1: 19 (4x+25=6x -13, solve for x, 25+13=6x-13, 38=2x, x=19, 4(19)+25=101 6(19)-13+101)
2: 23 (3x-8=2x+15)
3: Madison has $59 and Saul has $31 (together they have $90 so divided by 2 and you get 45, Madison has 6 dollars more so subtract 6 from Saul and ad 6 to Madison, make sure that when you ad them you get $90)
4 : 5 weeks ($3 each week, so you multiply 3x5 and get 15. 15+57=72 but the problem says more then so maybe it’s 6 weeks cause that’s $75)
5: 10 weeks (he has 1,800 and needs 2,150, subtract 1,800 from 2,150 for how much he needs. You get 350. He needs that and he makes 35 a week so that would be 350 divided by 35, which is 10)
6:20 miles (it’s 2.25 for that’s first mile. He spent 6.05 so subtract 2.25 from 6.05 and you get 3.80, 3.80 divided by 0.20 is 19, then you need to add that first mile and it’s 20)
7: 6 figures in a complete set (Charlie starts out with 2 and Patrick starts out with 20 individual figures. Charlie has 6 sets and Patrick has 3, i multiplied the number of sets they had by 6 then added their individual figures until they numbers matched up)
8: an orange has 45 calories and an apple has 75 calories (5x-150=3x, 5x-3x= -2x, -2x divided by -150 = 75, 75-30 is 45. 75 is the apple and because it’s 30 more you subtract 30 for orange)
Answer:
2x-2y
Step-by-step explanation:
2 times x = 2x
2 times -y = -2y
Answer:
Step-by-step explanation:
-3x -2 = 19
+3x +3x
-2 = 19 + 3x
-19 -19
-21 = 3x
Divided by 3
-7 = x
Answer: y= -4cos (pi/4x)-3
Step-by-step explanation:
Answer:
C. Test for Goodness-of-fit.
Step-by-step explanation:
C. Test for Goodness-of-fit would be most appropriate for the given situation.
A. Test Of Homogeneity.
The value of q is large when the sample variances differ greatly and is zero when all variances are zero . Sample variances do not differ greatly in the given question.
B. Test for Independence.
The chi square is used to test the hypothesis about the independence of two variables each of which is classified into number of attributes. They are not classified into attributes.
C. Test for Goodness-of-fit.
The chi square test is applicable when the cell probabilities depend upon unknown parameters provided that the unknown parameters are replaced with their estimates and provided that one degree of freedom is deducted for each parameter estimated.