Answer:
1 quarter, 1 dime, and 4 nickels
Step-by-step explanation:
First you subtract the 18 cents that were found in the pocket from the total amount of 73 cents to get the total amount of the 6 other coins:
73-18=55.
So the total amount of the 6 coins equals 55 cents.
The quarter = 25 cents
The dime = 10 cents
The 4 nickels = 5 + 5 + 5 + 5 = 20
1 quarter, 1 dime, and 4 nickels add up to the necessary total of 6 coins and when you add up the total cents of the quarter, dime, and nickels, 25 + 10 + 20 it comes to the sum of 55, which is how much the 6 coins are worth in total.
So therefore, the answer is 1 quarter, 1 dime, and 4 nickels.
Answer:
3/5
Step-by-step explanation:
3/5 x 2/5 = 6/25
Answer:
Step A) 250 times 3 months + 500 cause you already have that in your bank account= 1250
Step B) 12+12=24 so 250 times 24=6000+ 500 cause you already have that in your bank account= 6500
Step C) 8,300- 6500= 1800 then, 250 time 5 + 500= 1750, and 250 times 6 = 1500+500=2000 so you can go with whatever.
Step D) 3 years= 36 months so 250 times 36 months= 9000 so in three yrs you should have about 9500 bc you need to add the 500 on again. sry I could not do the graph but I hoped I helped. :)
Step-by-step explanation:
Answer:
There is a 0.82% probability that a line width is greater than 0.62 micrometer.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by
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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.
In this problem
The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so 
.
What is the probability that a line width is greater than 0.62 micrometer?
That is 
So
Z = 2.4 has a pvalue of 0.99180.
This means that P(X \leq 0.62) = 0.99180.
We also have that
There is a 0.82% probability that a line width is greater than 0.62 micrometer.
