Answer:
(a) It costs $55 at Happy Fruiterer
(b) It will cost $49.64
(c) It is cheaper at Happy Fruiterer the second week
Step-by-step explanation:
Since Fruitz shop sells as 10% less than Happy Fruiterer at $50
10% more of $50 = 50 × 10/100 + 50
= $5 + $50 = $55
(a) n Kg of grapes will cost $55 at Happy Fruiterer
(b) if Happy Fruits discounts $55 by 5%, we will have
55 × 5/100 = 2.75
that is $55 - $2.75 = $52.25
If it is discounted further at 5% more the following week, we will have
$52.25 × 5/100 = $2.6125
≈ $2.61
Then the new price for the 2nd week is $52.25 - $2.61
= $49.64
(c) at the second wee of discounting, Fruitz shop still sells at $50 will Happy Fruiterer now see at $49.64. Therefore, it is cheaper at Happy Fruiterer
Answer:
KArl should use 3/16 cups of chili powder
Step-by-step explanation:
Given that:
Recipe's requirement : 3/8 cups
Karl wants to use half : 1/2
The given question involves fractions. When the number of cups has to be divided into half, it will be multiplied with 1/2.
So,
Multiplying the chili powder requirement of the recipe to 1/2
Hence,
KArl should use 3/16 cups of chili powder
I believe that the last one is true
because the linear equations are
y=ax+b
which means that the point of y-axis is on one side and the slope, point on x-axis and y-intercept are on the other side
good luck
Answer:
The distance between both points would be of 17 units
Step-by-step explanation:
they have the same value through the y-axis. However if you count the spots from 3 to -14 their x values are 17 units apart.
Answer:
Step-by-step explanation:
The 90th percentile of a normally distributed curve occurs at 1.282 standard deviations. Similarly, the 10th percentile of a normally distributed curve occurs at -1.282 standard deviations.
To find the 
percentile for the television weights, use the formula:
, where 
is the average of the set, 
is some constant relevant to the percentile you're finding, and 
is one standard deviation.
As I mentioned previously, 90th percentile occurs at 1.282 standard deviations. The average of the set and one standard deviation is already given. Substitute 
, 
, and 
:
Therefore, the 90th percentile weight is 5.1282 pounds.
Repeat the process for calculating the 10th percentile weight:
The difference between these two weights is 
.
