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MrRissso [65]
3 years ago
7

Evaluate u^2 -2u+4 when u= -3

Mathematics
1 answer:
OLga [1]3 years ago
6 0

Answer:

Step-by-step explanation: Sub -3 into where all the u’s are. So it would be

(-3)^2 - 2(-3) + 4

Which would equal 19

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You want to test your newly created Web site, so you have 250 people access it from random locations at random times. Of the peo
Arte-miy333 [17]

Answer: 505

Step-by-step explanation:

The formula to find the sample size n , if the prior estimate of the population proportion (p) is known:

n= p(1-p)(\dfrac{z}{E})^2 , where E=  margin of error and z = Critical z-value.

Let p be the population proportion of crashes.

Prior sample size = 250

No. of people experience computer crashes = 75

Prior proportion  of crashes p=\dfrac{75}{250}=0.3

E= 0.04

From z-table , the z-value corresponding to 95% confidence interval = z=1.96

Required sample size will be :

n=0.3(1-0.3)(\dfrac{1.96}{0.04})^2 (Substitute all the values in the above formula)

n= (0.21)(49)^2= 0.21\times2401

n= 504.21\approx505   (Rounded to the next integer.)

∴ Required sample size = 505

6 0
3 years ago
Consider the initial value problem:
Inessa [10]

Answer:

\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2

Step-by-step explanation:

The given initial value problem is;

y''-5y'+6y=-5\sin(2t)---(1)\\y(0)=-5,y'(0)=2

Let

u(t)=y(t)----(2)\\v(t)=y'(t)----(3)

Differentiating both sides of equation (1) with respect to t, we obtain:

u'(t)=y'(t)---(4)\\ \\\implies u'(t)=v(t)---(5)

Differentiating both sides of equation (2) with respect to t gives:

v'(t)=y''(t)----(6)

From equation (1),

y''=5y'-6y-5\sin(2t)\\y''=5v(t)-6u(t)-5\sin(2t)\\\implies v'(t)=5v(t)-6u(t)-5\sin(2t)----(7)

Putting t=0 into equation (2) yields

u(0)=y(0)\\\implies u(0)=-5

Also putting t=0 into equation (3)

u'(0)=y'(0)\\u'(0)=2

The system of first order equations is:

\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2

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3 years ago
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The explaination is here
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