Answer:
Ten Peaches and five bananas.
Step-by-step explanation
To solve the problem increase the cost by adding more peaches and less by adding more bananas
7 peaches = 3.50
8 bananas = 3.20 + 3.50 = 6.70
We need 30 more cents so add three more peaches
10 peaches = 5
5 x 0.4 = 2
2 + 5 = 7
18 kg of 15% copper and 72 kg of 60% copper should be combined by the metalworker to create 90 kg of 51% copper alloy.
<u>Step-by-step explanation:</u>
Let x = kg of 15% copper alloy
Let y = kg of 60% copper alloy
Since we need to create 90 kg of alloy we know:
x + y = 90
51% of 90 kg = 45.9 kg of copper
So we're interested in creating 45.9 kg of copper
We need some amount of 15% copper and some amount of 60% copper to create 45.9 kg of copper:
0.15x + 0.60y = 45.9
but
x + y = 90
x= 90 - y
substituting that value in for x
0.15(90 - y) + 0.60y = 45.9
13.5 - 0.15y + 0.60y = 45.9
0.45y = 32.4
y = 72
Substituting this y value to solve for x gives:
x + y = 90
x= 90-72
x=18
Therefore, in order to create 90kg of 51% alloy, we'd need 18 kg of 15% copper and 72 kg of 60% copper.
Answer:
(-3, 5), (1, 4), (-2, 3), and (0, 2).
Step-by-step explanation:
When a function is inverted, the x-values become y-values, and the y-values become x-values.
The ordered pairs for the function are (5, -3), (4, 1), (3, -2), and (2, 0).
After inversion, the ordered pairs will be (-3, 5), (1, 4), (-2, 3), and (0, 2).
Hope this helps!
Yes... a half is equal to 10/20 and 11/20 is greater than 10/20 so 11/20 is greater than a half
The answer is 5.
Step-by-step explanation:
To find the interquartile range you must look at the median and the upper and lower halves of the data. The median (12) to the upper half of the data (15) is 3. Next, look at the median (12) to the lower half of the data (10) it is 2. We then add 2 and 3 to get the interquartile range of 5.
