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Morgarella [4.7K]
2 years ago
12

the cafeteria has 2 cases of tuna and 6 single cans of tuna. in all, there are 75 cans of tuna. How many cans of tuna are in a c

ase
Mathematics
2 answers:
Nastasia [14]2 years ago
7 0

Answer:63


Step-by-step explanation:


andreyandreev [35.5K]2 years ago
5 0
I think it may be 63?
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if 2 more students took the math test and both made a score of , what would the picture graph look like?
Sindrei [870]
I think it would be a dot graph
3 0
3 years ago
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Equivalent expression to 3(4m-2)-2(m+5
wolverine [178]

Answer:

10m - 16

Step-by-step explanation:

1.  Perform the indicated multiplication.  We get 12m - 6 - 2m - 10.

2.  Combine like terms:  We get 10m - 16

4 0
3 years ago
Based on the circle graph how many pounds of milk and eggs does the average American consume each year PLS EXPLAIN FOR BRAINLEST
Gnoma [55]

Answer:

H.  552

Step-by-step explanation:

Circle graph information:

  • Milk and eggs = x%
  • Fruits and vegetables = 38%
  • Cereals = 10%
  • Meat and fish = 10%
  • Sugar = 8%
  • Fats and oils = 4%

Total consumption = 1840 pounds per person

First, calculate the percentage of milk and eggs (the value of x).

Total percentages = 100%

⇒ x + 38 + 10 + 10 + 8 + 4 = 100

⇒ x + 70 = 100

⇒ x = 30

Therefore, 30% of food consumption is milk and eggs.

To calculate the number of pounds of milk and eggs the average American consumes each year, simply find 30% of the total consumption:

= 30% of 1840

= 30% × 1840

=\sf \dfrac{30}{100} \times 1840

= 552

Therefore, the average American consumes 552 pounds of milk and eggs each year.

5 0
2 years ago
Read 2 more answers
What equation represents the relationship between p and q shown in the table
mash [69]

q = 10p because all of them are multiplied by ten

6 0
3 years ago
In order to estimate the average time spent on the computer terminals per student at a local university, data were collected fro
VladimirAG [237]

Answer:

Margin of error  for a 95% of confidence intervals is 0.261

Step-by-step explanation:

<u>Step1:-</u>

 Sample n = 81 business students over a one-week period.

 Given the population standard deviation is 1.2 hours

 Confidence level of significance = 0.95

 Zₐ = 1.96

Margin of error (M.E) = \frac{Z_{\alpha  }S.D }{\sqrt{n} }

Given n=81 , σ =1.2 and  Zₐ = 1.96

<u>Step2:-</u>

<u />Margin of error (M.E) = \frac{Z_{\alpha  }S.D }{\sqrt{n} }<u />

<u />Margin of error (M.E) = \frac{1.96(1.2) }{\sqrt{81} }<u />

On calculating , we get

Margin of error = 0.261

<u>Conclusion:-</u>

Margin of error  for a 95% of confidence intervals is 0.261

<u />

4 0
3 years ago
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