Answer:
tiles
Step-by-step explanation:
there are many ways to solve this soo...
1. convert to centimetres (there are 100 centimetres in 1 metre)
2. the square tiles area = 
2. find the large area (4.8m x 3.8m or 480cm x 380cm)

2. find the two smaller areas (1.2m x 1.8m or 120cm x 180cm)

3. add to find total area

4. find how many tiles are needed
tiles
The probability of getting a piece of chocolate or a lollipop would be

. You can find this by first adding all of the values of the amount of chocolate, lollipops, and licorice. Then, add the value of the amount of chocolate and lollipops.
3(y - 2) + 6(x + 1) - 2 = 0
3y - 6 + 6x + 6 - 2 = 0
3y + 6x - 2 = 0
3y = -6x + 2
y = -2x + 2/3
slope = -2, y intercept = 2/3
keeping in mind that 4 months is not even a year, since there are 12 months in a year, 4 months is then 4/12 years.
let's assume is simple interest.
![\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$34300\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ t=years\to \frac{4}{12}\dotfill &\frac{1}{3} \end{cases} \\\\\\ A=34300\left[ 1+(0.035)\left( \frac{1}{3} \right) \right]\implies A= 34300(1.011\overline{6})\implies A=34700.1\overline{6}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%20%5C%2434300%5C%5C%20r%3Drate%5Cto%203.5%5C%25%5Cto%20%5Cfrac%7B3.5%7D%7B100%7D%5Cdotfill%20%260.035%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B4%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B1%7D%7B3%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D34300%5Cleft%5B%201%2B%280.035%29%5Cleft%28%20%5Cfrac%7B1%7D%7B3%7D%20%5Cright%29%20%5Cright%5D%5Cimplies%20A%3D%2034300%281.011%5Coverline%7B6%7D%29%5Cimplies%20A%3D34700.1%5Coverline%7B6%7D)
Answer:
The value is c = 21.1445.
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.
The weight distribution of parcels sent in a certain manner is normal with mean value 12lb and standard deviation 3.5 lb.
This means that 
What value of c is such that 99% of all parcels are at least 1 lb under the surcharge weight?
This 1 added to the value of X for the 99th percentile, which is X when Z has a p-value of 0.99, so X when Z = 2.327.




1 + 20.1445 = 21.1445
The value is c = 21.1445.