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2 years ago

Please answer the questions(in the pictures) as fast as possible.

Mathematics

1 answer:

andrew-mc [135]2 years ago

5 0

it is 2/3

0--4/6 =

4/6

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For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[

Ivenika [448]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

$1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

Solve point 1 that is $\sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\$ :

when,

$k= 1 \to s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\$

$= \frac{1}{2} - \frac{1}{3}\\\\$

$k= 2 \to s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\$

$= \frac{1}{3} - \frac{1}{4}\\\\$

$k= 3 \to s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\$

$= \frac{1}{4} - \frac{1}{5}\\\\$

$k= n^ \to s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\$

Calculate the sum $(S=s_1+s_2+s_3+......+s_n)$

$S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\$

When $s_n \ \ dt_{n \to 0}$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\$

$\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}$

In point 2: $\sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

when,

$k= 1 \to s_1 = \frac{1}{(1+6)(1+7)}\\\\$

$= \frac{1}{7 \times 8}\\\\= \frac{1}{56}$

$k= 2 \to s_1 = \frac{1}{(2+6)(2+7)}\\\\$

$= \frac{1}{8 \times 9}\\\\= \frac{1}{72}$

$k= 3 \to s_1 = \frac{1}{(3+6)(3+7)}\\\\$

$= \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\$

$k= n^ \to s_n = \frac{1}{(n+6)(n+7)}\\\\$

calculate the sum: $S= s_1+s_2+s_3+s_n\\$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\$

when $s_n \ \ dt_{n \to 0}$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066$

$\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}$

8 0

2 years ago

Fill in the blank 3*one sixth = 3/ __ or ___{fraction}

docker41 [41]

0.5 as a decimal
hope this partially helped:)
sorry couldn't answer completely
let me know if it helped a bit:)

7 0

3 years ago

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 419 gram setting. It is

Dahasolnce [82]

Answer:

The calculated t- value = 1.09 > 3.19 at 0.025 level of significance

Null hypothesis is rejected

There is sufficient evidence to support the claim that the bags are underfilled

Step-by-step explanation:

Step(i):-

Given the mean of the Population(μ) = 419 grams

The sample size 'n' =19

Given mean of the sample x⁻ = 412

Given variance of the sample S² = 784

S = √784 = 28

Step(ii):-

Null hypothesis: H₀: There is no sufficient evidence to support the claim that the bags are underfilled.

Alternative hypothesis: H₁:

There is sufficient evidence to support the claim that the bags are underfilled.

Test statistic

$t = \frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }$

$t = \frac{412 -419}{\frac{28}{\sqrt{19} } }$

t = -1.09

|t| = |-1.09|

Degrees of freedom ν = n-1 = 19-1 =18

t₀.₀₂₅ = 3.19

Final answer:-

The calculated t- value = 1.09 > 3.19 at 0.025 level of significance

Null hypothesis is rejected

There is sufficient evidence to support the claim that the bags are underfilled

8 0

3 years ago

What is the slope of a line that goes through (0,6) and (6,0)

vazorg [7]

(x1,y1)=(0,6)
(x2,y2)=(6,0)
slope=(0-6)/(6-0)=0

3 0

3 years ago

Use a table of values to estimate the value of the limit. Confirm your result graphically by graphing the function with a graphi

il63 [147K]

Hello,

$\lim_{x \to 0}\dfrac{ \sqrt{x+25}-5}{x}\\\\$

$=\lim_{x \to 0}\dfrac{(\sqrt{x+25}-5)*(\sqrt{x+25}+5)}{x*(\sqrt{x+25}+5)}\\\\$

$=\lim_{x \to 0}\dfrac{x+25-25)}{x*(\sqrt{x+25}+5)}\\\\$

$=\dfrac{1}{10}$

4 0

2 years ago