1.6*100/0.95=168% 1.6 to 0.95
0.95*100/1.6=59% 0.95 to 1.6
Complete question :
Cheddar Cheese
$3/lb
Swiss Cheese
$5/lb
Keisha is catering a luncheon. She has $30 to spend on a mixture of Cheddar cheese and Swiss cheese. How many pounds of cheese can Keisha get if she buys only Cheddar cheese? Only Swiss cheese? A mixture of both cheeses?
What linear equation in standard form can she use to model the situation?
Answer:
10 lbs of cheddar cheese
6 lbs of Swiss cheese
$3a + $5b = $30
Step-by-step explanation:
Given that :
Cheddar cheese = $3/lb
Swiss cheese = $5/lb
Total amount budgeted for cheese = $30
How many pounds of cheese can Keisha get if she buys only Cheddar cheese?
Pounds of cheedar cheese obtainable with $30
Total budget / cost per pound of cheddar cheese
$30 / 3 = 10 pounds of cheedar cheese
Only Swiss cheese?
Pounds of cheedar cheese obtainable with $30
Total budget / cost per pound of Swiss cheese
$30 / 5 = 6 pounds of Swiss cheese
A mixture of both cheeses?
What linear equation in standard form can she use to model the situation?
Let amount of cheddar cheese she can get = a
Let amount of Swiss cheese she can get = b
Hence,
(Cost per pound of cheddar cheese * number of pounds of cheddar) + (Cost per pound of Swiss cheese * number of pounds of Swiss cheese) = total budgeted amount
(3 * a) + (5 * b) = $30
$3a + $5b = $30
150750 × 100100 = 20100 = 20%
If you are using a calculator, simply enter 150÷750×100 which will give you 20 as the answer.
The data-set that places 22.6 as an outlier is given as follows:
2.4, 5.3, 3.5, 22.6, 1.8, 2.1, 4.6, 1.9
<h3>When a measure is considered an outlier in a data-set?</h3>
A measure is considered an outlier in a data-set if it is very far from other measures, especially in these two cases:
- If the measure is considerably less than the second smallest value.
- If the measure is considerably more than the second highest value.
In this problem, he data-set that places 22.6 as an outlier is given as follows:
2.4, 5.3, 3.5, 22.6, 1.8, 2.1, 4.6, 1.9.
The second highest value is 5.3, which is considerably less than 22.6, hence 22.6 is an outlier in the data-set.
More can be learned about statistical outliers at brainly.com/question/9264641
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