A regular pentagon is a polygon with five sides of equal length.
The perimeter of a regular polygon is the number of sides (n) multiplied by the length of one side (s).
P=ns
100=5s
s=20
The length of each side is 20 inches
The Experimental probability would be 1/2 or 50% it goes that way for both, hope that helps
To solve this problem you must follow the proccedure shown below:
1. Amount of lemonade in pints:
1 <span>quart of water=2 pints
1 pint of lemon
(9 ounces of honey)(0.0625 pints/1 ounce)=0.56 pints
Total in pints=2 pints+1 pint+0.56 pints
Total in pints=3.56 pints
2. </span>Amount of lemonade in<span> cups:
Total in cups=(3.56 pints)(2 cups/1 pint)
Total in cups=7.12 cups
3. </span>Amount of lemonade in<span> ounces:
Total in ounces=(7.12 cups)(8 ounces/1 cup)
Total in ounces=56.96 ounces</span>
Answer:
Here we have given two catogaries as degree holder and non degree holder.
So here we have to test the hypothesis that,
H0 : p1 = p2 Vs H1 : p1 not= p2
where p1 is population proportion of degree holder.
p2 is population proportion of non degree holder.
Assume alpha = level of significance = 5% = 0.05
The test is two tailed.
Here test statistic follows standard normal distribution.
The test statistic is,
Z = (p1^ - p2^) / SE
where SE = sqrt[(p^*q^)/n1 + (p^*q^)/n2]
p1^ = x1/n1
p2^ = x2/n2
p^ = (x1+x2) / (n1+n2)
This we can done in TI_83 calculator.
steps :
STAT --> TESTS --> 6:2-PropZTest --> ENTER --> Input all the values --> select alternative "not= P2" --> ENTER --> Calculate --> ENTER
Test statistic Z = 1.60
P-value = 0.1090
P-value > alpha
Fail to reject H0 or accept H0 at 5% level of significance.
Conclusion : There is not sufficient evidence to say that the percent of correct answers is significantly different between degree holders and non-degree holders.
