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iris [78.8K]
3 years ago
14

Caleb invested $6,200 in an account paying an interest rate of 7 1/2% compounded

Mathematics
1 answer:
Citrus2011 [14]3 years ago
7 0
Oh this is a hard one I will get back to you whenever I get it
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To simplify 8.3 − 4 15(6.1 + 12) , which operation must be performed first
Ivan
Do whats in the parenthesis first......add the 6.1 + 12
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9x9=<br> 8x8=<br> 8x7=<br> 5x4=<br> Vă rog să mă ajutați
sergiy2304 [10]

Answer:

9×9=72

8×8=64

8×7=56

5×4=20

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A ______ is a rule that pairs each element in one set with exactly one element from a second set.​
Cloud [144]

Answer:

function

Step-by-step explanation:

A <u>function</u> is a rule that pairs each element in one set with exactly one element from a second set.​

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3 years ago
Boxes of raisins are labled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is re
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Answer:

The 99 % confidence interval on the basis of mean is ( 21.7393  ;   22.0257)

Step-by-step explanation:

Mean = Sum of observations / Number of observations

Mean = 21.88 +21.76 +22.14 +21.63+ 21.81 +22.12+ 21.97+ 21.57+ 21.75+ 21.96 +22.20 +21.80/ 12

Mean =x`= 262.59/12=  21.8825

Standard Deviation = s= ∑x²/n - ( ∑x/n)²

∑x²/n= 478.7344 +473.4976 + 490.1796+467.8569+ 475.6761 + 489.2944+ 482.6809+ 465.2649+ 473.0625+ 482.2416 +492.84 + 475.24/ 12

∑x²/n= 5746.5689/12= 478.8807 = 478.881

Standard Deviation = s= ∑x²/n - ( ∑x/n)²

s= 478.881- (21.8825)²= 478.881-478.843= 0.037

The confidence limit 99% for the mean will be determined by

x` ± α(100-1) √s/n

Putting the values in the above equation

= 21.8825 ± 2.58 √0.037/12

Solving the square root

= 21.8825 ± 2.58 (0.05549)

Multiplying the square root with 2.58

=21.8825 ± 0.1432

Adding and subtracting would give

21.7393  ;   22.0257,

Hence the 99 % confidence interval on the basis of mean is ( 21.7393  ;   22.0257)

5 0
3 years ago
What is the slope of the line that passes<br> through<br> 0(-4, 8) and (5, 2)?
zlopas [31]

Answer:

(-8,-4) and (-2,5). at reflect on x axis

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3 years ago
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