Answer:
The weight of Paula's dog is 36 pounds.
Step-by-step explanation:
SO we would draw one box for the weight of Carla's dog. Then we would draw 3 boxes for the weight of Paula's dog.
To find how much does Paula's dog weigh we divide the total weight of the dogs by the total number of boxes.
So then you would calculate 48 divided by 4.
The weight of Carla's dog is 12 pounds.
Therefore the weight of Paula's dog is 3x12=3x10+3x2=30+6=36 pounds.
Answer:
Neither of the two advertising options meet the goal.
Step-by-step explanation:
Every $4,000 spent on cable TV will reach 160 fans.
Therefore, the goal of reaching 200000 fans will be fulfilled by
dollars i.e. $5 million which is greater than the budget of $3 million.
Now, every $2,750 spent on radio ads will reach 125 fans.
Therefore, the goal of reaching 200000 fans will be fulfilled by
dollars i.e. $4.5 million which is again greater than the budget of $3 million.
Therefore, neither of the two advertising options meet the goal. (Answer)
Answer:
P-value = 0.0455
Step-by-step explanation:
In this question, we are concerned with calculating the P- value in a test.
Mathematically we know that;
P-value = 2 * P(Z > |z|)
Please check attachment for complete solution and by step explanation
<span>opposite sides and _____ </span>and angles
Answer:
Step-by-step explanation:
a )
sample mean = sum total of given data / no of data
= 415.35 / 20 = 20.76
To calculate the median we arrange the data in ascending order and take the average of 10 th and 11 th term .
= 20.50 + 20.72 / 2
= 20.61
b ) To calculate the 10% trimmed mean , we neglect the largest 10% and smallest 10 % data and then calculate the mean . Here we neglect the first two smallest and last two greatest
(18.92 + 19.25 ..... + 22.43 + 22.85) / 16
= 20.74
c )
We can easily plot the data on number line from 17 to 24
d )
Maximum value of data set = 23.71 and minimum value is 18.04
mean is 20.76 , median is 20.61 and trimmed mean is 20.74
They are between maximum and minimum values of given data . Hence there is no outliers .