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1 year ago

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There are five envelopes on a table. Four of the envelopes are marked 8, 2, 4 and 1. You are told that the mean of the numbers o

n the envelopes is 4. What is the number on the fifth envelope?

1 answer:

Lostsunrise [7]1 year ago

3 0

Answer: 16

Step-by-step explanation:

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In a fruit cocktail, for every 18 ml of orange juice you need 3 ml of apple juice and 54 ml of coconut milk. What is the ratio o

Marizza181 [45]

Answer:

1:18:6

Step-by-step explanation:

3:54:18

This equals 1:18:6

7 0

2 years ago

A barrel of crude oil contains about 5.61 cubic feet of oil. How many barrels of oil contained in 1 mile (5280 feet) of a pipeli

8_murik_8 [283]

1)volume of the pipeline

The pipeline is a cylinder, therefore;
Volume (cylinder)=πr²h
r=radius
h=height of the cylinder

diameter=6 in*(1 ft / 12 in)=0.5 ft
raius=diameter / 2=0.5 ft / 2=0.25 ft.
height=5280 ft

Volume (pipeline)=π(0.25 ft)²(5280 ft)=330π ft³≈1036.73 ft³.

2) we calculate the number of barrel
1 mile of oil in this pipeline is 330π ft³ of oil.

1 barrel of crude------------------5.61 ft³
x----------------------------------330π ft³

x=(1 barrel*330π ft³) / 5.61 ft³=184.8 barrels

3) we calculate the price.

1 barrel---------------$100
184.8 barrels---------- x

x=(184.8 barrels * $100) / 1 barrel=$18,480

Solution: ≈$18,480

6 0

2 years ago

A rectangular garden has a width of 90 feet.The perimeter is 500 feet.What is the length of the garden?

Lorico [155]

Perimeter for a rectangle equals 2W + 2L where W - width and L - length

We are given that:
2 (90) + 2L = 500

or, 180 + 2L = 500

Subtracting 180 from both sides of the equation, we get

2L = 320

or, L = 160 ft

5 0

2 years ago

Read 2 more answers

A jar contains only pennies, nickels, dimes, and quarters. There are 21 pennies, 26 dimes, and 15 quarters. The rest of the

pshichka [43]

Answer:

14

Step-by-step explanation:

21+26+15=62 72-62=14 :)

8 0

1 year ago

Read 2 more answers

Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

3 0

2 years ago