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N76 [4]
3 years ago
7

Amelia used 1/7 of an ounce of butter to make 1/4 of a pound of Jello. How many ounces of butter are needed to make 1 pound of J

ello
Mathematics
2 answers:
son4ous [18]3 years ago
3 0

Answer:

4/7

Step-by-step explanation:

Gnesinka [82]3 years ago
3 0

Answer:

4/7 ounces of butter makes a pound of jello

Step-by-step explanation:

7 times 4 is 28, so 4/28 ounces of butter for 7/28 pounds of jello. so 4/28 times 4 is 16/28. 16/28 is our answer, then simplified is 4/7. Thats how we get our answer.

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Geometry question
brilliants [131]

Answer:

Step-by-step explanation:

12pl+B where p represents the perimeter of the base, l the slant height and B the area of the base.

6 0
3 years ago
Please help me with this problem ASAP, Thanks!!<br>​
Tema [17]

Answer:

x=28

y= -4

Step-by-step explanation:

9x+6y= -12

-x-4y= -12

multiply -x-4y=-12 by 9 so the x's are equal

9x+6y= -12

-9x-36y= -108

add the equations so the x's cancel out

-30y=120

y= -4

plug y into an equation to find x

-x-4(-4)=-12

-x+16= -12

-x= -28

x=28

6 0
3 years ago
What is the sum of -3 and -1
noname [10]

Answer:

-4

Step-by-step explanation:

8 0
4 years ago
Why is positive 3 ?
ycow [4]
3 is positive because it is more than 0. Anything more than zero is positive.
4 0
3 years ago
The lengths of pregnancies in a small rural village are normally distributed with a mean of 262 days and a standard deviation of
RideAnS [48]

Answer:

The range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].

Step-by-step explanation:

We are given that the lengths of pregnancies in a small rural village are normally distributed with a mean of 262 days and a standard deviation of 17 days.

Let X = <u><em>lengths of pregnancies in a small rural village</em></u>

SO, X ~ Normal(\mu=262,\sigma^{2} = 17^{2})

Here, \mu = population mean = 262 days

         \sigma = standard deviation = 17 days

<u>Now, the 68-95-99.7 rule states that;</u>

  • 68% of the data values lies within one standard deviation points.
  • 95% of the data values lies within two standard deviation points.
  • 99.7% of the data values lies within three standard deviation points.

So, middle 68% of most pregnancies is represented through the range of within one standard deviation points, that is;

[ \mu -\sigma , \mu + \sigma ]  =  [262 - 17 , 262 + 17]

                          =  [245 days , 279 days]

Hence, the range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].

5 0
3 years ago
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