For 36 3 6 13. For 48 4 8 24. I think there are more but I can't think of them
Year rank population
1790 3 18320
1800 4 24937
1810 4 33787
1820 4 43298
1830 4 61392
1840 5 93383
1850 3 136881
1860 5 177840
1870 7 250526
1880 5 362839
1890 6 448477
1900 5 560892
1910 5 670585
1920 7 748060
1930 9 781188
1940 9 770816
1950 10 801444
1960 13 697197
1970 16 641071
1980 20 562994
1990 20 574283
Answer:
The percentage of the bag that should have popped 96 kernels or more is 2.1%.
Step-by-step explanation:
The random variable <em>X</em> can be defined as the number of popcorn kernels that popped out of a mini bag.
The mean is, <em>μ</em> = 72 and the standard deviation is, <em>σ</em> = 12.
Assume that the population of the number of popcorn kernels that popped out of a mini bag follows a Normal distribution.
Compute the probability that a bag popped 96 kernels or more as follows:
Apply continuity correction:


*Use a <em>z</em>-table.
The probability that a bag popped 96 kernels or more is 0.021.
The percentage is, 0.021 × 100 = 2.1%.
Thus, the percentage of the bag that should have popped 96 kernels or more is 2.1%.
Answer: -7b² + 2b - 8
Step-by-step explanation:
<u>Given expression</u>
3 - b (7b + 2) + 3b - (11 - b)
<u>Expand parentheses and apply the distributive property if necessary</u>
=3 - b · 7b - b · 2 + 3b - 11 + b
=3 - 7b² - 2b + 3b - 11 + b
<u>Combine like terms</u>
=-7b² + (3b - 2b + b) + (3 - 11)
=
Hope this helps!! :)
Please let me know if you have any questions
Answer:
65.4
Step-by-step explanation:
60*.09 = 5.4
60+5.4 = 65.4