Answer:
25%
Step-by-step explanation:
Answer:
The expected value for a student to spend on lunch each day = $5.18
Step-by-step explanation:
Given the data:
Number of students______$ spent
2 students______________$10
1 student________________$8
12 students______________$6
23 students______________$5
8 students_______________$4
4 students_______________$3
Sample size, n = 50.
Let's first find the value on each amount spent with the formula:
Therefore,
For $10:
For $8:
l
For $6:
For $5:
For $4:
For $3:
To find the expected value a student spends on lunch each day, let's add all the values together.
Expected value =
$0.4 + $0.16 + 1.44 +$2.3 + $0.64 + $0.24
= $5.18
Therefore, the expected value for a student to spend on lunch each day is $5.18
Answer:
The complement event would be: "No rain today"
Its probability would be 70%
Step-by-step explanation:
Given the event:
A= rain today
The probability of rain today is 30%
P(A)=30%=0.3
The complement event has all the values that are not included in the event.
The complement event would be:
A'= No rain today
The probability of a event is the subtraction between 1 and the complement event:
P(A')=1-P(A)
Then, the probability of the event "No rain today" will be:
P(A')=1-0.3=0.7=70%
Answer:
Part A: 2.8 miles
Part B: 16.8 miles
Step-by-step explanation:
First, solve for how many miles Ted jogged in 1 day:
Variable x = number of miles
11.2/4 = x/1
Cross multiply
11.2 × 1 = 4 × x
11.2 = 4x
Divide both sides by 4
2.8 = x
He jogged 2.8 miles every day.
To find how many miles Ted would jog in 6 days, simply multiply his miles per day by the number of days he will jog:
2.8 miles × 6 days = 16.8 miles in 6 days