Answer:
Domain[0, ∞). Range (-∞,4).
Step-by-step explanation:
domain is your lowest x (zero) to your highest x (infinity)
range is the lowest y (negative infinity) to your highest y (four)
Answer:
{-29/4, -19/2, -47/4}
Step-by-step explanation:
Note that this is a decreasing arith. seq. Each term is smaller than the previous term, the result of the subtraction of 2 1/4 (equivalent to 9/4).
The fourth term is -11/4; subtracting 9/4 results in -20/4, or -5.
The fifth term is -20/4; subtracting 9/4 results in -29/4.
The 6th term is -29/4; subtracting 9/4 results in -38/4, or -19/2.
The 7th term is -38/4; subtracting 9/4 results in -47/4.
Thus, the "next three terms" following the first four terms are {-29/4, -19/2, -47/4}
Answer:
c. 0.088
Step-by-step explanation:
Let p(s) be the proportion of apples defected in the sample. The probability that p(s)<0.05 can be calculated by calculating z statistic of 0.05:
where
- p is the proportion of the apples in fact defected (0.08)
- N is the sample size (150) Then,
z(0.05)=
And P(z<-1.354)≈0.0879
Therefore the probability that a random sample of 150 among 8% defected apples can be accepted is 0.088.
Answer:
1) 2/11
2) 2/121
Step-by-step explanation:
1. Because there are 2 E's, you have to chances out of your 11 changes, thus why the fraction 2/11 is the solution.
2. You know that 2/11 is the first fraction, so now find R. R has a 1/11 chance of getting chosen, so you multiply 2/11 and 1/11 to get 2/121.
Answer:
The explicit rule of the geometric sequence
aₙ = 187.5 (0.8)ⁿ⁻¹
Step-by-step explanation:
<u><em> Step(i):-</em></u>
Given that the third term of the sequence = 120
tₙ = a rⁿ⁻¹
t₃ = a r³⁻¹ = ar²
120 = ar² ..(i)
Given that the fifth term of the given geometric sequence = 76.8
tₙ = a rⁿ⁻¹
t₅ = a r⁵⁻¹ = a r⁴
76.8 = a r⁴...(ii)
<u><em>Step(ii):</em></u>-
Dividing (ii) and (i)
r² = 0.64
r =√ 0.64 = 0.8
Substitute r= 0.8 in equation (i)
120 = ar²
120 = a(0.8)²
⇒
<u><em>Step(iii):-</em></u>
The explicit rule of the geometric sequence
aₙ = a rⁿ⁻¹
put a= 187.5 and r = 0.8
aₙ = 187.5 (0.8)ⁿ⁻¹