The pounds of alloy that contains 26% copper that would be used is 23.42 pounds.
The pounds of alloy that contains 69% copper that would be used is 29.58 pounds.
<h3>What are the linear equations that represent the question</h3>
0.26a + 0.69b = (53 x 0.5)
0.26a + 0.69b = 26.50 equation 1
a + b = 53 equation 2
Where:
- a =pounds of alloy that contains 26% copper
- b = pounds of alloy that contains 69% copper
<h3>How many pounds of each alloy should be in the third alloy?</h3>
Multiply equation 2 by 0.26
0.26a + 0.26b = 13.78 equation 3
Subtract equation 3 from equation 2
12.72 = 0.43b
b = 12.72 / 0.43
b = 29.58 pounds
a = 53 - 29.58 = 23.42 pounds
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Answer: X=216
so, you have 216 blueberries
Answer:
47.4 ;
50
Step-by-step explanation:
Given the data :
X ($) : 85 139 161 175 85 133 149 145 136 131 290 235 132 149 322 214 105 90 162 229 121 113 149 126139 118 156 214 172 87 172 230 195 126 128 142 118 139
The smallest class interval :
Range / number of classes
Number of classes to use = 5
Range = Maximum - Minimum = (322 - 85) =237
Hence, smallest class interval :
237 / 5 = 47.4
A better class interval would be, one without decimal, rounded to the nearest 10; this will be easier and make more statistical sense
Hence, smallest class interval rounded to the nearest 10 :
47.4 = 50 (nearest 10)
