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Answer:
There are 20 nickels and 10 dimes
The friend is correct, there are twice as many nickels as there are dimes
Step-by-step explanation:
Let
x -----> the number of nickels
y -----> the number of dimes
z ----> the number of quarters
Remember that
1 nickel=$0.05
1 dime=$0.10
1 quarter=$0.25
so
x+y+z=30 -----> equation A
0.05x+0.10y+0.25z=2
Multiply by 100 both sides
5x+10y+25z=200 -----> equation B
x=2y ----> equation C
substitute equation C in equation A and equation B
2y+y+z=30 -----> 3y+z=30 -----> equation D
5(2y)+10y+25z=200 ----> 20y+25z=200 ----> equation E
Solve the system of equations D and E by graphing
The solution is the point (0,10)
z=0, y=10
see the attached figure
Find the value of x
x=2y -----> x=2(10)=20
therefore
There are 20 nickels and 10 dimes
The friend is correct, there are twice as many nickels as there are dimes
1. The first step here is to arrange the data set's form least to greatest,
Sherelle: 26, 39, 56, 58, 60, 62, 65, 66, 66, 68, 71, 72, 72, 73, 74, 75, 81, 83, 84, 85
Venita: 44, 45, 51, 51, 53, 53, 55, 57, 58, 62, 65, 66, 69, 69, 70, 73, 75, 77, 78, 79
Now we can determine our 5 - number summary based on the numbers respective positions.
First Data Set,
<u>(Five - Number Summary) - Minimum : 26, Quartile 1 : 60, Median : 69.5, Quartile 3 : 75, Maximum : 85</u>
Second Data Set,
<u>(Five - Number Summary) - Minimum : 44, Quartile 1 : 53, Median : 63.5, Quartile 3 : 73, Maximum : 79</u>
2. This part is based on your drawings of the box and whisker plots, so you would have to figure that part out by yourself.
3. First off we know that our data set is composed of the years from 1900, so let's rewrite the set based off of the actual year -
Sherelle: 1926, 1939, 1956, 1958, 1960, 1962, 1965, 1966, 1966, 1968, 1971, 1972, 1972, 1973, 1974, 1975, 1981, 1983, 1984, 1985
Venita: 1944, 1945, 1951, 1951, 1953, 1953, 1955, 1957, 1958, 1962, 1965, 1966, 1969, 1969, 1970, 1973, 1975, 1977, 1978, 1979
( a ) Now in Sherelle's defence, she can say that the lowest coin date in her group is 1926, comparative to Venita's group - the lowest coin date in hers being 1944. Therefore, she is more likely to have the 1916 coin, after all that date is the lowest overall in both their data set.
( b ) In Venita defence, she can say that the mean of her data set is lower than the mean of Sherelle's data set. Take a look at the calculations below,
Sherella's Mean : =
= 1966.8,
Venita's Mean : =
= 1962.5
( c ) I would say Sherella's bag would most likely contain the 1916 coin. The mean is a prominent factor, but their mean(s) only differ by a very small quantity. That too, Sherella's bag contains the lowest coin in both their groups, and though that is not a prominent factor, it could be that she does have the 1916 coin.
