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Leona [35]
3 years ago
8

A group of friends is addressing wedding invitations. Casey can address 22 envelopes in one hour, Anne can address 46 envelopes

in two hours, and Paula can address 70 envelopes in three hours. About how long will it take them to address 450 envelopes?
A. 19.57 hours
B. 23.00 hours
C. 7.34 hours
D. 6.59 hours
Mathematics
2 answers:
DENIUS [597]3 years ago
7 0

Answer:

D. 6.59 hours

Step-by-step explanation:

First, divide 46 by 2.

46 / 2 = 23

Next, divide 70 by 3.

70 / 3 = 23.3333333... (rounded to 23.33)

Then, add them together and multiply by 6.59.

22 + 23 + 23.33 = 68.33

68.33 * 6.59 = 450.2947

Round it to the nearest ones place, and you get 450.

stepan [7]3 years ago
5 0

Answer:

6.30

Step-by-step explanation:

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miv72 [106K]

Answer:

B 98

Step-by-step explanation:

4*7=28  times 3 = 84

3.5*4=14

84+14=98

Hope this helps! :)

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2 years ago
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At 9 AM Monday morning Thomas feels a beaker with water and places it in the corner of the classroom. At 1 PM Tuesday Thomas exa
aev [14]

The level has dropped 11 mms in 22 hours

that is 0.5mm per hour

So the remaining 31 mm will be gone in 31 / 0.5 = 62 hours

62 hours after 11 am Wednesday brings us to   11 am Friday + 14 hours

=  1 am Saturday.   answer

5 0
4 years ago
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2. 8.SP.1.2
andrew11 [14]

Answer:

A

Step-by-step explanation:

A. because the line of best fit has to be at best fit between the points so is a good approximation of where all the points are heading when you try to predict something.

8 0
3 years ago
Given the formula PV = nRT , solve for the variable T
aleksley [76]
T=PV/Rn you just divide both sides by Rn 
8 0
3 years ago
Use the given values of n and p to find the minimum usual value μ−2σ and the maximum usual value μ+2σ. Round to the nearest hund
lbvjy [14]

Answer:

μ−2σ = 1,089.26

μ+2σ = 1,097.62

Step-by-step explanation:

The standard deviation of a sample of size 'n' and proportion 'p' is:

\sigma=\sqrt{\frac{p*(1-p)}{n} }

If n=1139 and p =0.96, the standard deviation is:

\sigma=\sqrt{\frac{p*(1-p)}{n}}\\\sigma = 0.001836

The minimum and maximum usual values are:

\mu-2\sigma = (p-2\sigma)*n\\\mu+2\sigma = (p+2\sigma)*n

\mu-2\sigma = (0.96-2*0.001836)*1139\\\mu-2\sigma = 1,089.26\\\mu+2\sigma = (0.96+2*0.001836)*1139\\\mu+2\sigma = 1,097.62

5 0
3 years ago
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