Pls help me with this!!!

Answer:

The quotient of 7 and 14 is 2..

Answer:

In the long run, ou expect to  lose $4 per game

Step-by-step explanation:

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.

Assuming X be the toss on which the first head appears.

then the geometric distribution of X is:

X [tex]\sim[/tex] geom(p = 1/2)

the probability function P can be computed as:

[tex]P (X = n) = p(1-p)^{n-1}[/tex]

where

n = 1,2,3 ...

If I agree to pay you $n^2 if heads comes up first on the nth toss.

this implies that , you need to be paid [tex]\sum \limits ^{n}_{i=1} n^2 P(X=n)[/tex]

[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)[/tex]

[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2[/tex]

[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2[/tex]        ∵  X [tex]\sim[/tex] geom(p = 1/2)

[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}[/tex]

[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}[/tex]

[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}[/tex]

[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}[/tex]

[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}[/tex]

[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}[/tex]

[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}[/tex]

[tex]\sum \limits ^{n}_{i=1} n^2 P(X=n) =6[/tex]

Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6

= $4

∴

In the long run, you expect to  lose $4 per game

Answer:

on [3, 4] = 0.30

on [4, 5] = 0.18

on [5, 6] = 0.12

Step-by-step explanation:

The average rate of change f, of a function f(x) on an interval [a, b] is given by;

[tex]f = \frac{f(b) - f(a)}{b - a}[/tex]              -------------(i)

In our case,

f(x) = log 2(3x - 6)

Now let's get the average rate of change of f(x) on;

(i) [3, 4]

Here, a = 3 and b = 4

f(a) = f(3)        [This is f(x) at x = 3]

=> f(3) = log[2(3(3) - 6)]

=> f(3) = log[2(9 - 6)]

=> f(3) = log[2(3)]

=> f(3) = log[6]

Also,

f(b) = f(4)        [This is f(x) at x = 4]

=> f(4) = log[2(3(4) - 6)]

=> f(4) = log[2(12 - 6)]

=> f(4) = log[2(6)]

=> f(4) = log[12]

Now substitute the values of a, b, f(a) and f(b) into equation (i) as follows;

[tex]f = \frac{log 12 - log 6}{4 - 3}[/tex]             [Remember that log m - log n = log (m / n)]

[tex]f = \frac{log (12 / 6)}{4 - 3}[/tex]

[tex]f = \frac{log (2)}{1}[/tex]

f = log 2 = 0.3010

f = 0.30          [to two decimal places]

∴ The average rate of change on [3, 4] = 0.30

(ii) [4, 5]

Here, a = 4 and b = 5

f(a) = f(4)        [This is f(x) at x = 4]

=> f(4) = log[2(3(4) - 6)]

=> f(4) = log[2(12 - 6)]

=> f(4) = log[2(6)]

=> f(4) = log[12]

Also,

f(b) = f(5)        [This is f(x) at x = 5]

=> f(5) = log[2(3(5) - 6)]

=> f(5) = log[2(15 - 6)]

=> f(5) = log[2(9)]

=> f(5) = log[18]

Now substitute the values of a, b, f(a) and f(b) into equation (i) as follows;

[tex]f = \frac{log 18 - log 12}{5 - 4}[/tex]             [Remember that log m - log n = log (m / n)]

[tex]f = \frac{log (18 / 12)}{5 - 4}[/tex]

[tex]f = \frac{log (1.5)}{1}[/tex]

f = log 1.5 = 0.176

f = 0.18          [to two decimal places]

∴ The average rate of change on [4, 5] = 0.18

(iii) [5, 6]

Here, a = 5 and b = 6

f(a) = f(5)        [This is f(x) at x = 5]

=> f(5) = log[2(3(5) - 6)]

=> f(5) = log[2(15 - 6)]

=> f(5) = log[2(9)]

=> f(5) = log[18]

Also,

f(b) = f(6)        [This is f(x) at x = 6]

=> f(6) = log[2(3(6) - 6)]

=> f(6) = log[2(18 - 6)]

=> f(6) = log[2(12)]

=> f(6) = log[24]

Now substitute the values of a, b, f(a) and f(b) into equation (i) as follows;

[tex]f = \frac{log 24 - log 18}{6 - 5}[/tex]             [Remember that log m - log n = log (m / n)]

[tex]f = \frac{log (24 / 18)}{6 - 5}[/tex]

[tex]f = \frac{log (1.33)}{1}[/tex]

f = log 1.33 = 0.124

f = 0.12         [to two decimal places]

∴ The average rate of change on [5, 6] = 0.12

Answer:CORRECT ANSWER ON PLATO

[3,4]= 1

[4,5]=0.59

[5,6]= 0.41

Answer:

The formula of area is Pi×R×R

and then you add all sides and answer of your question is 30cm

Answer:

a = 28

Step-by-step explanation: First you would have to multiply 7 x 8 which equals =56. Then you would have to divide 56 by 2 which is =28 and that would be your area if you are solving for a triangle.

