Answer:
99.87% of cans have less than 362 grams of lemonade mix
Step-by-step explanation:
Let the the random variable X denote the amounts of lemonade mix in cans of lemonade mix . The X is normally distributed with a mean of 350 and a standard deviation of 4. We are required to determine the percent of cans that have less than 362 grams of lemonade mix;
We first determine the probability that the amounts of lemonade mix in a can is less than 362 grams;
Pr(X<362)
We calculate the z-score by standardizing the random variable X;
Pr(X<362) = 
This probability is equivalent to the area to the left of 3 in a standard normal curve. From the standard normal tables;
Pr(Z<3) = 0.9987
Therefore, 99.87% of cans have less than 362 grams of lemonade mix
892.48
You have to subtract he given numbers.
Answer:
9/10
Step-by-step explanation:
the rule says
so in our case
a=3
b=4
c=5
d=6
so the division is
we can simplify
that's our answer
I don't know if you need step by step of the rule, if you need let me know
Since y has to be larger, the area is above the line.
Since the boundary is included (because of ≥ in stead of >) it should be drawn as a solid line.
Brings us to answer A, right?
See plot below.
Answer:
<h2>
min: 52</h2><h2>
max: 78</h2><h2>
median: 63</h2><h2>
lower quartile median: 54</h2><h2>
upper quartile median: 73</h2><h2>
interquartile range: 19</h2>
Step-by-step explanation:
the temperatures ordered from least to greatest
52, 53, 54, 54, 57, 62, 64, 69, 73, 73, 77, 78
the min is 52
the max is 78
median:
52, 53, 54, 54, 57, 62, 64, 69, 73, 73, 77, 78
(62+64)/2=126/2=63
thus the median is 63
now we don't include the median in finding the upper and lower quartile medians:
lower quartile median:
52, 53, 54, 54, 57, 62
54 (same #s no need to add up and divide by 2)
upper quartile median:
64, 69, 73, 73, 77, 78
73 (same #s no need to add up and divide by 2)
interquartile range is:
upper quartile median-lower quartile median:
73-54
19
