Answer:
Not an arithmetic sequence
Step-by-step explanation:
For the sequence to be arithmetic there must be a common difference between consecutive terms.
- 7 - (- 3) = - 7 + 3 = - 4
- 10 - (- 7) = - 10 + 7 = - 3
- 14 - (- 10) = - 14 + 10 = - 4
The differences are not common , hence not an arithmetic sequence
Answer:
Step-by-step explanation:
so 2/5 of 100 contain chocolate...
2/5 * 100 = 200/5 = 40 <== contain chocolate
Answer:
Sorry no such thing only flourless:(
Step-by-step explanation:
The range of the graph is would be (0,2)
Answer:
Using z score formula:
X = z ∂ + µ
= 157.833
Step-by-step explanation:
Solution:
Mean = µ = 1262
Standard deviation = ∂ = 117
(a) 28th percentile for the number of chocolate chip.
P( z < z) = 28%
= 0.28
P( z<- 0.58) = 0.28
Z = -0.58
By using z score formula:
Z = x - µ /∂
-0.58= x – 117 / 1262
X = (- 0.58)(117) + (1262)
= 1194.14
(b) Middle 97% of bag.
P(-z < z < z) = 97%
= 0.97
P( z < z) – p(z < -z) = 0.97
2p(z < z) -1 = 0.97
2p (z < z) = 1 + 0.97
P(z < z) = 1.97 / 2
= 0.99
P(z < 2.33) = 0.99
Z ± 2.33
By using z score formula:
Z = x - µ / ∂
X = z ∂ + µ
= - 2.33 x 117 + 1262
=989.39
Z = 2.33
X = z ∂ + µ
= 2.33 x 117 + 1262
=1533.61
(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate.
By using standard normal table,
The z dist’n formula:
P(z < z ) = 25%
=0.25
P(z < -0.6745) = 0.25
Z = 0.6745
Using z score formula:
X = z∂ + µ
= - 0.6745 x 117 + 1262
= 1183.0835
First quartile = Q1 =1183.0835
The third quartile is:
P(z<z) = 75%
= 0.75
P(z < 0.6745) = 0.75
Z = 0.6745
Using z score formula:
X = z ∂ + µ
= 0.6745 x 117 + 1262
= 1340.9165
IQR = Q3 – Q1
= 1340.9165 – 1183.0835
= 157.833
