You can drive 512 miles on one tank of gas if you have 16 gallons of tank how many miles can you drive in 1 gallon of gas

Integrate the following. ∫<img src="https://tex.z-dn.net/?f=5x%5E%7B4%7D%20dx" id="TexFormula1" title="5x^{4} dx" alt="5x^{4} dx

Vadim26 [7]

Answer:

$\displaystyle D) {x}^{5} + \rm C$

Step-by-step explanation:

we would like to integrate the following Integral:

$\displaystyle \int 5 {x}^{4} \, dx$

well, to get the constant we can consider the following Integration rule:

$\displaystyle \int c{x} ^{n} \, dx = c\int {x}^{n} \, dx$

therefore,

$\displaystyle 5\int {x}^{4} \, dx$

recall exponent integration rule:

$\displaystyle \int {x} ^{n} \, dx = \frac{ {x}^{n + 1} }{n + 1}$

so let,

$n = 4$

Thus integrate:

$\displaystyle = 5\left( \frac{ {x}^{4+ 1} }{4 + 1} \right)$

simplify addition:

$\displaystyle = 5\left( \frac{ {x}^{5} }{5} \right)$

reduce fraction:

$\displaystyle = {x}^{5}$

finally we of course have to add the constant of integration:

$\displaystyle \boxed{ {x}^{5} + \rm C}$

hence,

our answer is D)

7 0

3 years ago

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Given a soda can with a volume of 36 and a diameter of 4, what is the volume of a cone that fits perfectly inside the soda can?

spin [16.1K]

First we need to find the heigh of the soda can be rearanging the volume formula, . We can make that We know that V is 36 and radius is half of the diameter, so radius is 2.

h = 2.87

Now, we can use the height to figure out the volume of a cone. The volume of a cone is
R is 2 again and h is 2.87

12.56*.96 = 12.0576
So a cone with a volume of 12.0576 is the largest that will fit into the soda can

6 0

3 years ago

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In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

Marrrta [24]

Answer:

a) Bi [P ( X >=15 ) ] ≈ 0.9944

b) Bi [P ( X >=30 ) ] ≈ 0.3182

c) Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) Bi [P ( X >40 ) ] ≈ 0.0046

Step-by-step explanation:

Given:

- Total sample size n = 745

- The probability of success p = 0.037

- The probability of failure q = 0.963

Find:

a. 15 or more will live beyond their 90th birthday

b. 30 or more will live beyond their 90th birthday

c. between 25 and 35 will live beyond their 90th birthday

d. more than 40 will live beyond their 90th birthday

Solution:

- The condition for normal approximation to binomial distribution:

n*p = 745*0.037 = 27.565 > 5

n*q = 745*0.963 = 717.435 > 5

Normal Approximation is valid.

a) P ( X >= 15 ) ?

- Apply continuity correction for normal approximation:

Bi [P ( X >=15 ) ] = N [ P ( X >= 14.5 ) ]

- Then the parameters u mean and σ standard deviation for normal distribution are:

u = n*p = 27.565

σ = sqrt ( n*p*q ) = sqrt ( 745*0.037*0.963 ) = 5.1522

- The random variable has approximated normal distribution as follows:

X~N ( 27.565 , 5.1522^2 )

- Now compute the Z - value for the corrected limit:

N [ P ( X >= 14.5 ) ] = P ( Z >= (14.5 - 27.565) / 5.1522 )

N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 )

- Now use the Z-score table to evaluate the probability:

P ( Z >= -2.5358 ) = 0.9944

N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 ) = 0.9944

Hence,

Bi [P ( X >=15 ) ] ≈ 0.9944

b) P ( X >= 30 ) ?

- Apply continuity correction for normal approximation:

Bi [P ( X >=30 ) ] = N [ P ( X >= 29.5 ) ]

- Now compute the Z - value for the corrected limit:

N [ P ( X >= 29.5 ) ] = P ( Z >= (29.5 - 27.565) / 5.1522 )

N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 )

- Now use the Z-score table to evaluate the probability:

P ( Z >= 0.37556 ) = 0.3182

N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 ) = 0.3182

Hence,

Bi [P ( X >=30 ) ] ≈ 0.3182

c) P ( 25=< X =< 35 ) ?

- Apply continuity correction for normal approximation:

Bi [P ( 25=< X =< 35 ) ] = N [ P ( 24.5=< X =< 35.5 ) ]

- Now compute the Z - value for the corrected limit:

N [ P ( 24.5=< X =< 35.5 ) ]= P ( (24.5 - 27.565) / 5.1522 =<Z =< (35.5 - 27.565) / 5.1522 )

N [ P ( 24.5=< X =< 25.5 ) ] = P ( -0.59489 =<Z =< 1.54011 )

- Now use the Z-score table to evaluate the probability:

P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

N [ P ( 24.5=< X =< 35.5 ) ]= P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

Hence,

Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) P ( X > 40 ) ?

- Apply continuity correction for normal approximation:

Bi [P ( X >40 ) ] = N [ P ( X > 41 ) ]

- Now compute the Z - value for the corrected limit:

N [ P ( X > 41 ) ] = P ( Z > (41 - 27.565) / 5.1522 )

N [ P ( X > 41 ) ] = P ( Z > 2.60762 )

- Now use the Z-score table to evaluate the probability:

P ( Z > 2.60762 ) = 0.0046

N [ P ( X > 41 ) ] = P ( Z > 2.60762 ) = 0.0046

Hence,

Bi [P ( X >40 ) ] ≈ 0.0046

4 0

3 years ago

Solve for px thanks

koban [17]

Px + 12 = 17
px = 5
x = 5/p

ans) D

6 0

3 years ago

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A chance with probability 1 is certain to happen true or false

KIM [24]

Probability always adds up to 1 so, if there is only one choice, the chance that it will happen is 100% (1 out of 1) so true

Hope this helps! :)

3 0

3 years ago