Math 9

PLEASE HELP ILL MARK BRAINLIEST !!!

Alexus [3.1K]

Answer:

a) the common difference is 20

b) $x_8=115 , x_{12}=195$

c) the common difference is -13

d) $a_{12}=52, a_{15}=13$

Step-by-step explanation:

a) what is the common difference of the sequence xn

Looking at the table, we get x_3=16, x_4=36 and x_5= 56

Deterring the common difference by subtracting x_4 from x_3 we get

36-16 =20

So, the common difference is 20

b) what is x_8? what is x_12

The formula used is: $x_n=x_1+(n-1)d$

We know common difference d= 20, we need to find $x_1$

Using $x_3=16$ we can find $x_1$

$x_n=x_1+(n-1)d\\x_3=x_1+(3-1)d\\15=x_1+2(20)\\15=x_1+40\\x_1=15-40\\x_1=-25$

So, We have $x_1 = -25$

Now finding $x_8$

$x_n=x_1+(n-1)d\\x_8=x_1+(8-1)d\\x_8=-25+7(20)\\x_8=-25+140\\x_8=115$

So, $\mathbf{x_8=115}$

Now finding $x_{12}$

$x_n=x_1+(n-1)d\\x_{12}=x_1+(12-1)d\\x_{12}=-25+11(20)\\x_{12}=-25+220\\x_{12}=195$

So, $\mathbf{x_{12}=195}$

c) what is the common difference of the sequence $a_m$

Looking at the table, we get a_7=104, a_8=91 and a_9= 78

Deterring the common difference by subtracting a_7 from a_8 we get

91-104 =-13

So, the common difference is -13

d) what is a_12? what is a_15?

The formula used is: $a_n=a_1+(n-1)d$

We know common difference d= -13, we need to find $a_1$

Using $a_7=104$ we can find $x_1$

$a_n=a_1+(n-1)d\\a_7=a_1+(7-1)d\\104=a_1+7(-13)\\104=a_1-91\\a_1=104+91\\a_1=195$

So, We have $a_1 = 195$

Now finding $a_{12}$ , put n=12

$a_n=a_1+(n-1)d\\a_{12}=a_1+(12-1)d\\a_{12}=195+11(-13)\\a_{12}=195-143\\a_{12}=52$

So, $\mathbf{a_{12}=52}$

Now finding $a_{15}$ , put n=15

$a_n=a_1+(n-1)d\\a_{15}=a_1+(15-1)d\\a_{15}=195+14(-13)\\a_{15}=195-182\\a_{15}=13$

So, $\mathbf{a_{15}=13}$

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3 years ago

472 students are going on a field trip. 120 students will be returning home and the rest will be spending the night in cabins. a

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Step-by-step explanation:

472-120=352 sleep in a cabin

352÷8=44 cabins

4 0

3 years ago

Read 2 more answers

The area of the sector of a circle with a radius of 8 centimeters is 125.6 square centimeters. The estimated value of pi is 3.14

tangare [24]

$\bf \textit{area of a sector of a circle}\\\\
A=\cfrac{\theta\pi r^2}{360}\qquad 
\begin{cases}
\theta=\textit{angle in degrees}\\
r=radius\\
----------\\
r=8\\
A=125.6
\end{cases}
\\\\\\
360A=\theta\pi r^2\implies \cfrac{360A}{\pi r^2}=\theta\implies \cfrac{360\cdot 125.6}{\pi 8^2}=\theta$

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3 years ago

Three students want to estimate the mean word length of the same book. To do this, each student randomly chose 4 words from the

Firlakuza [10]

The solutions to the questions are given below

a)

sample(n) word length sample mean

1 5,4,4,2 3.75

2 3,2,3,6 3.5

3 5,6,3,3 4.25

b)R =0.75

c)

The mean of the sample means will tend to be a better estimate than a single sample mean.
The closer the range of the sample means is to 0, the more confident they can be in their estimate.

<h3>What is the students are going to use the sample means to estimate the mean word length in the book.?</h3>

The table below shows sample means in the table.

sample(n) word length sample mean

1 5,4,4,2 3.75

2 3,2,3,6 3.5

3 5,6,3,3 4.25

b)

Generally, the equation for is mathematically given as

variation in the sample means

R =maximum -minimum

R=4.25-3.5

R =0.75

c)

In conclusion, In most cases, the mean of many samples will provide a more accurate estimate than the mean of a single sample.

They may have a higher level of confidence in their estimate if the range of the sample means is closer to 0 than it is now.