A supermarket shop cans display consists of 5 levels of stacked cans. In the display, each level has 4 more cans than the level above it-the total number of cans in display are (b) 100
Step-by-step explanation:
Given that ,
The display consists of 5 levels of stacked cans- <u>which means that there are 5 levels in which the stacked cans are displayed .</u>
so,the no:of levels are 5
Now given that level has 4 more cans than the level above it.
so. the number of cans in a single level = 5*4=20
Now we know that the total number of level is 5 so we multiply 20 withe the number of levels of can stacked
==>20*5=100
<u>Thus we can say that the total number of cans in display are (b) 100</u>
For this case, the first thing we must do is define variables.
We have then:
t: number of hours
F (t): total charge
We write the function that models the problem:
Where,
b: represents an initial fee.
We must find the value of b.
For this, we use the following data:
Her total fee for a 4-hour job, for instance, is $ 32.
We have then:
From here, we clear the value of b:
Then, the function that models the problem is:
Answer:
the function's formula is:

1) Convert 4% to decimal = .04
2) Multiply the purchase by that decimal
.04(56.70) = $2.27
C
Answer:
Step-by-step explanation:
find attached the solution
Answer:
a.
b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349
Step-by-step explanation:
a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.
b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.
c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842
d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166
e. The z-score related to 6.4 kg is
and the z-score related to 7 kg is
, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194
f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349