Answer:
81.85%
Step-by-step explanation:
Given :
The average summer temperature in Anchorage is 69°F.
The daily temperature is normally distributed with a standard deviation of 7°F .
To Find:What percentage of the time would the temperature be between 55°F and 76°F?
Solution:
Mean =
Standard deviation =
Formula :
Now At x = 55
At x = 76
Now to find P(55<z<76)
P(2<z<-1)=P(z<2)-P(z>-1)
Using z table :
P(2<z<-1)=P(z<2)-P(z>-1)=0.9772-0.1587=0.8185
Now percentage of the time would the temperature be between 55°F and 76°F =
Hence If the daily temperature is normally distributed with a standard deviation of 7°F, 81.85% of the time would the temperature be between 55°F and 76°F.
Answer:
Which One ?
Step-by-step explanation:
She plays soccer and golf for a total of 125 minutes:
s + g = 125
She plays soccer 45 minutes more than she plays golf:
s = g + 45 this is for A
B= 40 minutes of golf everyday
C= Yes, because in fact, she plays 85 minutes of soccer everyday, and 85 is greater than 80, so yes she CAN play 80 minutes of soccer every day. Perhaps they meant possible to spend ONLY 80 minutes, then NO, that wouldn't be possible, because then by playing 45 minutes more soccer than golf, she wouldn't have played enough to have 125 total minutes of total play. Technically the way the question is worded, the answer is yes.
I hope this helps!
Answer: A B
Group A 0.25 0.75
Group B 0.45 0.55
Step-by-step explanation:
Since we have given that
Number of people of A in Group A = 15
Number of people of B in Group A = 45
Number of people of A in Group B = 20
Number of people of B in Group B = 25
Total number of people in Group A = 15+45 = 60
Total number of people in Group B = 20+25 = 45
So, Relative frequencies will be given below:
Relative frequency of A in Group A is given by
Relative frequency of B in Group A is given by
Relative frequency of A in Group B is given by
Relative frequency of B in Group B is given by
Hence, A B
Group A 0.25 0.75
Group B 0.45 0.55
The x variable, that is being multiplied.