
MathAll of these are answered from the same basic principle. The pvalue, in essence, gives the probability that the sample mean you obtained occurred by chance assuming that the null hypothesis is...

MathWe are asked to test the claim that the average wing span is greater than or equal to 64 inches. We have a sample of size 54 with a mean of 62 inches. The population standard deviation is 8 inches....

MathWe are asked to test the claim of the mean salary being $91,350 (`mu = 91350 ` ) using a sample of 60 salaries with a mean of $90,350 (` bar(x)=90350 ` ) and a population standard deviation of...

MathWe are asked to solve `x^(1/3)2x^(1/3)=1 ` with the hint to let `y=x^(1/3) ` : If we let `y=x^(1/3) ` we get: `y2y^(1)=1 ` since `x^(1/3)=(x^(1/3))^(1) ` or `y2/y=1 ` Multiplying by y...

MathHello! Denote the minimum number of weeks after which Carlos will attend all three classes during one week as `N.` Our task is to find that `N.` The next art classes will be after `4` weeks since...

MathHello! To answer this question we only need the fact `3^(376) approx 4*10^(180).` By the definition, raise some number `b` to a negative natural power `n` means 1) raise `b` to the positive...

MathGiven ` x^2+5x+7 ` : (i) Find the vertex: (a) Rewrite in vertex form: `x^2+5x+7 ` Add and subtract the square of 1/2 the linear term, `(5/2)^2`, to get ` =x^2+5x+25/425/4+7 ` `=(x+5/2)^2+3/4 ` is...

MathTo solve, let's apply the least common multiple (LCM) of a set of numbers. Pizza is served every sixth day. To get the days in which it will be served, take the multiples of 6. The multiples of 6...

MathGiven the equations y=0, x=6 and `y=\frac{x}{\sqrt{x+3}}` We have to find the area bounded by the three equations. The graph is shown below: The yellow region is the bounded area. Point (6,2) is...

MathGiven to solve , `int 4/(csc(theta)cot(theta)) d theta` just for the convenience let` x= theta` so,` int 4/(csc(theta)cot(theta)) d theta` =`int 4/(csc(x)cot(x)) d x` =`int...

Math`int (sin sqrt theta)/sqrt theta d theta` To solve, apply usubstitution method. `u=sqrt theta` `u= theta ^(1/2)` `du = 1/2 theta^(1/2) d theta` `du = 1/(2theta^(1/2))d theta` `du =1/(2 sqrt...

MathGiven to solve , `int cos(theta) / (1+cos(theta)) d theta` just for easy solving let `x=theta ` so the equation is given as `int cos(x) / (1+cos(x)) d x ` (1) let `u= tan(x/2)` ,=> then...

Math`int ( sin (theta))/(32cos(theta)) d theta` To solve, apply usubstitution method. `u = 32cos (theta)` `du = 2*(sin (theta)) d theta` `du = 2sin (theta) d theta` `1/2du= sin (theta) d theta`...

MathBasis (n=1) We will use integration by parts `int u dv=uvint v du` `int_0^infty xe^x dx=[u=x,dv=e^x dx],[du=dx,v=e^x]=` `xe^x_0^infty+int_0^infty e^x dx=(xe^xe^x)_0^infty=` `lim_(x...

Math`int_0^infty sin(x/2)dx=` Substitute `u=x/2` `=>` `du=dx/2` `=>` `dx=2du,` `u_l=0/2=0,` `u_l=infty/2=infty` (`u_l` and `u_u` are lower and upper bound respectively). `2int_0^infty sin u...

Math`int_0^infty cos (pi x)dx=` Substitute `u=pi x` `=>` `du=pi dx` `=>` `dx=(du)/pi,` `u_l=pi cdot 0=0,` `u_u=pi cdot infty=infty.` (`u_l` and `u_u` are lower and upper bound respectively)....

Math`int_0^infty e^x/(1+e^x)dx=` Substitute `u=1+e^x` `=>` `du=e^xdx,` `u_l=1+e^0=2,` `u_u=lim_(x to infty)(1+e^x)=infty.` `int_1^infty 1/u du=ln u_2^infty=lim_(u to infty)ln uln 2=infty` As...

