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HACTEHA [7]
3 years ago
11

45 centimeters is equivalent to how many inches?

Mathematics
2 answers:
Elza [17]3 years ago
8 0
It’s about 17.7 inches
Wittaler [7]3 years ago
7 0

2. 54 centimeters = 1 inch

Therefore  45 cms =  45 / 2.54

= 17.72 inches to nearest hundredth

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I have 30 points to give away. One person answers this question and takes 10 points. If another person takes 10 more points, how
Ghella [55]

Answer:

10 points

Step-by-step explanation:

plz brainleist

8 0
3 years ago
The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and
Firdavs [7]

Answer:

1a

  P(39 <  X < 48  ) = 0.8767

1b

    95% of all sample means will fall between 40.1  <  \mu < 49.9

1c

    \= x = 41. 795

2

   n =  25

Step-by-step explanation:

From the question we are told that

   The mean is n   =  45

   The population standard deviation is  \sigma =  10

   The sample size is n  =  16

Generally the standard error of the mean is mathematically represented as

       \sigma_{x} =  \frac{ \sigma}{\sqrt{n} }

=>    \sigma_{x} =  \frac{ 10 }{\sqrt{16 } }

=>    \sigma_{x} = 2.5

Generally the probability that the sample mean will be between 39 and 48 minutes is

    P(39 <  X < 48  ) =  P( \frac{ 39 - 45}{ 2.5} <  \frac{X - \mu }{\sigma } <  \frac{ 48 - 45}{ 2.5} )

=> P(39 <  X < 48  ) =  P(-2.4 < Z< 1.2 )

=> P(39 <  X < 48  ) =  P( Z< 1.2 ) - P(Z <  -2.4)

From the z table  the area under the normal curve to the left corresponding to  1.2  and  -2.4  is

=> P( Z< 1.2 ) = 0.88493

and  

    P( Z< - 2.4 ) = 0.0081975

So

   P(39 <  X < 48  ) = 0.88493 -0.0081975

=> P(39 <  X < 48  ) = 0.8767

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

      E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

=>   E = 1.96 * 2.5  

=>   E =4.9  

Generally the  95% of all sample means will fall between

      \mu  -E <  and   \mu   +E

=>   45  -4.9\   and \  45  + 4.9

Generally the value which  90% of sample means is  greater than is mathematically represented

      P( \= X >  \= x  ) = 0.90

=>   P( \= X >  \= x  ) =  P( \frac{\= X  - \mu }{ \sigma_x} >  \frac{\= x  -45 }{ 2.5}  ) = 0.90

=>  P( \= X >  \= x  ) =  P( Z >  z  ) = 0.90

Generally from the z-table  the critical  value  of  0.90  is  

      z = -1.282

      \frac{\= x  -45 }{ 2.5}  = -1.282

=>   \= x = 41. 795

Considering question 2

 Generally we are told that the standard deviation of the mean to be one fifth of the population standard deviation, this is mathematically represented as

         s = \frac{1}{5} \sigma

  Generally the standard deviation of the sample mean is mathematically  represented as

          s = \frac{\sigma }{ \sqrt{n} }

=>       \frac{1}{5} \sigma = \frac{\sigma }{ \sqrt{n} }

=>       n =  5^2

=>       n =  25

5 0
3 years ago
Using Laplace transforms, solve x" + 4x' + 6x = 1- e^t with the following initial conditions: x(0) = x'(0) = 1.
professor190 [17]

Answer:

The solution to the differential equation is

X(s)=\cfrac 1{6}  -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)

Step-by-step explanation:

Applying Laplace Transform will help us solve differential equations in Algebraic ways to find particular  solutions, thus after applying Laplace transform and evaluating at the initial conditions we need to solve and apply Inverse Laplace transform to find the final answer.

Applying Laplace Transform

We can start applying Laplace at the given ODE

x''(t)+4x'(t)+6x(t)=1-e^t

So we will get

s^2 X(s)-sx(0)-x'(0)+4(sX(s)-x(0))+6X(s)=\cfrac 1s -\cfrac1{s-1}

Applying initial conditions and solving for X(s).

If we apply the initial conditions we get

s^2 X(s)-s-1+4(sX(s)-1)+6X(s)=\cfrac 1s -\cfrac1{s-1}

Simplifying

s^2 X(s)-s-1+4sX(s)-4+6X(s)=\cfrac 1s -\cfrac1{s-1}

s^2 X(s)-s-5+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}

Moving all terms that do not have X(s) to the other side

s^2 X(s)+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}+s+5

Factoring X(s) and moving the rest to the other side.

