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beks73 [17]
3 years ago
9

Bob eats 4 bottles of of baby food per day. If his father buys him 38 bottles of baby food how many days would it take him to fi

nish it
Mathematics
2 answers:
dimaraw [331]3 years ago
8 0
4 x 9 = 36, so 4 x 10= 40 10 days to finish it all
belka [17]3 years ago
3 0

Answer:

There are 7 days in a week so 7 times 4 = 28

Step-by-step explanation:

38 minus 28 = 10 days

It will take him 10 days

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What is the first quartile, Q1, of the data set?
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3 years ago
1. Suppose half of all newborns are girls and half are boys. Hospital A, a large city hospital, records an average of 50 births
kicyunya [14]

Answer:

5%

Step-by-step explanation:

Hospital A (with 50 births a day), because the more births you see, the closer the proportions will be to 0.5.

Hospital B (with 10 births a day), because with fewer births there will be less variability.

The two hospitals are equally likely to record such an event, because the probability of a boy does not depend on the number of births

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3 years ago
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It's a 7th grade math problem ​
ANEK [815]

Answer:

1296 cm²

Step-by-step explanation:

36² = 1296 cm²

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5 0
2 years ago
A metalworker has a metal alloy that is 20​% copper and another alloy that is 60​% copper. How many kilograms of each alloy shou
puteri [66]
<h3>Answer:</h3>
  • <u>20</u> kg of 20%
  • <u>80</u> kg of 60%
<h3>Step-by-step explanation:</h3>

I like to use a little X diagram to work mixture problems like this. The constituent concentrations are on the left; the desired mix is in the middle, and the right legs of the X show the differences along the diagonal. These are the ratio numbers for the constituents. Reducing the ratio 32:8 gives 4:1, which totals 5 "ratio units". We need a total of 100 kg of alloy, so each "ratio unit" stands for 100 kg/5 = 20 kg of constituent.

That is, we need 80 kg of 60% alloy and 20 kg of 20% alloy for the product.

_____

<em>Using an equation</em>

If you want to write an equation for the amount of contributing alloy, it works best to let a variable represent the quantity of the highest-concentration contributor, the 60% alloy. Using x for the quantity of that (in kg), the amount of copper in the final alloy is ...

... 0.60x + 0.20(100 -x) = 0.52·100

... 0.40x = 32 . . . . . . . . . . .collect terms, subtract 20

... x = 32/0.40 = 80 . . . . . kg of 60% alloy

... (100 -80) = 20 . . . . . . . .kg of 20% alloy

6 0
3 years ago
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