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xenn [34]
3 years ago
9

Question 18(Multiple Choice Worth 1 points) (01.01) Period 1 has 15 students in it and a test average of 86%. Period 2 has 21 st

udents in it and an average of 88%. Period 3 has 12 students in it and an average of 95%. Use weighted means to find the overall average of the classes.
Mathematics
1 answer:
andrew11 [14]3 years ago
8 0

Answer:

89.125%

Step-by-step explanation:

The weighted mean is simply the multiplication of the number of student in each period times the respective average of said period, divided by the total number of students:

Average = \frac{15* 86 + 21*88 + 12*95}{15 + 21 + 12} = 89.125%

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What is 7 - (-4) ? I got -24
koban [17]

Answer:

11

Step-by-step explanation:

7-(-4) is 11. You must make the subtraction symbol in the middle to addition, since you have 2 negatives. Two negatives = postive. So after you do that it would be 7+4=11

Hope this helped!

6 0
3 years ago
Find d for the arithmetic series with S17=-170 and a1=2
Irina18 [472]
So, we know the sum of the first 17 terms is -170, thus S₁₇ = -170, and we also know the first term is 2, well

\bf \textit{ sum of a finite arithmetic sequence}\\\\
S_n=\cfrac{n(a_1+a_n)}{2}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
----------\\
n=17\\
S_{17}=-170\\
a_1=2
\end{cases}
\\\\\\
-170=\cfrac{17(2+a_{17})}{2}\implies \cfrac{-170}{17}=\cfrac{(2+a_{17})}{2}
\\\\\\
-10=\cfrac{(2+a_{17})}{2}\implies -20=2+a_{17}\implies -22=a_{17}

well, since the 17th term is that much, let's check what "d" is then anyway,

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=17\\
a_{17}=-22\\
a_1=2
\end{cases}
\\\\\\
-22=2+(17-1)d\implies -22=2+16d\implies -24=16d
\\\\\\
\cfrac{-24}{16}=d\implies -\cfrac{3}{2}=d
6 0
3 years ago
15 pts + brainliest to right/best answer
asambeis [7]

Split the second term 7x^2 - 8x - 12 into two terms

7x^2 + 6x - 14x - 12

Factor out common terms in the first two terms, then in the last two terms

x(7x + 6) - 2(7x + 6)

Factor out the common term; 7x + 6

<u>(7x + 6)(x - 2) </u>

4 0
3 years ago
Read 2 more answers
PLease help
Kipish [7]
It is longing because it is the definition of yearning
7 0
2 years ago
Read 2 more answers
All of the functions shown below are either exponential growth or decay functions.
AlekseyPX

Answer:

Step-by-step explanation:

If an exponential function is in the form of y = a(b)ˣ,

a = Initial quantity

b = Growth factor

x = Duration

Condition for exponential growth → b > 1

Condition for exponential decay → 0 < b < 1

Now we ca apply this condition in the given functions,

1). y=3.2(1+0.45)^{2x}

   Here, (1 + 0.45) = 1.45 > 1

   Therefore, It's an exponential growth.

2). y=(0.85)^{3x}

    Here, (0.85) is between 0 and 1,

    Therefore, it's an exponential decay.

3). y = (1 - 0.03)ˣ + 4

    Here, (1 - 0.03) = 0.97

    And 0 < 0.97 < 1

    Therefore, It's an exponential decay.

4). y = 0.5(1.2)ˣ + 2

    Here, 1.2 > 1

    Therefore, it's an exponential growth.

5 0
2 years ago
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