Answer:
T = 5c +5d, where unit of the equation is $.
Step-by-step explanation:
Cost of 1 large cheese pizza = $5
No. of large cheese pizza = c
therefore
cost of c large cheese pizza = c*Cost of 1 large cheese pizza = $5c
Cost of 1 large one topping pizza = $6
No. of large one topping pizza = d
therefore
cost of d large one topping pizza = d*Cost of 1 large one topping pizza = cost of d large one topping pizza = $6d
Total cost of c large cheese pizza and cost of d large one topping pizza
= $5c +$6d
total cost is represented by T
thus, T = $5c +$5d Answer
Answer:
Step-by-step explanation:
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Answer:
x = -7/2
Step-by-step explanation:
( x + 9 ) - ( x -2 )=
2x +7 =
2x = -7
x = -7/2
Answer:
IQ scores of at least 130.81 are identified with the upper 2%.
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.
Mean of 100 and a standard deviation of 15.
This means that 
What IQ score is identified with the upper 2%?
IQ scores of at least the 100 - 2 = 98th percentile, which is X when Z has a p-value of 0.98, so X when Z = 2.054.




IQ scores of at least 130.81 are identified with the upper 2%.
Answer: 5.37
Step-by-step explanation:
Let x = ACT scores.
Given: ACT scores have a mean of 20.8 and 9 percent of the scores are above 28. The scores have a distribution that is approximately normal.
i.e. P(X>28)=0.09 (i)
Now,
(ii)
One -tailed z value for p-value of 0.09 =1.3408 [By z-table]
From (i) and (ii)

Hence, the standard deviation = 5.37