Answer:
(x+4)(x-3) or A
Step-by-step explanation:
I'm going to assume that you meant to write x² and not x2
which means we have
x²+x-12
and we need to factor this
to do this we need two numbers that add up to 1 (b) but mulitply to -12 (a*c)
our two numbers are
4, -3
Which means the answer is
(x+4)(x-3)
Answer:
is there a picture of the question
The number of terms in the given arithmetic sequence is n = 10. Using the given first, last term, and the common difference of the arithmetic sequence, the required value is calculated.
<h3>What is the nth term of an arithmetic sequence?</h3>
The general form of the nth term of an arithmetic sequence is
an = a1 + (n - 1)d
Where,
a1 - first term
n - number of terms in the sequence
d - the common difference
<h3>Calculation:</h3>
The given sequence is an arithmetic sequence.
First term a1 =
= 3/2
Last term an =
= 5/2
Common difference d = 1/9
From the general formula,
an = a1 + (n - 1)d
On substituting,
5/2 = 3/2 + (n - 1)1/9
⇒ (n - 1)1/9 = 5/2 - 3/2
⇒ (n - 1)1/9 = 1
⇒ n - 1 = 9
⇒ n = 9 + 1
∴ n = 10
Thus, there are 10 terms in the given arithmetic sequence.
learn more about the arithmetic sequence here:
brainly.com/question/503167
#SPJ1
Disclaimer: The given question in the portal is incorrect. Here is the correct question.
Question: If the first and the last term of an arithmetic progression with a common difference are
,
and 1/9 respectively, how many terms has the sequence?
Answer:
80p + 48q - 8
Step-by-step explanation:
Answer:
Probability that a student who passed the test did not complete the homework = 0.07
Step-by-step explanation:
Given:
Total number of students = 28
Number of students who passed the test = 18
Number of students who completed the assignment = 23
Number of students who passed the test and also completed the assignment = 16
To find: probability that a student who passed the test did not complete the homework
Solution:
Probability refers to chances of occurrence of some event.
Probability = number of favorable outcomes/total number of outcomes
Let A denotes the event that students passed the test and B denotes the event that students completed the assignment
P(A only) = ![P(A)-P(A\cap B)](https://tex.z-dn.net/?f=P%28A%29-P%28A%5Ccap%20B%29)
Here,
![P(A)=\frac{18}{28}\,,\,P(A\cap B)=\frac{16}{28}](https://tex.z-dn.net/?f=P%28A%29%3D%5Cfrac%7B18%7D%7B28%7D%5C%2C%2C%5C%2CP%28A%5Ccap%20B%29%3D%5Cfrac%7B16%7D%7B28%7D)
So,
![P(A\,\,only)=\frac{18}{28}-\frac{16}{28}=\frac{2}{28}=\frac{1}{14}=0.07](https://tex.z-dn.net/?f=P%28A%5C%2C%5C%2Conly%29%3D%5Cfrac%7B18%7D%7B28%7D-%5Cfrac%7B16%7D%7B28%7D%3D%5Cfrac%7B2%7D%7B28%7D%3D%5Cfrac%7B1%7D%7B14%7D%3D0.07)
Therefore,
probability that a student who passed the test did not complete the homework = 0.07