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3 years ago

Using mapping notation to describe a translation up 8 units.

Mathematics

1 answer:

Ad libitum [116K]3 years ago

3 0

The answer would be “b” because in translations when the question says up it’s referring to y

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Find the circumference of this circle.

finlep [7]

Answer:

81.64

Step-by-step explanation:

2(3.14)=6.28

6.28*13=81.64

:))))

3 0

3 years ago

55÷11[120÷2{4+(10+5-7)}]

Lynna [10]

Answer:

3600

Step-by-step explanation:

First, solve the innermost brackets first.

=> 55/11 [120 /2{4 + (10 + 5 - 7)}]

=> 55/11 [120 /2{4 + (8)}]

=> 55/11 [120 /2{4+8}]

=> 55/11 [120 /2{12}]

=> 55/11[ 60 x 12]

=> 55/11 [720]

=> 5 [720]

=> 3600

8 0

3 years ago

Read 2 more answers

Delta pays Pete Rose $280 per day to work in the maintenance department at the airport. Pete became ill on Monday and went home

topjm [15]

Answer:

5x280=1400

Step-by-step explanation:

$1400

if he works on the weekends then add 580

s0 $1980

7 0

3 years ago

A fair 6-faced die is tossed twice. Letting E and F represent the outcomes of the each toss (which are independent), compute the

andreyandreev [35.5K]

Answer:

(a) $\frac{1}{18}$ (d) $\frac{1}{9}$

(b) $\frac{1}{2}$ (e) $\frac{1}{3}$

Step-by-step explanation:

On tossing a 6-face die twice the outcomes of E and F are:

{1, 2, 3, 4, 5 and 6}

And there are total 36 outcomes of the form (E, F).

(a)

The sample space of getting a sum of 11 is: {(5, 6) and (6, 5)}

The probability of getting a sum of 11 is:

$P(Sum 11) =\frac{2}{36} \\=\frac{1}{18}$

(b)

The sample space of getting an even sum is:

{(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6) (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6) (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}

The probability of getting an even sum is:

$P(Even Sum)=\frac{18}{36}\\ =\frac{1}{2}$

(c)

The sample space of getting an odd sum more than 3 is:

{(1, 4), (1, 6), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3) and (6, 5)}

The probability of getting an odd sum more than 3 is:

<em>P</em> (Odd Sum more than 3) =

$=\frac{16}{36}\\=\frac{4}{9}$

(d)

The sample space of (E, F) such that E is even and less than 6, and F is odd and greater than 1 is:

{(2, 3), (2, 5), (4, 3) and (4, 5)}

The probability such that E is even and less than 6, and F is odd and greater than 1 is:

$P(Even E1) =\frac{4}{36} \\=\frac{1}{9}$

(e)

The sample space of E and F such that E is more than 2, and F is less than 4 is:

E = {3, 4, 5 and 6} F = {1, 2 and 3}

Then the total outcomes of (E, F) will be 12.

The probability such that E is more than 2, and F is less than 4 is:

$P(E>2, F$

(f)

The sample space of (E, F) such that E is 4 and sum of E and F is odd is:

{(4, 1), (4, 3) and (4, 5)}

The probability such that E is 4 and sum of E and F is odd is:

$P(E=4, E+F = Odd)=\frac{3}{36} \\=\frac{1}{12}$

6 0

3 years ago

Straight-Line Depreciation A small delivery truck was purchased on January 1 at a cost of $21,920. It has an estimated useful li

lesantik [10]

Answer:

Rate of depreciation per year is 4310 $/ year and the depreciation schedule is given below.

Step-by-step explanation:

According to the question,

since, it is straight-line depreciation,

rate of depreciation per year = $\frac{21920 - 4680}{4}$ $/year

= 4310 $/ year

So we can prepare the following table. (year = 0 for the base period)

Year Depreciation Accumulated Book value

expense depreciation

( in $) (in $) (in $)

0 0 0 21920

1 4310 4310 17610

2 4310 8620 13300

3 4310 12930 8990

4 4310 17240 4680

6 0

3 years ago