Answer:
150
Step-by-step explanation:
This is a word problem which should be tackled accordingly.
The sum of 48 and itself
= 48+48
= 96
It's half and half of its half
= 48/2 and (48/2)/2
= 24+ 24/2
= 24+12
= 36
Is added to 18
= 96+36+18
= 150
From the description above the entire sum is 150.
Answer:
Infinite number of rational numbers exist between any two distinct rational numbers. We know that a rational number is a number which can be written in the form of qp where p and q are integers and q =0.
Answer:
I should help u am busy strongly with mine
The correct interpretation of the p-value is given by:
B If the average battery life really is 5 hours, then a sample of 10 observations having a sample mean of 4.25 hours or lower would only occur about 3.8% of the line.
<h3>How to find the p-value of a test?</h3>
It depends on the test statistic z, as follows:
- For a left-tailed test, it is the area under the normal curve to the left of z, which is the p-value of z.
- For a right-tailed test, it is the area under the normal curve to the right of z, which is 1 subtracted by the p-value of z.
- For a two-tailed test, it is the area under the normal curve to the left of -z combined with the area to the right of z, hence it is 2 multiplied by 1 subtracted by the p-value of z, which means that the p-value for a two-tailed test is twice the p-value of a one-tailed test.
In this problem, a left-tailed test is used, as we are testing if the mean is less than 5 hours.
The sample mean from the 10 times was of 4.25, and the p-value is of 0.038, which means that the area to the left of Z under the normal curve is of 0.038, that is, a sample mean of 4.25 hours or lower would only occur about 3.8% of the line, hence option B is correct.
You can learn more about p-values at brainly.com/question/13873630
The equation for a circle is
(x-h)² + (y-k)² = r²
h = your given x
y = your given k
Let's plug everything in!
(x - 5) ² + (y - (-1)) ² = 12²
Your final equation is
(x - 5) ² + (y + 1) ² = 144
<em>Hope I helped! Comment or message me if you have any questions :) </em>