Answer:
C. Test for Goodness-of-fit.
Step-by-step explanation:
C. Test for Goodness-of-fit would be most appropriate for the given situation.
A. Test Of Homogeneity.
The value of q is large when the sample variances differ greatly and is zero when all variances are zero . Sample variances do not differ greatly in the given question.
B. Test for Independence.
The chi square is used to test the hypothesis about the independence of two variables each of which is classified into number of attributes. They are not classified into attributes.
C. Test for Goodness-of-fit.
The chi square test is applicable when the cell probabilities depend upon unknown parameters provided that the unknown parameters are replaced with their estimates and provided that one degree of freedom is deducted for each parameter estimated.
Answer:
2.03 ft
Step-by-step explanation:
132 divided by 48 is 2.75 and 5.6 divided by 2.75 is 2.03
Answer:
138$
Step-by-step explanation:
120$. Increased by 15%.
Since 15% equals to 0.15, you multiply 0.15 by the price.
0.15*120=18.
Then you ADD the 15%.
120+18=138
The new price is now 138$
Answer:
The probability of randomly picking a red candy is 1/20 and In percentage value 5%
Step-by-step explanation:
There are 5 blue candies, 6 red candies, and 9 yellow candies.
Now, Total
5 + 6 + 9 = 20 candies
Now, The probability of randomly picking a red candy is 1/20
Now, In percentage
1/20 × 100% = 5%
Thus, The probability of randomly picking a red candy is 1/20 and In percentage value 5%.
<u>-TheUnknownScientist</u>
