Divide the limit by the weight of one book to find the number of books allowed.
70/12 = 5.833 books
You cannot send a part of a book, so round down. The answer is 5 books.
Answer:
1/6
1/12
1/2
Step-by-step explanation:
There are three possible outcomes in the left spinner, and four possible outcomes in the right spinner. So there are a total of 3×4=12 possible combinations. We can show that by making a grid:
![\left[\begin{array}{cccc}&R&B&G\\R&RR&BR&GR\\B&BR&BB&BG\\P&PR&BP&GP\\Y&RY&BY&GY\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26R%26B%26G%5C%5CR%26RR%26BR%26GR%5C%5CB%26BR%26BB%26BG%5C%5CP%26PR%26BP%26GP%5C%5CY%26RY%26BY%26GY%5Cend%7Barray%7D%5Cright%5D)
Of these 12 combinations, 2 show both spinners landing on the same color (RR and BB). So the probability is 2/12 = 1/6.
There is only 1 outcome in which the first spinner lands on R <em>and</em> the second spinner lands on P (PR), so the probability is 1/12.
There are 6 outcomes in which the first spinner lands on R <em>or</em> the second spinner lands on P (RR, BR, PR, RY, BP, GP). So the probability is 6/12 = 1/2.
Answer:
6
Step-by-step explanation:
Answer:
Step-by-step explanation:
The quadrilateral has 4 sides and only two of them are equal.
A) to find PR, we will consider the triangle, PRQ.
Using cosine rule
a^2 = b^2 + c^2 - 2abcos A
We are looking for PR
PR^2 = 8^2 + 7^2 - 2 ×8 × 7Cos70
PR^2 = 64 + 49 - 112 × 0.3420
PR^2 = 113 - 38.304 = 74.696
PR = √74.696 = 8.64
B) to find the perimeter of PQRS, we will consider the triangle, RSP. It is an isosceles triangle. Therefore, two sides and two base angles are equal. To determine the length of SP,
We will use the sine rule because only one side,PR is known
For sine rule,
a/sinA = b/sinB
SP/ sin 35 = 8.64/sin110
Cross multiplying
SPsin110 = 8.64sin35
SP = 8.64sin35/sin110
SP = (8.64 × 0.5736)/0.9397
SP = 5.27
SR = SP = 5.27
The perimeter of the quadrilateral PQRS is the sum of the sides. The perimeter = 8 + 7 + 5.27 + 5.27 = 25.54 cm
