Answer:
Step-by-step explanation:
We would use the t- distribution.
From the information given,
Mean, μ = 2950
Standard deviation, σ = 115
number of sample, n = 25
Degree of freedom, (df) = 25 - 1 = 24
Alpha level,α = (1 - confidence level)/2
α = (1 - 0.98)/2 = 0.01
We will look at the t distribution table for values corresponding to (df) = 24 and α = 0.01
The corresponding z score is 2.492
We will apply the formula
Confidence interval
= mean ± z ×standard deviation/√n
It becomes
2950 ± 2.492 × 115/√25
= 2950 ± 2.492 × 23
= 2950 ± 57.316
The lower end of the confidence interval is 2950 - 57.316 =2892.68
The upper end of the confidence interval is 2950 + 57.316 = 3007.32
The solution is correct.
Answer:
Step-by-step explanation:
each part makes how you can show by what to do on mathematics by fractions and unit fractions with multiplication.
If you mean what goes in the blanks, it is 5 for the first empty box and 10 for the second empty box.
Answer:
d
Step-by-step explanation:
x/7 <2
Multiply both sides if the inequality by 7
x/7×7 < 2×7
Then cancel out the 7s from x/7×7 to get x by itself
now
x < 14
<u>Answer:</u>
The value in 3x + 2 = 15 for x using the change of base formula is 0.465 approximately and second option is correct one.
<u>Solution:</u>
Given, expression is ![3^{(x+2)}=15](https://tex.z-dn.net/?f=3%5E%7B%28x%2B2%29%7D%3D15)
We have to solve the above expression using change of base formula which is given as
![\log _{b} a=\frac{\log a}{\log b}](https://tex.z-dn.net/?f=%5Clog%20_%7Bb%7D%20a%3D%5Cfrac%7B%5Clog%20a%7D%7B%5Clog%20b%7D)
Now, let us first apply logarithm for the given expression.
Then given expression turns into as, ![x+2=\log _{3} 15](https://tex.z-dn.net/?f=x%2B2%3D%5Clog%20_%7B3%7D%2015)
By using change of base formula,
x + 2 = 2.4649
x = 2.4649 – 2 = 0.4649
Hence, the value of x is 0.465 approximately and second option is correct one.