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Butoxors [25]
3 years ago
12

Please help with 12 through 14

Mathematics
1 answer:
BabaBlast [244]3 years ago
4 0
The formula to find the area of a circle is A=r^2pie
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Help please I need help with this asap ​
solniwko [45]

Answer:

The answer is A.

Step-by-step explanation:

Hope fully this helps.

4 0
3 years ago
8TH GRADE MATH HELP PLEASE
s2008m [1.1K]

Answer:

55 is the answer hope it helps...

6 0
2 years ago
Read 2 more answers
Could you check my answer, whether I done it right or wrong?
Iteru [2.4K]

your almost right check once and you wil get especially alternate angles

4 0
3 years ago
How do i solve 51/2x-7=62
Nataly [62]

51/2x-7=62

  1. Multiply both sides by 2x-7

\frac{51}{2x-7} ×(2x-7) = 62(2x-7)

    2. Switch sides

62(2x-7) = 51

     3. Divide each side by 62

\frac{62(2x-7)}{62} = \frac{51}{62}

     4. Simplify

2x-7 = 51/62

     5. Add 7 on both sides

2x-7+7 = (51/62)+7

     6. Simplify

2x = 485/62

      7. Divide both sides by 2

2x/2= 485/62/2

x= 485/124

7 0
3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73
faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62      [from the standard t table]

Therefore,

CI = (Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})

Thus,

Lower limit of CI =  (Mean - \frac{t\times\sigma}{\sqrt{n}})

or

Lower limit of CI =  (98.3 - \frac{2.62\times0.73}{\sqrt{103}})

or

Lower limit of CI = 98.11

and,

Upper limit of CI =  (Mean + \frac{t\times\sigma}{\sqrt{n}})

or

Upper limit of CI =  (98.3 + \frac{2.62\times0.73}{\sqrt{103}})

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and  the mean temperature could very possibly be 98.6°F

7 0
3 years ago
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