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Finger [1]
3 years ago
15

I am making oatmeal cookies and one batch calls for 2 3/4 cups of flour. If I am going to make 4 batches, how many cups of flour

will I need?
Mathematics
2 answers:
Lynna [10]3 years ago
7 0
The easiest way to figure this out is by multiplying improper fractions. So, the first step will be turning both into improper fractions:

2 3/4 = 11/4
4 = 4/1

Then, multiply the numerators and denominators.

11 x 4 = 44
1 x 4 = 4

This gives you 44/4. Lastly, simplify your fraction with basic division. You're answer should be 11.
AlekseyPX3 years ago
6 0
I would say 11 cups of flour is needed for 4 batches. i am sorry if i am wrong.
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Is this set of ratios equal?: 4/7 and 9/13
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Answer:

NO this set is not equivalent

Step-by-step explanation:

Okay the way you find out if a pair of ratios are equal is you would divide the denominator and nominator with each other, so: 13/7 and 9/4. But sense the answer would be in decimal form and cant be simplified this wouldnt be considered equivalent.  Also if you divide 7 by 4 you get a different answer than when you do 13 divided 9.

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3 years ago
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Will get Brainliest plz help
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Answer: B

Step-by-step explanation:

An equation is y=mx+b for regular equation stander form. C would automatically be eliminated that leaves you with A and B... a you don't have a y intercept so you would start at zero.

You have to pay an entrance fee plus any other souvenirs you would want. so the entrance fee would be the y intercept in "B"

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2 years ago
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A random sample of 77 fields of corn has a mean yield of 26.226.2 bushels per acre and standard deviation of 2.322.32 bushels pe
PSYCHO15rus [73]

Answer:

Therefore the  95% confidence interval is (25,707.480 < E < 26,744.920)

Step-by-step explanation:

n = 77

mean u = 26,226.2  bushels per acre

standard deviation s = 2,322.32

let E = true mean

let A = test statistic

Find 95% Confidence Interval

so

let  A =  (u - E) *  (\sqrt{n}  / s)   be the test statistic

we want      P( average_l <  A  < average_u )  = 95%

look for  lower 2.5%  and the upper 97.5%  Because I think this is a 2-tail test

average_l =  -1.96  which corresponds to the 2.5%

average_u = 1.96

P(  -1.96  <  A  <  1.96)  =  95%

P(  -1.96  <  (u - E) *  (\sqrt{n}  / s)  <  1.96)  =  95%

Solve for the true mean E ok

E   <   u + 1.96* (s  / \sqrt{n})

from  -1.96  <  (u - E) *  (\sqrt{n}  / s)

E < 26,226.2 +  1.96*( 2,322.32 / \sqrt{77} )

E < 26,226.2 +  1.96*( 2,322.32 / \sqrt{77} )

E < 26,226.2 +  518.7197348105429466

upper bound is 26,744.9197

or

u - 1.96* (s  / \sqrt{n})  < E

26,226.2 -  518.7197348105429466  < E

25,707.48026519  < E

lower bound is 25,707.48026519

Therefore the  95% confidence interval is (25,707.480 < E < 26,744.920)

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2 years ago
X + (-x) = 0 help me plz​
andriy [413]

x+-x = 0

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What is 54/12 in a mixed number?
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12 can go into 54 only 4 times because 4•12 is 48. Now 56-48=6 so this is the remainder 6/12 wish can be simplified to 1/2 so your answer is 4 1/2 or 4.5. Hope this helps <333
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3 years ago
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