sixth person work per week=47 hours
Average:
The average is defined as the mean value which is equal to the ratio of the sum of the number of a given set of values to the total number of values present in the set
Given :
the number of hours worked per week for five of her employees is
30, 30, 38, 35, and 30.
the average work of 6 employees is 35 hours per week
Solution :
Let, Total work per week= S
sixth person work per week=x
Total work per week= Number of people x average working hour
S= 35(6) = 210..(1)
S = 30+30+38+35+30+x
S= 163+x..(2)
By equation 1 and equation 2
210= 163+x
210-163=x
47=x
sixth person work per week=47 hours
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Answer:
3
Step-by-step explanation:
it resembles a staircase when graphed
Answer:
2 and -3
Step-by-step explanation:
because a coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression (e.g. 4 in 4x).
so in 2x-3y+5, 2 and -3 are the coefficients.
Answer:
a) For the 90% confidence interval the value of 
and 
, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:
b) For the 99% confidence interval the value of 
and 
, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:
Step-by-step explanation:
Previous concepts
The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".
The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.
The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."
Solution to the problem
Part a
For the 90% confidence interval the value of and , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:
Part b
For the 99% confidence interval the value of and , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:
