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BlackZzzverrR [31]

3 years ago

5

Tammy is participating in a 5 -day cross-country biking challenge. She biked for 63 , 50 , 63 , and 60 miles on the first four d

ays. How many miles does she need to bike on the last day so that her average (mean) is 60 miles per day?

1 answer:

mojhsa [17]3 years ago

3 0

On the last day she needs to travel 64 miles

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Please help i am failing this class currently!!

lana [24]

Answer: A- $5,250
B- $7,750

3 0

3 years ago

Find pb, if pj=14 and jb= 28

anyanavicka [17]

Answer:

42

Step-by-step explanation:

Based on the information given, point j is the midpoint, as it is shared by both line segments. Set the equation:

pb = pj + jb

Note:

pj = 14

jb = 28

Plug in the corresponding numbers to the corresponding variables:

pb = pj + jb

pb = 14 + 28

pb = 42

pb = 42 is your answer.

~

3 0

3 years ago

Read 2 more answers

Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????

LenaWriter [7]

Answer:

$E[X^2]= \frac{2!}{2^1 1!}= 1$

$Var(X^2)= 3-(1)^2 =2$

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

$\phi(t) = E[e^{tX}]$

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

$\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx$

And we have that the moment generating function can be write like this:

$\phi(t) = e^{\frac{t^2}{2}$

And we can write this as an infinite series like this:

$\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...$

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

$E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]$

$E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...$

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

$\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}$

And then we have this:

$E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...$

And then we can find the $E[X^2]$

$E[X^2]= \frac{2!}{2^1 1!}= 1$

And we can find the variance like this :

$Var(X^2) = E[X^4]-[E(X^2)]^2$

And first we find:

$E[X^4]= \frac{4!}{2^2 2!}= 3$

And then the variance is given by:

$Var(X^2)= 3-(1)^2 =2$

7 0

3 years ago

Given: AD = BC and AD || BC

bearhunter [10]

Answer:

ABCD is a parallelogram.

Step-by-step explanation:

A parallelogram is a quadrilateral that has two parallel and equal pairs of opposite sides.

From the given diagram,

Given: AD = BC and AD || BC, then:

i. AB = DC (both pairs of opposite sides of a parallelogram are congruent)

ii. <ADC = < BCD and < DAB = < CBA

thus, AD || BC and AB || DC (both pairs of opposite sides of a parallelogram are parallel)

iii. < BAC = < DCA (alternate angle property)

iv. Join BD, line AC and BC are the diagonals of the quadrilateral which bisect each other. The two diagonals are at a right angle to each other.

v. <ADC + < BCD + < DAB + < CBA = $360^{0}$ (sum of angles in a quadrilateral equals 4 right angles)

Therefore, ABCD is a parallelogram.

4 0

3 years ago

What is the approximate value of t?

Maslowich

Answer:

8.48

Step-by-step explanation:

36 + 36 = 72

t^2 = 72

t = 6 $\sqrt{2}$ = 8.48

5 0

2 years ago

Read 2 more answers