Answer:
1.) He still owes $100 of debt
2.) 1 rose would be .75 cents and 1 carnation would be .90 cents
3.) C 8 and 3/8
Step by Step explanation
(/ means divided by in the first 2 equations but / in the last question is for the fractions.)
1.) _(1,500 / 3 =500. 1,500 - 500 =1,000 / 5 =200 x 3 =600. 700 - 600 =$100)
2.)_(1.50/2 =.75 cents. 2.70/3 =.90 cents.)
3.) i first added the fractions ( i equalized them) then i added the whole numbers (the u in the equation was to separate the whole num. and fraction )
Before i equalized it: (6u3/4 + 1u1/8 = 7u1/8 + 1/8 = 8u3/8)
After i equalized it: (6u6/8 + 1u1/8 = 7u1/8 + 1/8 = 8u3/8)
Answer:
1.93433
Step-by-step explanation:
Since all triangles have an interior angle sum of 180, the second angle is 45. this also means that x and y are equal. Using the pythagorean theorem, we can produce the equation
. dividing both sides by two we get
. when we take the square root of root 14 we get 1.934335
Answer:
Give 3 what?
Step-by-step explanation:
I would like to help but I need more information
Answer:
A customer who sends 78 messages per day would be at 99.38th percentile.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Average of 48 texts per day with a standard deviation of 12.
This means that 
a. A customer who sends 78 messages per day would correspond to what percentile?
The percentile is the p-value of Z when X = 78. So



has a p-value of 0.9938.
0.9938*100% = 99.38%.
A customer who sends 78 messages per day would be at 99.38th percentile.
Show me the choices...................