Beth has just received an inheritance of $500,000 and would like to be able to make quarterly withdrawals over the next 20 years. She decides on an annuity that pays 6.4%, compounded quarterly. How much will her quarterly payments be to draw the account down to zero at the end of 20 years? Quarterly Payments =​

Answer:

percentage of paper used= 18%

Number of papers printed= 90

Total number of papers in the paper package= 'y'

total number of papers

= [tex]\frac{18}{100}[/tex] × y =90

transpose to the other side,

y= 90 ×[tex]\frac{100}{18}[/tex]

y=10 x100

y=1000

hence total number of papers in the package is 1000

Hope this helps

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Answer:

[tex]27,\!405[/tex] if:

Step-by-step explanation:

[tex]\displaystyle \genfrac{(}{)}{0}{}{30}{4} = \frac{30 \times 29 \times 28 \times 27}{4 \times 3 \times 2 \times 1} = 27,\!405[/tex].

Assume for now that the ordering of the four cards does matter. Hands like [tex]\verb!A!\, \verb!B!\, \verb!C!\, \verb!D![/tex] and [tex]\verb!A!\, \verb!B!\, \verb!D!\, \verb!C![/tex] would then be considered different from one another.

There would [tex]30[/tex] choices for the first card. Since the first card was not returned to the pile, there would be only [tex]29[/tex] choices for the second card. Likewise, there would be [tex]28[/tex] choices for the third card and [tex]27[/tex] for the fourth.

By this reasoning, there would be [tex]30 \times 29 \times 28 \times 27 = 657,\!720[/tex] different ways to draw a hand of four cards from this set when the ordering of these four cards do matter.

However, in many card games, once a hand of cards is drawn, the ordering of cards within that hand does not matter. In other words, hands like [tex]\verb!A!\, \verb!B!\, \verb!C!\, \verb!D![/tex] and [tex]\verb!A!\, \verb!B!\, \verb!D!\, \verb!C![/tex] would not be considered as distinct from one another.

In that case, the [tex]30 \times 29 \times 28 \times 27 = 657,\!720[/tex] ways of drawing cards would include a large number of duplicates.

There are be [tex]4 \times 3 \times 2 \times 1 = 24[/tex] ways to arrange a hand of four cards when the order matter. Hence, when the ordering within a hand no longer matters, each hand of four cards would have been counted [tex]24[/tex] times among those [tex]30 \times 29 \times 28 \times 27 = 657,\!720[/tex] ways.

Therefore, when the ordering of cards within a set does not matter, [tex]\displaystyle \frac{30 \times 29 \times 28 \times 27}{4 \times 3 \times 2 \times 1} = 27,\!405[/tex] would give the number of distinct ways to draw a hand of four out of this set of thirty distinct cards.

Answer:

D. Vertical translation of 4 units.

Step-by-step explanation:

This transformation keeps x the same but shifts y up 4 units (adding 4 to it). This is a vertical shift up 4, which we can also call a vertical "translation."

Answer:

17

Step-by-step explanation:

f(x) = 3x + 2

f(5) = 3(5) + 2

f(5) = 15 + 2

f(5) = 17

Answer:

17

Step-by-step explanation:

f(x) = 3x + 2

Plug in:

3(5) + 2

15 + 2 = 17

Hope this helped.

Answer:

[tex]x = 30[/tex]

Step-by-step explanation:

If 3, 9, 10, x are in proportion. Then,

[tex]x \times 3 = 10 \times 9[/tex]

[tex]x \times 3 = 90[/tex]

[tex]x = 30[/tex]

Step-by-step explanation:

Given that 3, 9, 10, x are in proportion <br> product of exterms = Product of means <br>

<br>

3x=9×10

X=(9x10)/3

X=90/3

X=30

Answer:

703.7 in²

Step-by-step explanation:

SA = 2πrh+2πr²

= 2×π×7×9+2×π×7²

= 224π

= 703.7 in² (rounded to the nearest tenth)

Answer:

224π

in²

Step-by-step explanation:

the models aren't given..

Answer: no models given

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

I took a test in school and this was the answer...at least for my class.

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Answer:

8:15

Step-by-step explanation:

For drawing the graph for y = 3/4x + 7, a person can draw a point at x of 0 and y of __7_, a second point by going over 3 and up __8.25__, and then draw a line through the points.

