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Gelneren [198K]
4 years ago
11

A birthday celebration meal is $46.80 including​ tax, but not the tip. find the total cost if a 10% tip is added to the cost of

the meal.
Mathematics
1 answer:
irina1246 [14]4 years ago
8 0
$51.48 should be the answer
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Write an equation to represent the N'th term of the sequence (2,-1.-4,-7,...)
mr_godi [17]

Answer: a(n) = 5 - 3n

the sequence has:

a1 = 2

a2 = -1

a3 = -7

.........

we can see that: a2 - a1 = a3 - a2 = -3

=> the sequence is a  arithmetic sequence

=> the distance between the numbers is d = -3

because this sequence is a  arithmetic sequence

=> a(n) = a1 + (n - 1)d = 2 + (n - 1).(-3) = 5 - 3n

Step-by-step explanation:

7 0
3 years ago
In the diagram, AC=BD. Show that AB=CD
Debora [2.8K]
I believe you just have to draw 2 little lines in between AB and CD.
8 0
3 years ago
Your sock drawer has 3 pairs of black socks, 4 pairs of brown socks, 1 pair of yellow socks and 5 pairs of blue socks. you reach
erik [133]
All socks together: 13
Black socks: 3

Probability to pick out black socks: 3/13

After one pick out (You picked out a pair of black socks):

Probability to pick out black socks: 2/12

Multiply them:

2/12 * 3/13 = 1/26 ≈ 0,038

8 0
4 years ago
Help me plz branniest to whoever right
Iteru [2.4K]

Answer:

vertex: (-1,25)

aos; -1

left x int: -6

right x int: 4

(not sure)

range is less than of equal to 25

4 0
3 years ago
The two dot plots below compare the forearm lengths of two groups of schoolchildren:
Angelina_Jolie [31]

Answer:

Group A, because seven children have a forearm length longer than 10 inches

Step-by-step explanation:

Create the dot plots based on the given information (please see attached images).

Group A = blue dots

Group B = red dots

From inspection of the attached dot plots, we can see that Group A appears to have the longer average forearm length.  This is because 7 children have a forearm length of longer than 10 inches.

To calculate the average forearm length, sum all data values and divide by the total number of data values.

\textsf{Average of Group A}=\dfrac{ 3 \times 10 + 4 \times 11 + 3 \times 12}{10}=11

\textsf{Average of Group B}=\dfrac{ 3 \times 9 + 5\times 10+ 2\times 11}{10}=9.9

Thus proving that Group A has the longer average former length.

7 0
2 years ago
Read 2 more answers
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