Many couples believe that it is getting too expensive to host an average wedding in the United States. According to a statistics study in the U. S., the average cost of a wedding in the U. S. in 2014 was $25,200. Recently, in a random sample of 35 weddings in the U. S. it was found that the average cost of a wedding was $24,224 with a standard deviation of $2,210. If a 95% confidence interval for the mean for the wedding sample is ($23465, $24983), does this mean that the sample results are significantly different from the claimed value for the mean of $25,200?

We conclude that the population mean is different from $25,200

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $25,200

Sample mean, =  $24,224

Sample size, n = 35

Alpha, α = 0.05

Sample standard deviation, σ = $2,210

95% confidence interval:

First, we design the null and the alternate hypothesis

We use two-tailed z test to perform this hypothesis.

The 95% confidence interval tells that we are 95% confidence that the population mean lies within the given interval.

Thus, at 0.05 significance level we fail to accept the null hypothesis as the population mean does not lie in the confidence interval.

Since the claimed population mean is outside of the 95% confidence interval, we conclude that the population mean is different from $25,200.

ad = ae + ed = y + 7

similar triangles therefor

ac/ab = ad/ae

6/4 = (y + 7) / y

cross multiply

6y = 4(y + 7)

6y = 4y + 28

2y = 28

y = 14

step-by-step explanation:

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)