Answer:
- The sequence is an Arthemtic Progression
An=A1+(n-1)d
A1 is first term, An is nth term, n is number of term, and d is common difference
therefore
A4=35, A1= -17
A4=A1+(4-1)d
35= -17+3d
35+17=3d
52=3d
52/3=3d/3
14=d
common diffrence(d)=14
- The general solution is given by
An= -17+(n-1)14
An= -17+14n-14
An= -31+14n
<u>An= 14n-31</u>
A14 term, means n=14
From An=A1+(n-1)d
A14= -17+(14-1)14
= -17+(13×14)
= -17+182
= 165.
<u>Therfore, the 14th term is 165.</u>
2. A sequence has a CR of 4/5 and its eighth term (a8) is (393216/3125). What is its general equation? Its 3rd term?
<u>solution</u>
common ratio(r)=4/5
eighth term(G8)=393216/3125
From Gn= G1r^(n-1)
G8 means n=8
G8=G1r^(n-1)
393216/3125=G1(4/5)^(8-1)
393216/3125=G1(4/5)^7
G1=(393216/3125)/(4/5)^7
G1=600
<u>The first term is given by G1=600</u>
The General equation is given by
The General equation is given by Gn= 600(4/5)^(n-1)
3rd term (G3)
G3= G1(4/5)^(3-1) where n=3,
=600(4/5)^2
=600(16/25)
=384
<u>Therefore, the 3rd term is given by G3= </u><u>3</u><u>8</u><u>4</u><u>.</u>
<u>I</u><u> </u><u>h</u><u>a</u><u>v</u><u>e</u><u> </u><u>m</u><u>a</u><u>d</u><u>e</u><u> </u><u>s</u><u>o</u><u>m</u><u>e</u><u> </u><u>C</u><u>o</u><u>r</u><u>r</u><u>e</u><u>c</u><u>t</u><u>i</u><u>o</u><u>n</u><u>s</u><u> </u><u>i</u><u> </u><u>m</u><u>e</u><u>s</u><u>s</u><u>e</u><u>d</u><u> </u><u>u</u><u>p</u><u> </u><u>s</u><u>o</u><u>m</u><u>e</u><u>w</u><u>h</u><u>e</u><u>r</u><u>e</u><u>.</u>
Answer:
There is not enough information to tell the answer to this problem.
Step-by-step explanation:
sorry
Find the vertex.
x=b/-2a
12/-2(1)
x=-6
then plug into equation
-6^2+12(-6)+6
36-72+6=-30
(-6,-30)
Round each coin weight to the nearest gram first, penny 3g, nickel 5g, dime 2g, quarter 6g, half dollar 11g. Now add those numbers together as he has one of each coin 3+5+2+6+11=27 grams
Answer:
The corresponding p-value, is p = 1
Step-by-step explanation:
The maximum score SAT score, n = 1,600
The mean of the district's total SAT score distribution = 1,200
The claim of one of the districts principal, is the that mean of the district's total SAT score distribution ≠ 1,200
Using proportions, we have;
p = 1,200/1,600 = 0.75
q = 1 - p = 0.25
The margin of error, E = Z√(p·q/n)
∴ E = 5% = Z×√((0.75 × 0.25)/1,600)
z = 0.05/(√((0.75 × 0.25)/1,600)) ≈ 4.61880
Therefore, the corresponding p-value, p = 1