All the points (and only those points) which lie on the graph of the function satisfy its equation.

Thus, if a point lies on the graph of a function, then it must also satisfy the function.

For this case, the equation given to us is:

[tex]y = \dfrac{3}{4}x + 7[/tex]

Any equation of the form [tex]y = mx + c[/tex] where m and c are constants and x and y are variables is the equation of a straight line.

For a straight line to be characterized, only two points are sufficient.

For x = 0, the y-coordinate would be such that it would satisfy the equation [tex]y = \dfrac{3}{4}x + 7[/tex]

Putting x = 0, we get:

[tex]y = \dfrac{3}{4} \times 0 + 7 = 7[/tex]

Thus, y-coordinate of the point on this line whose x-coordinate is 0 is 7. Thus, (0,7) is one of the point's coordinate on the considered line.

Putting x = 3, we get:

[tex]y = \dfrac{3}{4} \times 3 + 7 = 9.25[/tex]

Thus, y-coordinate of the point on this line whose x-coordinate is 0 is 9.25 . Thus, (0,9.25) is another of the point's coordinate on the considered line.

Thus, for drawing the graph for y = 3/4x + 7, a person can draw a point at x of 0 and y of __7_, a second point by going over 3 and up __8.25__, and then draw a line through the points.

Learn more about points lying on graph of a function here:

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Answer:

=32÷40%

=32÷0.4

=80

the owner have total 80 stores

Step-by-step explanation:

th

Answer:

80.

Step-by-step explanation:

40 % = 32 stores, so:

10% = 32/4 = 8 stores, and

100% = 8*10 = 80 stores.

Answer:

0.1088 or 10.88%

Step-by-step explanation:

q = 1 - 0.35 = 0.65

P(X=2) = 12C2 × (0.35)² × (0.65)¹⁰

= 0.1088

Answer:

106.25km

Step-by-step explanation:

8.5cm = 0.000085km

map:real life

1:1250000

0.000085km : 0.000085km * 1250000 = 106.25km

Answer:

26 and 2

Step-by-step explanation:

Given,

Sum of two even numbers = 54

let one number be x

another number be x +2

x + x +2 = 54

2x + 2 = 54

2x = 54 - 2

2x = 52

x = 52/2

x = 26

Therefore, the numbers are 26 and 26 + 2 = 28..

Refer to the attaachment

Answer:

5+4 = 9

6+4 = 10

7+4 =11

Step-by-step explanation:

Answer:

(5,9)

(6,10)

(7,11)

Step-by-step explanation:

w = m + 4

when m = 5,

w = 5 + 4 = 9

when m = 6

w = 6 + 4 = 10

when m = 7

w = 7 + 4 = 11

It seems like you want to find the sum of 38 to 115:

[tex] \displaystyle \large{38 + 39 + 40 + 41 + ... + 114 + 115}[/tex]

If we notice, this is arithmetic series or the sum of arithmetic sequences.

To find the sum of the sequences, there are three types of formulas but I will demonstrate only one and the best for this problem.

[tex] \displaystyle \large{S_n = \frac{n(a_1+a_n) }{2} }[/tex]

This formula only applies to the sequences that have the common difference = 1.

Given that a1 = first term of sequence/series, n = number of terms and a_n = last term

We know the first term which is 38 and the last term is 115. The problem here is the number of sequences.

To find the n, you can use the following formula.

[tex] \displaystyle \large{n = (a_n - a_1) + 1}[/tex]

Substitute an = 115 and a1 = 38 in the formula of finding n.

[tex] \displaystyle \large{n = (115 - 38) + 1} \\ \displaystyle \large{n = (77) + 1} \\ \displaystyle \large{n = 78}[/tex]

Therefore the number of sequences is 78.

Then we substitute an = 115, a1 = 38 and n = 78 in the sum formula.

[tex] \displaystyle \large{S_{78} = \frac{78(38+115) }{2} } \\ \displaystyle \large{S_{78} = \frac{39(38+115) }{1} } \\ \displaystyle \large{S_{78} = 39(153) } \\ \displaystyle \large \boxed{S_{78} = 5967}[/tex]

Hence, the sum is 5967.

