Thompson realizes there are more than 24 two week periods in a year, so he revises his estimate using 26×______ to get the numbe
r $_______ for his estimate.
Here is the full question and the chart to reference attached.
Here is the full question and the chart to reference attached.
1 answer:
You might be interested in
Answer:
hey hope this helps
<h3 /><h3>Comparing sides AB and DE </h3>AB =
DE
So DE = 2 × AB
and since the new triangle formed is similar to the original one, their side ratio will be same for all sides.
<u>scale factor</u> = AB/DE
= 2
It's been reflected across the Y-axis
<em>moved thru the translation of 3 units towards the right of positive x- axis </em>
for this let's compare the location of points B and D
For both the y coordinate is same while the x coordinate of B is 0 and that of D is 3
so the triangle has been shifted by 3 units across the positive x axis
Answer:
Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.
Step-by-step explanation:
We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.
So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;
Z = ~ N(0,1)
where, = average age of the random sample of horses with colic = 12 yrs
= average age of all horses seen at the veterinary clinic = 10 yrs
= standard deviation of all horses coming to the veterinary clinic = 8 yrs
n = sample of horses = 60
So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(
12)
P(
12) = P(
) = P(Z
1.94) = 1 - P(Z < 1.94)
= 1 - 0.97381 = 0.0262
Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.