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NeX [460]
2 years ago
11

The following were the recorded birth weights for babies born July 16, 2011: 8.1 lbs., 6.0 lbs., 4.7 lbs., 6.9 lbs., 5.6 lbs., 7

.7 lbs., 6.3 lbs., 7.8 lbs., 6.1 lbs., and 9.2 lbs. What was the average birth weight on the day? Round to two decimal places.
Mathematics
1 answer:
rodikova [14]2 years ago
6 0
8.1 + 6.0 + 4.7 + 6.9 + 5.6 + 7.7 + 6.3 + 7.8 + 6.1 + 9.2 = 68.4/9 = 7.6
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Of 3/1=6/2 then 3-1/1=
Oksi-84 [34.3K]
I'm not sure what you want with the first part (3/1=6/2)

well PEMDAS or parenthasees exponents, mult or division whichever comes first, addition subtraction, whichever comes first so

3-1/1 would be divison first
1/1=1
3-1=2


5 0
2 years ago
Antonio bought a shirt for 1/2 off. He paid $21.75 for the shirt s. Draw a bar diagram to find the original cost of the shirt.
Anon25 [30]

Answer:

21.75 times 2 is 43.5

8 0
3 years ago
Rewrite the fraction pairs with common denominators
Nataliya [291]

Answer:

a) \frac{3}{20} and \frac{16}{20} = \frac{19}{20}

b) \frac{8}{21} and \frac{18}{21} = 1\frac{5}{21}

c) \frac{18}{72} and \frac{32}{72} = \frac{25}{36}

Step-by-step explanation:

a) \frac{3}{20} and \frac{4}{5}

\frac{4}{5} x 4 = \frac{16}{20}

\frac{3}{20} and \frac{16}{20} = \frac{19}{20}

b) \frac{8}{21} and \frac{6}{7}

\frac{6}{7} x 3 = \frac{18}{21}

\frac{8}{21} and \frac{18}{21} = 1\frac{5}{21}

c) \frac{2}{8} and \frac{4}{9}

\frac{2}{8} x 9 = \frac{18}{72}

\frac{4}{9} x 8 = \frac{32}{72}

\frac{18}{72} and \frac{32}{72} = \frac{25}{36}

3 0
2 years ago
S= 2LW + 2LH + 2WH<br> solve for W
Darina [25.2K]

Rewrite the equation as

2lw+2lh+2wh=a.2lw+2lh+2wh=a

Subtract

2lh

from both sides of the equation.

2lw+2wh=a−2lh

Factor

2w out of 2lw+2wh.

2w(|+h)=w−2lh

Divide each term by

2(l+h)

and simplify.

.w=a−2lh2(l+h)

4 0
2 years ago
According to TMUSS Quarterly (Totally Made-Up Sports Statistics) April 2017, the probability that the Tampa Bay Buccaneers will
dsp73

Answer: 0.31

Step-by-step explanation:

Let A denotes the event of Tampa Bay Buccaneers will score a touchdown on their opening drive and B denote the event that their defense will have 3 or more sacks in the game.

Given : P(A)=0.14     P(B) = 0.31     P(A or B)=0.14

Formula : P(A and B)= P(A) + P(B) - P(A or B)

Now, the probability that they will both score a touchdown on the opening drive and have 3 or more sacks in the game will be :-

P(A and B)= 0.14 + 0.31 - 0.14=0.31

Hence, the required probability : 0.31

7 0
3 years ago
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