Answer:

n + 3

Step-by-step explanation:

6 is 3 + 3, 11 is 8 + 3 and 12 is 9 + 3 so the output seems to be the input plus 3. Therefore, when the input is n, the output is n + 3.

Answer:

  output = n+3

Step-by-step explanation:

When the input changes by 5 units (from 3 to 8), the output changes by 5 unis (from 6 to 11). So the rate of change of the output with respect to the input is 1. However, each output is 3 more than the input, so ...

for input n

output = n+3

Answer:

the SA is 40

Step-by-step explanation:

the formula for SA of a rectangular prism is 2(lw + lh + wh)

so it would be: 2(2x2+2x2+4x2)

can i get brainliest please? i only need two more

Answer:

203

Step-by-step explanation:

Square of 101 = 101^2= 10201

Square of 102 = 102 ^2=10404

No. of numbers between both of them =10404-10201=203

Answer:

203

Step-by-step explanation:

✓101 kare = 10.201

✓102 kare = 10.404

✓ 10.404- 10.201 = 203

#Başarılar

#Türkiye

Answer:

  (1/10)∛100 ≈ 0.4642

Step-by-step explanation:

For a cube of volume V, the edge length is ...

  s = ∛V

and the area is ...

  A = 6s² = 6V^(2/3)

Then the ratio of area to volume is ...

  r1 = A/V = 6V^(2/3)/V = 6V^(-1/3)

If the value of V is increased by a factor of 10, the ratio of area to volume is now ...

  r2 = 6(10V)^(-1/3)

The factor by which the ratio changed is ...

  r2/r1 = (6(10V)^(-1/3))/(6V^(-1/3)) = 10^(-1/3) = (1/10)∛100

The surface area per unit volume changes by a factor of (∛100)/10.

___

Example

For a cube of side length 2, the volume is 2³ = 8 and the surface area is 6·2² = 24. The ratio of surface area to volume is 24/8 = 3.

Multiplying the edge length by ∛10, we now have a volume of (2∛10)³ = 80, and a surface area of 6(2∛10)² = 24(10^(2/3)). The ratio of surface area to volume is ... 24(10^(2/3))/80 = 0.3∛100.

The ratio of area to volume changed by a factor of (0.3∛100)/3 = (∛100)/10, as above.

You cannot divide any number by zero since it is meaningless to say that we are dividing a number or object in zero groups or in 0 size parts

Answer: opposite

Step-by-step explanation:

Answer:

opposite

Step-by-step explanation:

the opposite of -4 is 4.

the absolute value of -4 is 4.

the reciprocal of -4 is -1/4.

Answer:

1 2/8

Step-by-step explanation:

3 1/8 = 25/8

25/8 - 15/8 = 10/8 = 1 2/8

Answer:

see below (I hope this helps!)

Step-by-step explanation:

∠VZY and ∠VXW form a linear pair and are supplementary, and since m∠VZY = 90°, m∠VXW = 180 - 90 = 90°. Since m∠WXU + m∠UXV = ∠VXW, we can write the following equation:

44 + 5x - 4 = 90

5x + 40 = 90

5x = 50

x = 10°

Therefore, m∠UXV = 5 * 10 - 4 = 46°

Answer:

x = 10 m(UXV)=46°

Step-by-step explanation:

m(UXV) + m(UXW) = 90° because m(VXY)= 90°

m(UXW) = 44°

m(UXV) + 44° = 90°

m(UXV) = 46°

5x - 4 = 46°

5x = 50

5x/5 = 50/5

x = 10

Hope this helps ^-^

Answer:

Distribute the 3

Step-by-step explanation:

order of operations mate

Answer:

The cordinates are:

y = 3

x = 4

Step-by-step explanation:

If you draw a line in the middle of the triangle

It Will go through the base at point V

point V can be marked at y=3 and x=4 the line drawn will intersect 0

Answer:

$7,563

Step-by-step explanation:

We will round to the nearest whole number to round to the nearest dollar

$7,563.27

since we have 27 cents ( .27), that is less than 50 cents, we leave it alone

$7,563.00 or $7,563

Answer:

$7,563

Step-by-step explanation:

Rounding to the nearest dollar is just like rounding to the nearest whole number.

The number we are given is:

$7,563.27

To round to the nearest dollar, look to the tenths place. If it is less than or equal to 4, round down. If it is greater than or equal to 5, round up.

The tenths place has a 2. Therefore, we leave the number as is.

