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Alexus [3.1K]
3 years ago
6

kali trains for a race by running round a track. each lap is 400 meters. she run 4 kilometers each day. how many lapsaround he t

rack does kali run in 3 days
Mathematics
1 answer:
Levart [38]3 years ago
3 0
1 km = 1000m
so 4 km = 4(1000) = 4000 m

so Kali runs 4000 m per day
4000/400 = 10 laps per day

for 3 days = 3(10) = 30 laps for 3 days
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Answer my math question I asked so many times I lost most of points by
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Answer:

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Step-by-step explanation:

equation: y = x - 6 + x^2

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I figured this out by looking at the point (6,0). To get 0 (y), you have to subtract 6. Knowing this, I subtracted 6 from the rest of the coordinates, leaving me with numbers that are able to be squared to get y. This led me to the equation x - 6 + x^2.

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The Kroger Company is one of the largest grocery retailers in the United States with over two thousand grocery stores across the
Ilia_Sergeevich [38]

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1) Categorical

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Step-by-step explanation:

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The data collected by Kroger in this example categorical or quantitative identified below:

From the given information, Kroger uses an online customer opinion to obtain the data about its products and services. All the questions based on yes or no type questions. Here the questions are ‘products that have a brand name, products that are environmentally friendly, products that are organic’ these type of questions cannot be expressed numerically so the data collected by Kroger Company is categorical variable because these answers of the questions cannot be counted.

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The measurement scale is identified below:

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In a triangle ABC, with angles A, B, and C and sides AB, BC, and AC, angle B is a right (90°) angle. If the sin of angle A is 0.
rewona [7]
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5 0
3 years ago
Bacteria of species A and species B are kept in a single environment, where they are fed two nutrients. Each day the environment
DiKsa [7]

Answer:

We require 4,550 of species A and 1,460 of species B that can coexist in the environment so that all the nutrients are consumed each day

Step-by-step explanation:

Let n₁ be the population of A required and n₂ be the population of B required.

Now we require 2 units of the first nutrient for species A and one unit of the first nutrient for species B. The total nutrients required by species A is 2n₁ and that by species B is 1n₂ = n₂. So, the total nutrients required by both species A and B is 2n₁ + n₂. Since this equals the quantity of the first nutrient which is 10,560, then  2n₁ + n₂ = 10,560 (1)

Now we require 5 units of the second nutrient for species A and 6 units of the second nutrient for species B. The total nutrients required by species A is 5n₁ and that by species B is 6n₂. So, the total nutrients required by both species A and B is 5n₁ + 6n₂. Since this equals the quantity of the first nutrient which is 31,510, then  5n₁ + 6n₂ = 31,510 (2).

So, we have two simultaneous equations which we would solve to find the populations of A and B which satisfy both equations.

2n₁ + n₂ = 10,560  (1)

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Substituting equation (3) into (2), we have

5n₁ + 6(10,560 - 2n₁) = 31,510

expanding the brackets, we have

5n₁ + 63,360 - 12n₁ = 31,510

collecting like terms, we have

5n₁ - 12n₁ = 31,510 - 63,360

simplifying, we have

- 7n₁ = -31,850

dividing both sides by -7, we have

n₁ = -31,850/-7

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Substituting n₁ = 4,550 into (3), we have

n₂ = 10,560 - 2(4,550)

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So, we require 4,550 of species A and 1,460 of species B that can coexist in the environment so that all the nutrients are consumed each day

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2 years ago
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