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den301095 [7]
3 years ago
5

during a trip to the supermarket, Jim bought the following items bread $1.45 beans $0.88 toothpaste $2.90 milk $3.55 dog food $6

.50 The state in which he lives levies a 6% sales tax on all items except food for human beings. What was Jim's total bill?
Mathematics
2 answers:
mr Goodwill [35]3 years ago
5 0
15.28 I think it is the answer bc it said total so u add all the numbers. I added them and it gave me 15.28
borishaifa [10]3 years ago
3 0

Answer:

15.84 would be the correct answer


6.50 x 6%= .39 + 6.50= 6.89

2.90 x 6%=.17 + 2.90= 3.07

then add all togeather


Step-by-step explanation:


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Spiral Review At a restaurant, Brianna and Naomi each order an appetizer for $6.50 and an entree for $9.75. They decide to leave
Licemer1 [7]

Answer: 4.55

Step-by-step explanation:

6.50x2 = 13 + 9.75 = 20% = 4.55

hoped this helped lol

4 0
3 years ago
Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-
Shtirlitz [24]

Answer:

The number c is 2.

Step-by-step explanation:

Mean Value Theorem:

If f is a continuous function in a bounded interval [0,4], there is at least one value of c in (a,b) for which:

f(c) = \frac{1}{b-a}\int\limits^a_b {f(x)} \, dx

In this problem, we have that:

f(x) = x, a = 0, b = 4

So f(c) = c

----------

f(c) = \frac{1}{b-a}\int\limits^a_b {f(x)} \, dx

c = \frac{1}{4-0}\int\limits^0_4 {x} \, dx

c = \frac{1}{4-0}*8

c = 2

The number c is 2.

3 0
3 years ago
What is the solution to the inequality +4 50?
kicyunya [14]

Answer:

d<-1.75

Step-by-step explanation:

4 0
2 years ago
Ppppppppeeeelelelelele help
kirill115 [55]
11/3*6/11=66/33=2
6 24/20-2 15/20=2 9/20
5 0
3 years ago
In an experiment, a fair coin is tossed 13 times and the face that appears (H for head or T for tail) for each toss is recorded.
zalisa [80]

Answer:

1) 1 element

2) 13 elements

3) 22 elements

4) 40 elements

Step-by-step explanation:

1) Only one element will have no tails: the event that all the coins are heads.

2) 13 elements will have exactly one tile. Basically you have one element in each position that you can put a tail in.

3) There are {13 \choose 2} = 78 elements that have exactly 2 tails. From those elements we have to remove the only element that starts and ends with a tail and in the middle it has heads only and the elements that starts and ends with a head and in the 11 remaining coins there are exactly 2 tails. For the last case, there are {11 \choose 2} = 55 possibilities, thus, the total amount of elements with one tile in the border and another one in the middle is 78-55-1 = 22

4) We can have:

  • A pair at the start/end and another tail in the middle (this includes a triple at the start/end)
  • One tail at the start/end and a pair in the middle (with heads next to the tail at the start/end)

For the first possibility there are 2 * 11 = 22 possibilities (first decide if the pair starts or ends and then select the remaining tail)

For the second possibility, we have 2*9 = 18 possibilities (first, select if there is a tail at the end or at the start, then put a head next to it and on the other extreme, for the remaining 10 coins, there are 9 possibilities to select 2 cosecutive ones to be tails).

This gives us a total of 18+22 = 40 possibilities.

5 0
3 years ago
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