X(s)(s^2 +4s+6)=\cfrac 1s -\cfrac1{s-1}+s+5

X(s)=\cfrac 1{s(s^2 +4s+6)} -\cfrac1{(s-1)(s^2 +4s+6)}+\cfrac {s+5}{s^2 +4s+6}

Partial fraction decomposition method.

In order to apply Inverse Laplace Transform, we need to separate the fractions into the simplest form, so we can apply partial fraction decomposition to the first 2 fractions. For the first one we have

\cfrac 1{s(s^2 +4s+6)}=\cfrac As + \cfrac {Bs+C}{s^2+4s+6}

So if we multiply both sides by the entire denominator we get

1=A(s^2+4s+6) +  (Bs+C)s

At this point we can find the value of A fast if we plug s = 0, so we get

1=A(6)+0

So the value of A is

A = \cfrac 16

We can replace that on the previous equation and multiply all terms by 6

1=\cfrac 16(s^2+4s+6) +  (Bs+C)s

6=s^2+4s+6 +  6Bs^2+6Cs

We can simplify a bit

-s^2-4s=  6Bs^2+6Cs

And by comparing coefficients we can tell the values of B and C

-1= 6B\\B=-1/6\\-4=6C\\C=-4/6

So the separated fraction will be

\cfrac 1{s(s^2 +4s+6)}=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6}

We can repeat the process for the second fraction.

\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac A{s-1} + \cfrac {Bs+C}{s^2+4s+6}

Multiplying by the entire denominator give us

1=A(s^2+4s+6) + (Bs+C)(s-1)

We can plug the value of s = 1 to find A fast.

1=A(11) + 0

So we get

A = \cfrac1{11}

We can replace that on the previous equation and multiply all terms by 11

1=\cfrac 1{11}(s^2+4s+6) + (Bs+C)(s-1)

11=s^2+4s+6 + 11Bs^2+11Cs-11Bs-11C

Simplifying

-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C

And by comparing coefficients we can tell the values of B and C.

-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C\\-1=11B\\B=-\cfrac{1}{11}\\5=-11C\\C=-\cfrac{5}{11}

So the separated fraction will be

\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac {1/11}{s-1} + \cfrac {-s/11-5/11}{s^2+4s+6}

So far replacing both expanded fractions on the solution

X(s)=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6} -\cfrac {1/11}{s-1} -\cfrac {-s/11-5/11}{s^2+4s+6}+\cfrac {s+5}{s^2 +4s+6}

We can combine the fractions with the same denominator

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {-s/6-4/6+s/11+5/11+s+5}{s^2 +4s+6}

Simplifying give us

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{s^2 +4s+6}

Completing the square

One last step before applying the Inverse Laplace transform is to factor the denominators using completing the square procedure for this case, so we will have

s^2+4s+6 = s^2 +4s+4-4+6

We are adding half of the middle term but squared, so the first 3 terms become the perfect  square, that is

=(s+2)^2+2

So we get

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{(s+2)^2 +(\sqrt 2)^2}

Notice that the denominator has (s+2) inside a square we need to match that on the numerator so we can add and subtract 2

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61(s+2-2)/66+316 /66}{(s+2)^2 +(\sqrt 2)^2}\\X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66+194 /66}{(s+2)^2 +(\sqrt 2)^2}

Lastly we can split the fraction one more

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66}{(s+2)^2 +(\sqrt 2)^2}+\cfrac {194 /66}{(s+2)^2 +(\sqrt 2)^2}

Applying Inverse Laplace Transform.

Since all terms are ready we can apply Inverse Laplace transform directly to each term and we will get

\boxed{X(s)=\cfrac 1{6}  -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)}

6 0
4 years ago
103 + 21 = _ + 103 please help
lorasvet [3.4K]
This is the commutative property of addition. It basically says, x+y=y+x.
To answer your question, 103+21=21+103
6 0
4 years ago
Read 2 more answers
Which equation has the correct sign on the product?
eimsori [14]
Answer: b

Explanation: A and C are wrong because a negative times a positive is a negative
And D is wrong because a positive times a positive is a positive
7 0
3 years ago
Read 2 more answers
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