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3 years ago

Find the eighth term of the

Mathematics

1 answer:

Pie3 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

The n th term of a geometric sequence is

$a_{n}$ = a₁ $(r)^{n-1}$

where a₁ is the first term and r the common ratio, hence

$a_{8}$ = 6 × $(-1/3)^{7}$ = 6 × - $\frac{1}{3^{7} }$ = 6 × - $\frac{1}{2187}$ = - $\frac{2}{729}$

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Help me plz only number 22

VikaD [51]

Just multiply the fractions by 32

3 0

3 years ago

Point T is the midpoint of JH. The coordinate of T is (0, 5) and the coordinate of J is (0, 2). The coordinate of H is:

aev [14]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&J&(~ 0 &,& 2~) 
% (c,d)
&H&(~ x &,& y~)
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left(\cfrac{x+0}{2}~~,~~\cfrac{y+2}{2} \right)=\stackrel{midpoint~T}{(0~,~5)}\implies 
\begin{cases}
\cfrac{x+0}{2}=0\\\\
\boxed{x=0}\\
------\\
\cfrac{y+2}{2}=5\\\\
y+2=10\\
\boxed{y=8}
\end{cases}$

6 0

3 years ago

Erin had 55 stuffed bears. She took out her 7 favorite and equally divided the rest among her 3 sisters. How many bears did each

m_a_m_a [10]

Answer:

Step-by-step explanation:

55-7= 48

48/3 = 16

5 0

2 years ago

Please answer this correctly

lozanna [386]

Answer:

no it to low

Step-by-step explanation:

the answer was 4998

8 0

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$