Answer:

i) AUB={1,2,3,4,5,6}

ii) BUC={1,2,3,4,5,6,7}

iii)(AUB)UC={1,2,3,4,5,6,7}

iv) {1,2,3,4}

v) {4}

vi){1,2,3,4,5,6,7}

Step-by-step explanation

AUB means all the elements belongs to a and b

AnB means elements common to both sets

Answer:

(g÷f) (x) (1.8) ³x²+⁷x+2

Step-by-step explanation:

Im glad to help you

Answer:

Its the 3 one

Step-by-step explanation:

Answer:

wdym R6

Step-by-step explanation:

Answer:

12

Step-by-step explanation:

6*2

The equation of the line through (0, 1) and (c, 0) is

y - 0 = (0 - 1)/(c - 0) (x - c)   ==>   y = 1 - x/c

Let L denote the given lamina,

L = {(x, y) : 0 ≤ x ≤ c and 0 ≤ y ≤ 1 - x/c}

Then the center of mass of L is the point [tex](\bar x,\bar y)[/tex] with coordinates given by

[tex]\bar x = \dfrac{M_x}m \text{ and } \bar y = \dfrac{M_y}m[/tex]

where [tex]M_x[/tex] is the first moment of L about the x-axis, [tex]M_y[/tex] is the first moment about the y-axis, and m is the mass of L. We only care about the y-coordinate, of course.

Let ρ be the mass density of L. Then L has a mass of

[tex]\displaystyle m = \iint_L \rho \,\mathrm dA = \rho\int_0^c\int_0^{1-\frac xc}\mathrm dy\,\mathrm dx = \frac{\rho c}2[/tex]

Now we compute the first moment about the y-axis:

[tex]\displaystyle M_y = \iint_L x\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}x\,\mathrm dy\,\mathrm dx = \frac{\rho c^2}6[/tex]

Then

[tex]\bar y = \dfrac{M_y}m = \dfrac{\dfrac{\rho c^2}6}{\dfrac{\rho c}2} = \dfrac c3[/tex]

but this clearly isn't independent of c ...

Maybe the x-coordinate was intended? Because we would have had

[tex]\displaystyle M_x = \iint_L y\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}y\,\mathrm dy\,\mathrm dx = \frac{\rho c}6[/tex]

and we get

[tex]\bar x = \dfrac{M_x}m = \dfrac{\dfrac{\rho c}6}{\dfrac{\rho c}2} = \dfrac13[/tex]

The center of mass for a uniform triangular shape is on its centroid. The y-coordinate of the center of mass of the lamina is 1/3 (independent of c).

If the surface is plane triangle approximately and mass is uniformally distributed, then its center of mass will lie on the centroid of that triangle.

The point of intersection of a triangle's medians is its centroid (the lines joining each vertex with the midpoint of the opposite side).

If the triangle has its vertices as  [tex](x_1, y_1), (x_2, y_2) , \: (x_3, y_3)[/tex], then the coordinates of the centroid of that triangle is given by:

[tex](x,y) = \left( \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{y_1 + y_2 + y_3}{3} \right)[/tex]

For this case, the  triangular lamina has vertices (0, 0), (0, 1) and (c, 0)

Assuming its mass is spread regularly, the coordinates of its center of mass would be:

[tex](x,y) = \left( \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{y_1 + y_2 + y_3}{3} \right)\\\\(x,y) = \left( \dfrac{0+0+c}{3} + \dfrac{0+1+0}{3} \right) = (c/3, 1/3)[/tex]

Thus, the y-coordinate of the center of mass of the lamina is 1/3 (independent of c).

Learn more about centroid here:

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The t-value that satisfies the condition is t = 2.797 (rounded to three decimal places).

To determine the t-value such that 0.01 of the area under the curve is to the right of it, we need to use a t-distribution table or a statistical software.

The t-distribution is characterized by its degrees of freedom, which in this case is given as 27.

To find the t-value corresponding to an upper tail probability of 0.01, we need to locate the critical value in the t-distribution table.

Since we're looking for the right tail, we want to find the t-value that corresponds to a cumulative probability of 1 - 0.01 = 0.99.

Let's denote this critical value as t(critical).

Using the t-distribution table or a statistical software, we find that the t-value for a cumulative probability of 0.99 with 27 degrees of freedom is approximately 2.797.

Therefore, the t-value that satisfies the condition is t = 2.797 (rounded to three decimal places).

Learn more about t-distribution click;

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Answer:

A=13×13

=169sqr.units