40

6 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;

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