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3 years ago

How to do long division problem 5 devided by 98

Mathematics

2 answers:

murzikaleks [220]3 years ago

4 0

98 ÷ 5 = 19.6

19 × 5 = 95

98 - 95 = 3

98 <span>÷ 5 = 19 R3</span>

aleksley [76]3 years ago

4 0

1 9 r3
5 (98
5 /
-48
45
- 3

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Question 3 Unsaved

bazaltina [42]

Answer:

Step-by-step explanation:

DE given is

y''+8y'+15y=0, y(0)=0, y'(0)=1

Take Laplace on the DE

We get

$s^2 Y(s) -sY(0) -y'(0)+8(sY(s)-y(0))+15Y(s) =0\\s^2 Y(s) -s(0) -1+8(sY(s)-0)+15Y(s) =0\\Y(s)(s^2+8s+15)-1=0\\Y(s) = \frac{1}{s^2+8s+15}$

Simplify to get

Y(s) = $\frac{1}{2}[ {\frac{1}{s+3} -\frac{1}{s+5} }]$

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3 years ago

A man and a woman agree to meet at a cafe about noon. If the man arrives at a time uniformly distributed between 11:40 and 12:10

kompoz [17]

Answer:

0.3888

Step-by-step explanation:

To solve the exercise we need a neutral time, so we define at 12:00 as time 0.

If the man arrives in a uniformly distributed time between 11:40 and 12:10, the time is distributed from -20 to 10.

If the woman arrives in a time evenly distributed between 11:45 and 12:20, the time is distributed from -15 to 20.

Assuming that each one arrives at 12: X and 12: Y for men and women, then the space of our time is defined as

Man [-20,10]

Woman [-15,20]

So,

$f_x(x)=\frac{1}{10-(-20)}=\frac{1}{30}\\f_y(y)=\frac{1}{20-(-15)}=\frac{1}{35}$

Then,

$f_{xy}(x,y)=f_x(x)f_y(y)=\frac{1}{30} \frac{1}{35} = \frac{1}{1050}$

The probability of finding an arrival less than 5 minutes is

$P(|X-Y|\leq 5) = \int\limit^20_{-15} \int\limit^{x+5}_{x-5} f(x,y)dydx$

$P(|X-Y|\leq 5)= \frac{1}{1050}\int\limit^{20}_{-15} y|^{x+5}_{x-5}dx$

$P(|X-Y|\leq 5)= \frac{35*10}{900}=0.3888$

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