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Sphinxa [80]
2 years ago
12

A can of soda in 1972 costs $0.15, by 2012 it cost $1.35 find the rate at which the cost of a can of soda increases over this ti

me period
Mathematics
1 answer:
fenix001 [56]2 years ago
8 0

Answer:

800%

Step-by-step explanation:

<h2><u>Percentage increase</u></h2><h3><u>formula:</u></h3>

change/ original * 100

change is 1.35 - 0.15 = 1.2

original is the amount in the beginning which is 0.15

= 1.2/0.15 * 100

=800 %

<em><u>the percentage increase is by 800%</u></em>

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Sergei runs a bakery. He needs at least 175 kilograms of flour in total to complete the holiday orders he's received. He only ha
aliya0001 [1]

Answer:

Kindly check explanation

Step-by-step explanation:

Kilogram of flour needed to complete holiday order is atleast 175kg

Number of kilograms available = 34kg

Flour comes in bags each contain 23kg of flour

He wants to buy the smallest number of bags as possible and get the amount of flour he needs.

Let F = number of bags of flours Sergei needs to buy

F = (total number of kilograms needed - number of kilograms available) / kilograms per bag

F = (≥ 175 - 34) / 23

F = ≥ 141 / 23 = ≥ 6.1304347

Since flour is purchased per bag, the smallest number of bags he can possibly buy and still get the amount of flour he need = 7 bags

8 0
3 years ago
Can anybody answer this?
kirza4 [7]
Should be 180 my g, I’m pretty sure
7 0
2 years ago
Read 2 more answers
In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any
Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

E \cap F = \emptyset,\quad E \cup F = \Omega

where \Omega is the whole sample space.

This implies that

P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day d_1.

The second footballer can be born on any other day, so he has 364 possible birthdays:

d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}

the probability for the first two footballers to be born on two different days is thus

1 \cdot \dfrac{364}{365} = \dfrac{364}{365}

Similarly, the third footballer can be born on any day, except d_1 and d_2:

d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}

so, the probability for the first three footballers to be born on three different days is

1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

So, the probability that at least two footballers are born on the same day is

1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

since the two events are complementary.

8 0
3 years ago
How to solve this question ?
Ratling [72]

Step-by-step explanation:

eight times the difference of 2 and a number would be 8 × (2-x)

3 0
2 years ago
What is a factor of 15xy-45x-6y+18?<br><br> A. Y-2<br> B. 5x-3<br> C. 5x-2<br> D. Y-6
AfilCa [17]
Factor the following:
15 x y - 45 x - 6 y + 18
Factor 3 out of 15 x y - 45 x - 6 y + 18:
3 (5 x y - 15 x - 2 y + 6)
Factor terms by grouping. 5 x y - 15 x - 2 y + 6 = (5 x y - 2 y) + (6 - 15 x) = y (5 x - 2) - 3 (5 x - 2):
3 y (5 x - 2) - 3 (5 x - 2)
Factor 5 x - 2 from y (5 x - 2) - 3 (5 x - 2):
Answer:  3 (5 x - 2) (y - 3)
8 0
2 years ago
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