Math`int_0^infty 1/(e^x+e^x)dx=` Multiply both numerator and the denominator by `e^x.` `int_0^infty e^x/(1+e^(2x))dx=` Substitute `u=e^x` `=>` `du=e^x dx,``u_l=e^0=1,``u_u=lim_(x to infty)e^x=infty...

Math`int_0^infty x^3/(x^2+1)^2 dx=` Substitute `u=x^2+1` `=>` `du=2x dx` `=>` `x dx=(du)/2,` `u_l=0^2+1=1,` `u_u=lim_(x to infty)x^2+1=infty.` `u_l` and `u_l` denote new lower and upper bounds...

Math`int_infty^infty 4/(16+x^2)dx=` Divide both numerator and denominator by 16. `int_infty^infty (1/4)/(1+x^2/16)dx=int_infty^infty (1/4 dx)/(1+(x/4)^2)=` Substitute `u=x/4` `=>` `du=1/4 dx.`...

Math`int_1^infty (ln x)/x dx=` Substitute `u=lnx` `=>` `du=1/x dx,` `u_l=ln 1=0,` `u_u=ln infty=infty` (`u_l` and `u_u` denote lower and upper bound respectively). `int_0^infty u...

Math`int_4^infty 1/(x(ln x)^3)dx=` Substitute `u=ln x` `=>` `du=1/x dx,` `u_l=ln 4,` `u_u=ln infty=infty` (`u_l` and `u_u` denote lower and upper bound respectively). `int_(ln 4)^infty1/u^3...

MathWe will use integration by parts (twice): `int u dv=uvint v du` `int_0^infty e^x cos x dx=[u=e^x,dv=cos x dx],[du=e^x dx,v=sin x]=` `e^x sin x+int_0^infty e^x sin x dx=[u=e^x,dv=sin x...

MathWe will use integration by parts `int u dv=uvint v du` We will need to apply integration by parts two times in order to eliminate `x^2` from under the integral. `int_0^infty x^2e^x...

MathWe will use integration by parts `int udv=uvint vdu` `int_0^infty xe^(x/3)dx=[u=x,dv=e^(x/3)dx],[du=dx,v=3e^(x/3)]=` `3xe^(x/3)_0^infty+3int_0^infty e^(x/3)dx=`...

MathWe will use integration by parts `int udv=uvint vdu` `int_infty^0 xe^(4x)dx=[u=x,dv=e^(4x)dx],[du=dx,v=1/4e^(4x)]=` `1/4xe^(4x)_infty^0+1/4int_infty^0 e^(4x)dx=`...

Math`int_1^infty 4/root(4)(x)dx=4int_1^infty x^(1/4)dx=4x^(3/4)/(3/4)_1^infty=16/3x^(3/4)_1^infty=` `16/3(lim_(x to infty) x^(3/4)1)=16/3(infty1)=infty` As we can see the value if the integral is...

Math`int_1^infty 3/root(3)(x)dx=` Rewrite the integral using the following formula `root(n)(x^m)=x^(m/n).` `int_1^infty 3x^(1/3)dx=3x^(2/3)/(2/3)_1^infty=^(9x^(2/3))/2_1^infty=9/2( lim_(x to infty)...

MathAn integral in which one of the limits of integration is infinity is an improper integral. Because we cannot find the definite integral using infinity (since it is not an actual value), we will...

MathThe function under the integral is continuous and bounded on any interval `[1, A].` The only point that makes the integral improper is `x=+oo.` Therefore this integral is the limit `lim_(A>+oo)...

MathAny integral with infinite bounds is an improper integral therefore this is an improper integral. `int_infty^0 e^(3x) dx=` Substitute `u=3x` `=>` `du=3dx,` `u_l=3cdot(infty)=infty,`...

MathIntegral is improper if we have to take limit in order to calculate it. This can happen if we have infinite values of integration or if the interval if integration contains point(s) where the...

