Answer:
A) sample mean = $1.36 million
B) standard deviation = $0.9189 million
C) confidence interval = ($1.93 million , $0.79 million)
*since the sample size is very small, the confidence interval is not valid.
Step-by-step explanation:
samples:
- $2.7 million
- $2.4 million
- $2.2 million
- $2 million
- $1.5 million
- $1.5 million
- $0.5 million
- $0.5 million
- $0.2 million
- $0.1 million
sample mean = $1.36 million
the standard deviation:
- $2.7 million - $1.36 million = 1.34² = 1.7956
- $2.4 million - $1.36 million = 1.04² = 1.0816
- $2.2 million - $1.36 million = 0.84² = 0.7056
- $2 million - $1.36 million = 0.64² = 0.4096
- $1.5 million - $1.36 million = 0.14² = 0.0196
- $1.5 million - $1.36 million = 0.14² = 0.0196
- $0.5 million - $1.36 million = -0.86² = 0.7396
- $0.5 million - $1.36 million = -0.86² = 0.7396
- $0.2 million - $1.36 million = -1.16² = 1.3456
- $0.1 million - $1.36 million = -1.26² = 1.5876
- total $8.444 million / 10 = $0.8444 million
standard deviation = √0.8444 = 0.9189
95% confidence interval = mean +/- 1.96 standard deviations/√n:
$1.36 million + [(1.96 x $0.9189 million)/√10] = $1.36 million + $0.57 million = $1.93 million
$1.36 million - $0.57 million = $0.79 million
Answer:
74.73%
Step-by-step explanation:
First, we're gonna find out her total amount
We're gonna use the count interest formula: P = A(1 + r)ⁿ
P = final amount
A = starting amount (1300)
r = rate (0.06)
n = years (5)
P = 1300(1 + 0.06)⁵
P = 1739.693251
Now divide the starting amount by the total amount
1300 ÷ 1739.693251 = 0.7472582
So $60/8= $7.5 which means each ticket is $7.5
$195/$7.5= 26
That means that 26 students can buy tickets with $195
Y = mx + c
If y = 4, and x = 0, then:
4 = -1 (0) + c
c = 4
If c = 4:
y = -1 (2) + 4
y = -2 +4
y = 2
If you plug in the different values of x in each equation, the answer should be:
2,0,-2
Answer:
and 
Step-by-step explanation:
Let
x -----> the number of polyphonic ringtones
y ----> the number of standard ringtones
we know that
-----> equation A
-----> equation B
Solve the system of equations by substitution
Substitute equation A in equation B and solve for y
Find the value of x