$7, 563

$7,563 rounded to the nearest dollar is $7,563.00

Answer:=6x3−9c

Step-by-step explanation:

3x3+4x2+3x3−4x2−9c

=3x3+4x2+3x3+−4x2+−9c

Combine Like Terms:

=3x3+4x2+3x3+−4x2+−9c

=(3x3+3x3)+(4x2+−4x2)+(−9c)

=6x3+−9c

Answer:

(-9) still means -9 it doesn't really mean anything thou....but if u have another number besides it then it means to multiply like maybe 2(-9)then it will be -18...... but if it is just (-9) then it doesn't mean anything.....

Answer:

The ratio will be 48/109

Step-by-step explanation:

Answer:

Upstream speed= 6 mile per hour

Downstream speed= 9 miles per hour

Step-by-step explanation:

Distance traveled during upstream

= x miles

Distance traveled during downstream

= y

Time traveling upstream

= 45 minutes.

Time traveling downstream

= 30 minutes

Speed traveling upstream

= a miles/hour

Speed traveling downstream

= b miles/hour

But

b = a+3

Distance= speed*time

X=45a

a= x/45

Y= 30(a+3)

Y= 30a +90

a= (y-90)/30

Equating both a

X/45= (y-90)/30

30x= 45(y-90)

X= 1.5y-135

But x= y

Y= 1.5y-135

135= 0.5y

270 miles = y

Y= x= 270 miles

X= 45a

270/45= a

6 mile/hour = a

b = a+3

b= 6+3

b = 9 miles/hour

Answer:

Step-by-step explanation:

From the given information:

The present value of the house = 128000

interest rate compounded monthly r = 7.8% = 0.078

number of months in a year n= 12

duration of time t = 30 years

To find their regular monthly payment, we have:

[tex]PV = P \begin {bmatrix} \dfrac{1 - (1 + \dfrac{r}{n})^{-nt}}{\dfrac{r}{n}} \end {bmatrix}[/tex]

[tex]128000 = P \begin {bmatrix} \dfrac{1 - (1 + \dfrac{0.078}{12})^{- 12*30}}{\dfrac{0.078}{12}} \end {bmatrix}[/tex]

128000 = 138.914 P

P = 128000/138.914

P = $921.433

∴ Their regular monthly payment P = $921.433

To find the unpaid balance when they begin paying the $1400.

when they begin the payment ,

t = 30 year - 5years

t= 25 years

[tex]PV= 921.433 \begin {bmatrix} \dfrac{1 - (1 - \dfrac{0.078}{12})^{25*30}}{\dfrac{0.078}{12}} \end {bmatrix}[/tex]

PV = $121718.2714

C) In order to estimate how many payments of $1400  it will take to pay off the loan, we have:

[tex]121718.2714 = \begin {bmatrix} \dfrac{1300 (1 - \dfrac{12.078}{12}))^{-nt}}{\dfrac{0.078}{12}} \end {bmatrix}[/tex]

[tex]121718.2714 = 200000 \begin {bmatrix} (1 - \dfrac{12.078}{12}))^{-nt} \end {bmatrix}[/tex]

[tex]\dfrac{121718.2714}{200000 } = \begin {bmatrix} (1 - \dfrac{12.078}{12}))^{-nt} \end {bmatrix}[/tex]

[tex]0.60859 = \begin {bmatrix} (1 - \dfrac{12}{12.078}))^{nt} \end {bmatrix}[/tex]

[tex]0.60859 = (0.006458)^{nt}[/tex]

[tex]nt = \dfrac{0.60859}{0.006458}[/tex]

nt = 94.238 payments is required to pay off the loan.

How much interest will they save by paying the loan using the number of payments from part (c)?

The total amount of interest payed on $921.433 = 921.433 × 30(12) years

= 331715.88

The total amount paid using 921.433 and 1300 = (921.433 × 60 )+( 1300 + 94.238)

= 177795.38

The amount of interest saved = 331715.88  - 177795.38

The amount of interest saved = $153920.5

Answer: a or c one of those

Step-by-step explanation:

Answer:

3

Step-by-step explanation:

As per table, the mass is:

So the atom 3 has the highest mas of 132

Answer choice 3 is correct

Answer:

3

Step-by-step explanation:

Add protons and neutrons.

Answer:

[tex]y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}[/tex]

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

[tex]m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7[/tex]

So, the solution of the equation is

[tex]y = Ae^{m_{1} t} + Be^{m_{2} t}[/tex]

where m₁ = 3 and m₂ = -7.

So,

[tex]y = Ae^{3t} + Be^{-7t}[/tex]

Also,

[tex]y' = 3Ae^{3t} - 7e^{-7t}[/tex]

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

[tex]y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1 (1)[/tex]

[tex]y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}[/tex]

Substituting A into (1) above, we have

[tex]\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1 \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}[/tex]

Substituting B into A, we have

[tex]A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}[/tex]

Substituting A and B into y, we have

[tex]y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}[/tex]

So the solution to the differential equation is

[tex]y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}[/tex]