MathThe integral is improper because the function under the integral `f(x)=1/(x3)^(3/2)` is not defined at 3 (for `x=3` denominator is equal to zero). `int_3^4 1/(x3)^(3/2)dx=` Substitute `u=x3`...

MathThe integral is improper because the function under the integral is not defined at zero (see the image below). `1/sqrt0=1/0` `int_0^4 1/sqrt x dx=2sqrt x_0^4=2(sqrt 4sqrt0)=4` The integral...

MathThe integral `int_0^(pi/4)csc x dx` is improper because cosecant is not defined on zero (more generally `csc x` is not defined for `x in {k pi, k in ZZ}.`) and the interval of integration includes...

MathAny integral with infinite bounds is an improper integral, hence integral `int_infty^infty sin x/(4+x^2)dx` is an improper integral. It can be shown that `int_infty^infty sin x/(4+x^2)dx=0` The...

MathAny integral with infinite bounds is an improper integral. Hence `int_0^infty cos x dx` is an improper integral. Also, the cosine function is a periodic function and `lim_(x to infty)cos x` does...

MathThe integral does not have infinite bounds and the function is well defined over the whole interval of integration so there is no need to use limits. Therefore, the integral is not improper....

MathAny integral with infinite bounds is an improper integral, therefore `int_1^infty ln(x^2)dx` is an improper integral. Moreover, since the function under the integral is positive over the given...

Math`int_0^1 (2x5)/(x^25x+6) dx` The function `f(x) = (2x5)/(x^25x+6)` is continuous on the interval `(oo, 2) uu (2,3) uu (3,oo)` . Since the limit of the integral is from x=0 to x=1, then the...

MathIntegral `int_1^2 dx/x^3` is not an improper integral because the function `f(x)=1/x^3` is well defined (with finite values) over the whole interval of integration `[1,2].` ` ` The only point where...

MathAn integral is improper if we have to take limit in order to calculate it. This can happen if we have infinite values of integration or if the interval if integration contains point(s) where the...

MathGivne to solve, `lim_(x>1^(+)) (int_1^x cos(theta) d theta ) / (x1)` =`lim_(x>1^(+)) ([sin(theta)]_1^x) / (x1)` =`lim_(x>1^(+)) ([sin(x)sin(1)]) / (x1)` when `x> 1+` then...

Math`lim_(x>oo) (int_1^x ln(e^(4t1)) dt )/ x` = `lim_(x>oo) (int_1^x (4t1) dt )/ x` = `lim_(x>oo) ( [4(t^2)/2t]_1^x )/ x` = `lim_(x>oo) ( [2x^2 x][2(1^2)1] )/ x` = `...

MathGiven to solve, `lim_(x>0) x/arctan(2x)` as `x>0` then the `x/arctan(2x) =0/0` form so upon applying the L 'Hopital rule we get the solution as follows, as for the general equation it is as...

MathGiven to solve, `lim_(x>0)arctanx/sinx` as `x>0 ` on substituting we get `arctanx/sinx = 0/0 ` so by using the L'hopital rule we get the solution as follows, as for the general equation it...

Math`lim_(x>1) (ln(x))/(sin(pix))` To solve, plugin x = 1. `lim_(x>1) (ln(x))/(sin(pix)) = (ln(1))/(sin(pi*1)) = 0/0` Since the result is indeterminate, to find the limit of the function as x...

Math`lim_(x>0) (sin (5x))/(tan(9x))` To solve, plugin x = 0. `lim_(x>0) (sin(5*0))/(tan(9*0)) = 0/0` Since the result is indeterminate, to find the limit of the function as x approaches zero,...

MathGiven to solve , `lim_(x>oo) e^(x/2)/x` As `x` thends to ` oo` we get `e^(x/2)/x = oo/oo` L'Hopital's Rule says if `lim_(x>a) f(x)/g(x) = 0/0` or `(+oo)/(+oo)` then the limit is:...

MathGiven to solve, `lim_(x>oo) e^x/(x^4)` as `x>oo` then the ` e^x/(x^4) =oo/oo` form so upon applying the L 'Hopital rule we get the solution as follows, as for the general equation it is as...
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