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tester [92]
3 years ago
6

Help,Please go step by step​

Mathematics
1 answer:
irinina [24]3 years ago
6 0

Answer:

D) \displaystyle 100π\:in^2

Step-by-step explanation:

In the circumference, 20 represents the diameter, and the radius is <em>half</em> the diameter, so take half of it to get 10. Then use it in the area formula:

\displaystyle πr^2 = A \\ \\ π10^2 = 100π

I am joyous to assist you anytime.

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Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus
quester [9]
Hi! 

To compare this two sets of data, you need to use a t-student test:

You have the following data:

-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph

-Wednesday n2=20;  </span>x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510

To calculate the degrees of freedom you need to use the following equation:

df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.



5 0
4 years ago
HEEELEP ASAP PLZ 15 POINTS + BRAINLIEST
skad [1K]
1334 / 23 = 58
mean (average) = 58 miles per hour

answer
 58 miles per hour
3 0
3 years ago
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