Answer: 0.2743
Step-by-step explanation:
Let X be a random variable that represents the weight of bags of grasecks chocolate candoes.
X that follows normal distribution with, Mean = 4.3 ounces, Standard devaition = 0.05 ounces
The probability that a bag of these chocolate candies weighs less than 4.27 ounces :![P(X](https://tex.z-dn.net/?f=P%28X%3C4.27%29%20%3DP%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%3C%5Cdfrac%7B4.27-4.3%7D%7B0.05%7D%29)
![=P(z](https://tex.z-dn.net/?f=%3DP%28z%3C-0.6%29%5C%20%5C%20%5C%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-P%28Z%3C0.6%29%5C%5C%5C%5C%3D1-0.7257%5C%20%20%5B%5Ctext%7B%20By%20p-value%20table%7D%5D%5C%5C%5C%5C%3D0.2743)
Hence, the required probability = 0.2743
Answer:
The interest is
<h2>$ 769.50</h2>
Step-by-step explanation:
Simple interest is given by
![I = \frac{P \times R \times T}{100}](https://tex.z-dn.net/?f=I%20%3D%20%20%20%5Cfrac%7BP%20%5Ctimes%20%20R%20%20%5Ctimes%20T%7D%7B100%7D%20)
where
P is the principal
R is the rate
T is the time given
From the question
The principal / P = $ 5700
The rate / R = 4.5%
The time given / T = 3 years
So the interest is
![I = \frac{5700 \times 4.5 \times 3}{100}](https://tex.z-dn.net/?f=I%20%20%3D%20%20%5Cfrac%7B5700%20%20%5Ctimes%204.5%20%5Ctimes%203%7D%7B100%7D%20)
![I = \frac{76950}{100}](https://tex.z-dn.net/?f=I%20%3D%20%20%5Cfrac%7B76950%7D%7B100%7D%20)
We have the final answer as
I = $ 769.50
Hope this helps you
Answer:
i don't know
Step-by-step explanation:
Answer:
Dylan delivered 140 parcels on Wednesday.
Step-by-step explanation:
On Wednesday:
On Wednesday, he delivered x parcels.
Thursday:
10% more than Wednesday, so 100 + 10 = 110% of x = 1.1x
Friday:
50% pless than on Thursday, so 100 - 50 = 50% of 1.1x = 0.5*1.1*x.
THis is equals to 77. So
![0.5*1.1x = 77](https://tex.z-dn.net/?f=0.5%2A1.1x%20%3D%2077)
![x = \frac{77}{0.5*1.1}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B77%7D%7B0.5%2A1.1%7D)
![x = 140](https://tex.z-dn.net/?f=x%20%3D%20140)
Dylan delivered 140 parcels on Wednesday.