To find the z-score for a weight of 196 oz., use

A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
Answer:
3
Step-by-step explanation:
mhm for your brother
anyways...
2x-3+6x = 21
Adding like terms
8x -3 = 21
Adding 3 on both sides
8x = 24
Dividing both sides by 8
x = 3
Multiply both sides by 10.
2b=990
divide both sides by 2
b= 495
Answer: 2%, second option is correct.
Step-by-step explanation:
To state 1/50 in percent, divide 1 by 50, then multiply by 100
=( 1 ÷ 50) x 100
= 0.02 x 100
= 2%
I hope this helps, please mark as brainliest.
A book is sold in Belize for 80.39 Belize dollars.
In Brazil, it sells for 63.65 Brazilian reals.
The exchange rate of US Dollars to Belize Dollars is 1:1.9246
<span>The exchange rate of US Dollars to Brazilian Real is 1:1.7880
</span>
If we convert the two into dollars,
Belize Dollars
1 : 1.9246 = X : <span>80.39
1.9246X = 80.39
X = $ 41.77
</span>Brazilian Real
1 : 1.7880 = X : <span>63.65 </span><span>
</span>1.7880X = <span>63.65 </span><span>
X = $ 35.60
</span>
Difference = $ 41.77 - $ 35.60
Difference = $ 6.17
So, in Belize the book is more expensive by $ 6.17