ASAP Brainliest !!!! Answer both!!
1 answer:
Answer:
question 20 ) 19.1 feet
question 21) 24 clips
Step-by-step explanation:
<h2><u>Question 20:</u></h2><h3><u>Circumference of a circle </u></h3>formula:
therefore we can find the radius (r) , also include the circumference of 60 feet.
where
2 * 3.14 * r = 60
r =
r = 9.554....
<h3>to find the diameter we can use the formula :</h3><em>diameter = radius * 2</em>
diameter = 9.554 * 2
diameter = 19.108
to 3 sf
<h3><u>diameter = 19.1 feet </u></h3><u />
<h2><u>Question 21</u></h2>large binder clips each cost $ 1.50
amount for purchases is $ 36.00
therefore the number of clips bought can be regarded as (x)
the formula would be :
1.50 * x = 36
x = 36/1.5
<u>x = 24</u>
<u></u>
<h3><u>hence the mom bought 24 binder clips</u></h3>You might be interested in
Answer:
The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.
Step-by-step explanation:
20-year mean snowfall in the Denver/Boulder region is 28.76 inches. Test if the snowfall for the 1993-1994 winters has higher than the previous 20-year average.
At the null hypothesis, we test if the average was the same, that is, of 28.76 inches. So
At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
28.76 is tested at the null hypothesis:
This means that
Standard deviation of 7.5 inches. However, for the winter of 1993-1994, the average snowfall for a sample of 32 different locations was 33 inches.
This means that .
Value of the test statistic:
P-value of the test and decision:
The p-value of the test is the probability of finding a sample mean above 33, which is 1 subtracted by the p-value of z = 3.2. In this question, we consider the standard level .
Looking at the z-table, z = 3.2 has a p-value of 0.9993.
1 - 0.9993 = 0.0007
The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.