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Semmy [17]
2 years ago
5

At Prairie Elementary School, students are asked to pick their lunch ahead of time so the kitchen staff will know what to prepar

e. On Monday, 6 times as many students chose hamburgers as chose salads. The number of students who chose lasagna was one third the number of students who chose hamburgers. If 225 students ordered lunch, how many students chose each option if hamburger, salad, and lasagna were the only three options?
Mathematics
1 answer:
finlep [7]2 years ago
5 0

Answer:

6 x 225 is the answer

Step-by-step explanation:

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A store sold 1800 bags in 1985 and 1500 bags in 1986. If they averaged selling 2100 bags per year through 1987, how many
iris [78.8K]

Answer:

3000 bags

Step-by-step explanation:

In this case we must apply the formula of the mean, which would be the sum of the values divided by the amount of data is the mean:

m = (a1 + a2 ... an) / n

In this case we know the mean, we must know the 1987 data, therefore:

Let x be the number of bags sold in 1987, replacing in the previous equation:

2100 = (1800 + 1500 + x) / 3

2100 * 3 = 3300 + x

x = 6300 - 3300

x = 3000

Therefore, for the average to be a total of 2100 bags in those 3 years, the amount of bag sold in 1987 must have been 3000 bags

8 0
3 years ago
Shawna purchased a new laptop.  The sale advertised 25% off the regular price.  If the regular purchase price is $575.00, what w
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6 0
2 years ago
If C is the current value and S is the starting value, write an
LekaFEV [45]
Is there a diagram with this question?
7 0
3 years ago
What why did you say you had $80.50 and you add that up
soldier1979 [14.2K]

Answer:

im not sure i understand the question

Step-by-step explanation:

80.50 plus........plus what?

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8 0
3 years ago
19. If the pg", and "terms of an A.P. are a,b, and c respectively. prove that: p(b-c) + (ca) + rſa-b) = 0 50 Approved by Curricu
professor190 [17]

Answer:

Step-by-step explanation:a = m + (p-1)*d

b = m + (q-1)*d

c = m + (r-1)*d

p(b-c) = p*(q-r)*d

q(c-a) = q*(r-p)*d

r(a-b) = r*(p-q)*d

p(b-c)+q(c-a)+r(a-b)

= p*(q-r)*d + q*(r-p)*d +r*(p-q)*d

= (pq-pr+qr-pq+rp-qr)*d

= 0*d = 0

So i prove p(b-c)+q(c-a)+r(a-b)=0 hope this is helpfull

4 0
3 years ago
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