Answer:
C
Explanation:
he be looking better than cashmoney
Design and implement an application that reads a string from the user, then determines and prints how many of each lowercase vowel (a, e, i, o, and u) appear in the entire string . Have a separate counter for each vowel. Also count and print the number of nonvowel characters .
SPECIFICATION OF PROMPTS, LABELS AND OUTPUT : Your code should use the prompt "enter string : ". After the input is read, there are six lines of output , each starting with a different label: "a: ", "e: ", "i: ", "o: ", "u: ", "other: " in that order. After each label is the required count.
For example: if "aardvark heebie jeebies" were read in,
Answer:
<u>Hypothesis</u><u> </u>
H o : No difference between proportions
H 1 : There is a significant difference between proportions
n = 1228
b = 1228 - 856
b = 375
Since n ≥ 25
b = {( a + 0.5 ) - n/2 }\ ( √n )/2
= {( 372 + 0.5 ) - 1228/2 }\ ( √1228 )/2
= - 13.789
- given significance level is α = 0.01
- p-value is less than significance level hence there is enough evidence to reject null hypothesis
- Which means there is a significance difference between the proportion of lawsuits that go to trail which should be dismissed
Answer:
see explaination
Explanation:
a)
customerRecord.lastName
b)
customerPtr->lastName or (*customerPtr).lastName
c)
customerRecord.firstName
d)
customerPtr->firstName or (*customerPtr).firstName
e)
customerRecord.customerNumber
f)
customerPtr->customerNumber or (*customerPtr).customerNumber
g)
customerRecord.personal.phoneNumber
h)
customerPtr->personal.phoneNumber or (*customerPtr).personal.phoneNumber
i)
customerRecord.personal.address
j)
customerPtr->personal.address or (*customerPtr).personal.address
k)
customerRecord.personal.city
l)
customerPtr->personal.city or (*customerPtr).personal.city
m)
customerRecord.personal.state
n)
customerPtr->personal.state or (*customerPtr).personal.state
o)
customerRecord.personal.zipCode
p)
customerPtr->personal.zipCode or (*customerPtr).personal.zipCode
