Answer:
q1=11
q3= 33
Step-by-step explanation:
The data set has 44 number of students. The first quartile is 25 % of the numbers in the data set . So
25 % of 44 = 25/100 * 44= 0.25 *44 = 11
So the first quartile lies at 11.
Similarly the third quartile lies at the 75 % of the numbers of the data set . So
75 % of 44 = 75/100 * 44= 0.75 *44 = 33
So the third quartile lies at 33.
<em><u>Two</u></em> of the four statements that Ana wrote are <em><u>correct</u></em>. Number 1, "AB is a diameter" is incorrect, and so is number 3, "SQ = 12 cm". AB is not a diameter because it is instead a chord. "ST , SP and SQ are radii" is correct because they are straight lines from the center of the circle to the circumference of the circle, which is the exact definition of radius. The third statemet is incorrect because since ST is a radius and it equals 6, that means all radii are equal to 6. SQ is a radius of circle S, so it should also equal 6, not 12. The last statement is correct because PQ is a diameter of circle S. By rule, the diameter is always equal to double of the radius. The radius is 6, so 6 x 2= 12.
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
Answer:
Step-by-step explanation:
D
Answer: B. 89
Step-by-step explanation:
-29.202x7= -204.414
-204.414+293.5= 89.086
So the answer is B. 89