Answer:
Stratified Sampling
Step-by-step explanation:
Since Keri divides the day into different strata and each unit is selected from each strata randomly. So, it is Stratified Sampling.
Further, In Stratified Sampling population is divided into several groups such that within the group it is homogeneous and between the group it is heterogeneous. And now a selection of each stratum and unit has an equal chance of selection.
A. (−3, 3)
<span>3x – 4y = 21
</span>3(-3) - 4(3) = 21
-21 = 21 >>>>> not equal
B. (−1, −6)
<span>3(-1) - 4(-6) = 21
</span>21 = 21 >>>>>>>>>>Equal
C. (7, 0)
<span>3(7) - 4(0) = 21
</span>21 = 21>>>>>>>>>>equal
D. (11, 3)
<span>3(11) - 4(3) = 21
</span>21 = 21 >>>>>>>>>equal
Answer:
16
Step-by-step explanation:
Putting a + in front of a number indicates that the number is positive. so we are just going to add it up.
10+6=16
10:6=5:3
Hope it helps:)
Step-by-step explanation:
1. use the equation
current amount X interest to the power of years
commpound interest is 2.3%
The starting value is £7900
So: £7900*1.023^5=£8851.26
Therefore, the answer is £8851.26 after 5 years of interest added
Hope this helps!!!
Have a nice day :)
<h3>
Answer: Mean = 218.9.</h3><h3>Median = 229</h3><h3>Mode = Zero mode.</h3>
Step-by-step explanation:
Given blood cholesterol level was measured for each of 8 adults (in mg/dL) are:
264, 191, 160, 148, 262, 212, 268, 246
In order to find the mean, we need to add all those 8 numbers and divide by 8.
Therefore, mean = ![\frac{264+191+160+148+262+212+268+246}{8} =\frac{1751}{8} =218.9.](https://tex.z-dn.net/?f=%5Cfrac%7B264%2B191%2B160%2B148%2B262%2B212%2B268%2B246%7D%7B8%7D%20%3D%5Cfrac%7B1751%7D%7B8%7D%20%3D218.9.)
<h3>Mean = 218.9.</h3>
In order to find the median, we need to arrange them in ascending order:
148, 160, 191, 212, 246, 262, 264, 268.
The middle most two values are 212 and 246.
Therefore, median = ![\frac{212+246}{2} =\frac{458}{2} = 229.](https://tex.z-dn.net/?f=%5Cfrac%7B212%2B246%7D%7B2%7D%20%3D%5Cfrac%7B458%7D%7B2%7D%20%3D%20229.)
<h3>Median = 229.</h3>
![\mathrm{The\:mode\:is\:the\:term\:in\:the\:data\:set\:that\:appears\:the\:most.}](https://tex.z-dn.net/?f=%5Cmathrm%7BThe%5C%3Amode%5C%3Ais%5C%3Athe%5C%3Aterm%5C%3Ain%5C%3Athe%5C%3Adata%5C%3Aset%5C%3Athat%5C%3Aappears%5C%3Athe%5C%3Amost.%7D)
![\mathrm{If\:there\:is\:more\:than\:one\:term\:that\:appears\:the\:most,\:then\:there\:is\:no\:mode.}](https://tex.z-dn.net/?f=%5Cmathrm%7BIf%5C%3Athere%5C%3Ais%5C%3Amore%5C%3Athan%5C%3Aone%5C%3Aterm%5C%3Athat%5C%3Aappears%5C%3Athe%5C%3Amost%2C%5C%3Athen%5C%3Athere%5C%3Ais%5C%3Ano%5C%3Amode.%7D)
![\mathrm{Count\:the\:number\:of\:times\:each\:element\:appears\:in\:the\:list}](https://tex.z-dn.net/?f=%5Cmathrm%7BCount%5C%3Athe%5C%3Anumber%5C%3Aof%5C%3Atimes%5C%3Aeach%5C%3Aelement%5C%3Aappears%5C%3Ain%5C%3Athe%5C%3Alist%7D)
![\begin{pmatrix}148&160&191&212&246&262&264&268\\ 1&1&1&1&1&1&1&1\end{pmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bpmatrix%7D148%26160%26191%26212%26246%26262%26264%26268%5C%5C%201%261%261%261%261%261%261%261%5Cend%7Bpmatrix%7D)
![\mathrm{The\:most\:common\:element\:is\:not\:unique,\:so\:there\:is\:no\:mode}](https://tex.z-dn.net/?f=%5Cmathrm%7BThe%5C%3Amost%5C%3Acommon%5C%3Aelement%5C%3Ais%5C%3Anot%5C%3Aunique%2C%5C%3Aso%5C%3Athere%5C%3Ais%5C%3Ano%5C%3Amode%7D)
<h3>Therefore, Mode = Zero mode.</h3>