God loves all of you, remember that oK <3 32>2h+6h is
Answer:
Option E.
Step-by-step explanation:
Given information:
Air conditioning = 88
Automatic transmission = 96
power steering = 76
All three = 4
None of these extras = 19
Only air conditioning = 23
Only automatic transmissions = 61
Only power steering = 28
Both automatic transmission and power steering = 11
Place these values and draw a venn diagram as shown below.
Calculation for missing values:
From the given diagram it is clear that the number of cars that had air conditioning and automatic transmission but not power steering is 24.
Therefore, the correct option is E.
Answer: 36 hotdogs
Step-by-step explanation: You need to find the least common multiple of 9 and 12.
You can check by multiplying 9 and 12 by consecutive whole numbers until you find a number that they match.
Example:
9*1= 9 , 9*2=18, 9*3= 27, 9*4=36, 9*5=45
12*1=12, 12*2=24, 12*3 =36
They both share 36.
Answer:
infinite solutions
Step-by-step explanation:
replace x with 3y + 9 in the second equation
9 - (3y + 9) = -3y ➡ 9 -3y -9 = -3y ➡9-9 = 3y-3y ➡ 0 = 0
this means whatever numver you write instead of y there will be a solution.
Answer:
75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5 pounds.
Step-by-step explanation:
The question is missing. It is as follows:
Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains had the following weights (pounds): 69 104 125 129 60 64
Assume that the population of x values has an approximately normal distribution.
Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)
75% Confidence Interval can be calculated using M±ME where
- M is the sample mean weight of the wild mountain lions ()
- ME is the margin of error of the mean
And margin of error (ME) of the mean can be calculated using the formula
ME= where
- t is the corresponding statistic in the 75% confidence level and 5 degrees of freedom (1.30)
- s is the standard deviation of the sample(31.4)
Thus, ME= ≈16.66
Then 75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5