Answer:
14 pages per day
Step-by-step explanation:
To do this, you must divide:
43 (pages)/3(days)= 14 (pages per day)
Hope this helps! :)
I got D.
There's a few ways to solve it; I prefer using tables, but there are functions on a TI-84 that'll do it for you too. The logic here is, you have a standard normal distribution which means right away, the mean is 0 and the standard deviation is 1. This means you can use a Z table that helps you calculate the area beneath a normal curve for a range of values. Here, your two Z scores are -1.21 and .84. You might notice that this table doesn't account for negative values, but the cool thing about a normal distribution is that we can assume symmetry, so you can just look for 1.21 and call it good. The actual calculation here is:
1 - Z-score of 1.21 - Z-score .84 ... use the table or calculator
1 - .1131 - .2005 = .6864
Because this table calculates areas to the RIGHT of the mean, you have to play around with it a little to get the bit in the middle that your graph asks for. You subtract from 1 to make sure you're getting the area in the middle and not the area of the tails in this problem.
The chef can make 42 casseroles no remainder
She can buy 3 tubes be cause 8.50 time 2 is less then 20$ but 8.50 times three is more than 20$ its 25.5
Answer: 0.9649
Step-by-step explanation:
Let A denote the event that the days are cloudy and B denotes the event that the days are rainy.
Given : For the month of March in a certain city, the probability that days are cloudy :
Also in the month of March in the same city,, the probability that the days are cloudy and rainy :
Now by using the conditional probability, the probability that a randomly selected day in March will be rainy if it is cloudy will be :-

![\Rightarrow\ P(B|A)=\dfrac{0.55}{0.57}\\\\=0.964912280702\approx0.9649\ \ \text{[Rounded to four decimal places.]}](https://tex.z-dn.net/?f=%5CRightarrow%5C%20P%28B%7CA%29%3D%5Cdfrac%7B0.55%7D%7B0.57%7D%5C%5C%5C%5C%3D0.964912280702%5Capprox0.9649%5C%20%5C%20%5Ctext%7B%5BRounded%20to%20four%20decimal%20places.%5D%7D)
Hence, the probability that a randomly selected day in March will be rainy if it is cloudy = 0.9649