Answer:
7,173
x = 7
Step-by-step explanation:
Within the question it gives two examples.
x = 0 corresponds to 2000
x = 1 corresponds to 2001
If you pay close attention the x-value is always the same as the last digits in the year. So with 2007 the last digit is 7, which then leads us to determine that
x = 7
f (x) = -327 (7) + 9462
f (x) = -2289 + 9462 = 7173
Answer:
They can be seated in 120 differents ways.
Step-by-step explanation:
Taking into account that there are 3 couples and every couple has an specific way to sit, for simplify the exercise, every couple is going to act like 1 option and it's going to occupy 1 Place. If this happens we just need to organize 5 options (3 couples and 2 singles) in 5 Places (3 for a couple and 2 for the singles)
It means that now there are just 5 Places in the row and 5 options to organized. So the number of ways can be calculated using a rule of multiplication as:
<u> 5 </u>*<u> 4 </u>* <u> 3 </u> * <u> 2 </u> * <u> 1 </u> = 120
1st place 2nd Place 3rd place 4th Place 5th Place
Because we have 5 options for the 1st Place, the three couples and the 2 singles. Then, 4 options for the second Place, 3 options for the third place, 2 for the fourth place and 1 option for the 5th place.
Finally, they can be seated in 120 differents ways.
20 thousands would be 200 hundreds
750 sq ft
Step-by-step explanation:
since the fencing is going around the perimeter, you have to have two sides with the width of 25 feet, you would use 50 feet of fencing for 2 of the sides. You will have 60 feet of fencing left, divided by 2 (length of the garden) = 30 feet long.
The area (length x width) or (25 x 30) = 750 sq ft of garden space. Draw a rectangle with 25 ft on two parallel sides and 30 ft on the other two sides and show these equation= 25 x 2= 50, 110 - 50 = 60
60 ÷ 2 = 30, 30 x 25 = 750
Answer: chi square test goodness of fit
Step-by-step explanation:
Is used to investigate whether or not there is a significant difference between a distribution generated from a sample and a hypothesize population distribution
