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Serhud [2]
2 years ago
12

This list shows the age at which 43 U.S. Presidents began their terms. 57 61 50 54 54 54 56 54 61 54 48 49 42 51 61 57 68 65 50

51 60 52 57 51 52 47 56 62 69 58 49 56 55 55 43 64 57 64 46 55 51 55 46 What is the mode for presidents’ age when they begin their term? a. 51 c. 55 b. 54 d. 57
Mathematics
2 answers:
tamaranim1 [39]2 years ago
8 0

Answer:

B) 54

Step-by-step explanation:

A mode of a number set is the number the appears the most.

In the list given, 54 was the number that appeared the most.

ValentinkaMS [17]2 years ago
3 0

Answer:

B

Step-by-step explanation:

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The cost of a car is $15,570. You plan to make a down payment of $1,500, and a monthly payment of $338.08 for 60 months. What is
Helga [31]
Hi there
1) b
15,570−1,500
=14,070
2) a
338.08×60
=20,284.8
3)d
20,284.8−14,070
=6,214.8
4) c
338.08×60+1,500
=21,784.8

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2 years ago
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An oblique prism with a square base of edge length x units has a volume of One-halfx3 cubic units. Which expression represents t
lara [203]

Answer:one-half x units

Step-by-step explanation:

Given

Prism has a square base with length x\ units

If the volume of prism V=\frac{1}{2}x^3\ units

We know

Volume=base\ area\times height

Base area =x\times x=x^2\ units

height=\frac{Volume}{area}

height=\frac{\frac{1}{2}x^3}{x^2}

height=\frac{x}{2}\ units

7 0
2 years ago
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Noah read 6 books in 9 months. What was his rate of reading in books per month?​
MatroZZZ [7]

a quarter of a book a month

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2 years ago
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Evaluate the line integral by the two following methods. xy dx + x2y3 dy C is counterclockwise around the triangle with vertices
nadezda [96]

Answer:

a)

\frac{2}{3}

b)

\frac{2}{3}

Step-by-step explanation:

a) The first part requires that we use line integral to evaluate directly.

The line integral is

\int_C xydx +  {x}^{2}  {y}^{3} dy

where C is counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 2)

The boundary of integration is shown in the attachment.

Our first line integral is

L_1 = \int_ {(0,0)}^{(1,0)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is y=0, x varies from 0 to 1.

When we substitute y=0 every becomes zero.

\therefore \: L_1 =0

Our second line integral is

L_2 = \int_ {(1,0)}^{(1,2)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is:

x = 0 \implies \: dx = 0

y varies from 1 to 2.

We substitute the boundary and the values to get:

L_2 = \int_ {1}^{2}1 \cdot y(0) +  {1}^{2}   \cdot \: {y}^{3} dy

L_2 = \int_ {1}^2 {y}^{3} dy =  \frac{8}{3}

The 3rd line integral is:

L_3 = \int_ {(1,2)}^{(0,0)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is

y = 2x \implies \: dy = 2dx

x varies from 0 to 1.

We substitute to get:

L_3 = \int_ {1}^{0} x \cdot \: 2xdx +  {x}^{2}  {(2x)}^{3}(2 dx)

L_3 = \int_ {1}^{0} 8 {x}^{5}  + 2 {x}^{2} dx  =  - 2

The value of the line integral is

L = L_1 + L_2 + L_3

L = 0 +  \frac{8}{3}  +  - 2 =  \frac{2}{3}

b) The second part requires the use of Green's Theorem to evaluate:

\int_C xydx +  {x}^{2}  {y}^{3} dy

Since C is a closed curve with counterclockwise orientation, we can apply the Green's Theorem.

This is given by:

\int_C \: Pdx +Q  \: dy =  \int \int_ R \: Q_y -  P_x \: dA

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int \int_ R \: 3 {x}^{2}  {y}^{2}  -  y \: dA

We choose our region of integration parallel to the y-axis.

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \int_ 0^{2x}  \: 3 {x}^{2}  {y}^{2}  -  y \: dydx

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \:  {x}^{2}  {y}^{3}  -   \frac{1}{2}  {y}^{2} |_ 0^{2x}  dx

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \:  8{x}^{5} -  2 {x}^{2}   dx =  \frac{2}{3}

8 0
2 years ago
What is another way to group the factors (3x2)x5
tia_tia [17]

<u>Answer</u>

3×(2×5)

<u>Explanation</u>

Multiplication of numbers is associative. For example,

(a×b)×c = a×(b×c)

This is also called grouping. We multiply more than 2 numbers by grouping.

For the equation given above, (3x2)x5, it can also be grouped as 3×(2×5).

5 0
2 years ago
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