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
9514 1404 393
Answer:
x = 5°
Step-by-step explanation:
The marked angles are "alternate exterior angles," hence congruent.
2x +90° = x +95°
x = 5° . . . . . . . . . . . . subtract (x+90°) from both sides
Hi Malik.
2x+4>14 or -(2x+4)>14
solve each inequality individually
you get
x>5 or x<-9
Answer:
0.7823 = 78.23% probability that the response time is between 3 and 9 minutes.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 7.2 minutes and a standard deviation of 2.1 minutes.
This means that 
For a randomly received emergency call, find the probability that the response time is between 3 and 9 minutes.
This is the pvalue of Z when X = 9 subtracted by the pvalue of Z when X = 3.
X = 9
has a pvalue of 0.8051
X = 3
has a pvalue of 0.0228
0.8051 - 0.0228 = 0.7823
0.7823 = 78.23% probability that the response time is between 3 and 9 minutes.
