The sum of arithmetic series is given by:
Sn=n/2(a1+an)
where:
n=number of terms
a1=first term
an=nth term
but
n=18, an=275, Sn=4185
plugging the values in the formula we get:
4185=18/2(a1+275)
simplifying this we get:
4185=9(a1+275)
dividing through by 9 we get:
465=a1+275
thus
a1=465-275
a1=190
Answer: first number is 190
Answer:
Step-by-step explanation:
- −5.93 + (−8.62) + 5.93 =
- −5.93 - (8.62 - 5.93) = B
- −(5.93 + 8.62) + 5.93 = C
- −5.93 −8.62 - (- 5.93) = D
- -8.62 A
Answer: D
Step-by-step explanation:
All would add up to 241 instead of 2.41 because you are not using decimals for the constants (Example: 25q not .25q)
Its not possible because if you try to multiply back and forth it wont work.
Answer:
The sample consisting of 64 data values would give a greater precision.
Step-by-step explanation:
The width of a (1 - <em>α</em>)% confidence interval for population mean μ is:

So, from the formula of the width of the interval it is clear that the width is inversely proportion to the sample size (<em>n</em>).
That is, as the sample size increases the interval width would decrease and as the sample size decreases the interval width would increase.
Here it is provided that two different samples will be taken from the same population of test scores and a 95% confidence interval will be constructed for each sample to estimate the population mean.
The two sample sizes are:
<em>n</em>₁ = 25
<em>n</em>₂ = 64
The 95% confidence interval constructed using the sample of 64 values will have a smaller width than the the one constructed using the sample of 25 values.
Width for n = 25:
Width for n = 64:
![\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{64}}=\frac{1}{8}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]](https://tex.z-dn.net/?f=%5Ctext%7BWidth%7D%3D2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7B64%7D%7D%3D%5Cfrac%7B1%7D%7B8%7D%5Ccdot%20%5B2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Csigma%5D)
Thus, the sample consisting of 64 data values would give a greater precision