T is a linear function of C ==> T = mC + b
Each additional chirp C corresponds to an increase in T of 0.25 ==> m = 0.25
T = 52 when C = 20 ==> 52 = 0.25(20) + b ==> b = 52 - 5 = 47
So T = 0.25C + 47
When C = 100 then T = 0.25(100) + 47 = 25 + 47 = 72
When C = 120 then T = 0.25(120) + 47 = 30 + 47 = 77 so the
temperature range is between 72 and 77 when there are between 100 and
120 chirps per minute
Answer:
The conclusion that both groups of overweight and non - overweight got cardiovascular benefit from playing DDR games requires Inferential statistics.
Step-by-step explanation:
Inferential statistics is simply a type of research statistic whereby a generalized conclusion is made about a larger group based on representative observations
Now,in the given question, we see that both group hearts were above the minimum recommended for cardiovascular exercise. Now we can infer that the DDR games played by both groups gave them cardiovascular benefits. This conclusion is an example of Inferential statistics where we generalize about a large population based on observations from a small sample.
Thus the conclusion that both groups of overweight and non-overweight got cardiovascular benefit from playing DDR games requires Inferential statistics.
B. Associative Property
, you're working out the numbers in the parentheses before anything else.
Answer:
a. 
b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349
Step-by-step explanation:
a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.
b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.
c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842
d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166
e. The z-score related to 6.4 kg is 
and the z-score related to 7 kg is 
, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194
f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349
Answer:
30-100/1000*120
Step-by-step explanation:
Source: Khan Academy
