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tester [92]
3 years ago
13

Find the sum to n terms of the series 0.6+0.66+0.666+.....................

Mathematics
1 answer:
RSB [31]3 years ago
8 0
<span>We have
</span>0.6+0.66+0.666+0.6666

Let's take 6 out from the series
6(0.1+0.11+0.111+0.1111+....

<span>Now multiply and divide by 9
</span>\frac{6}{9}(0.9+0.99+0.999+0.9999+....

\frac{6}{9}(1-0.1+1-0.01+1-0.001+1-0.0001+....&#10;\\&#10;\\\frac{6}{9}(1-10^{-1}+1-10^{-2}+1-10^{-3}+...

<span>all 1s add up to n
</span>\frac{6}{9}(n-\frac{1}{10}\times \frac{1-(\frac{1}{10})^{n+1}}{1-\frac{1}{10}})

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Point E is the midpoint of AB and point F is the midpoint of CD .
Gwar [14]

Answer:

# AB is bisected by CD

# AE = 1/2 AB

# CE + EF = FD

Step-by-step explanation:

* Lets talk about the mid point

- The mid-point of a segment is divided the segment into two

  equal parts

- The figure has line segment AB

- E is the mid-point of AB

∴ E divides the line segment AB into two equal parts

∴ AE = EB

∴ AE = 1/2 AB ⇒ (1)

- Any line passes through the point E will bisects the line segment AB

∴ AB is bisected by CD ⇒ (2)

∵ F is the mid-point of CD

∴ F divides the line segment CD into two equal parts

∴ CF = FD

∵ Point E lies on CF

∴ CE + EF = CF

∵ CF = FD

∴ CE + EF = FD ⇒ (3)

* There are three statements must be true (1) , (2) , (3)

# AB is bisected by CD

# AE = 1/2 AB

# CE + EF = FD

6 0
3 years ago
Read 2 more answers
There are 8 swimmers in the swimming club. The club bought c swimming caps to split among its members. Write an expression that
love history [14]

Answer:

c/8

Step-by-step explanation:

This is division. This can be figured out by reading "split". If something is to be split, it is the dividend.

5 0
2 years ago
Read 2 more answers
What is the conjugate and modulus of 2 – 6i?
Serjik [45]

Answer:

2 + 6i

Step-by-step explanation:

It’s the same numbers with an opposite sign

5 0
2 years ago
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How many edges does a rectanglar prism have?
natta225 [31]

It has 12 edges in total

3 0
3 years ago
Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two​ bakeries' specialty pi
DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>

Step-by-step explanation:

The given data are;

Bakery A

\overline x_1<em> </em>= 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

\overline x_2<em> </em>= 1,711 cal

s₂ = 192 cal

n₂ = 10

\left (\bar{x}_1-\bar{x}_{2}  \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2}  \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, t_c = 1.706

\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}

\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469

\hat \sigma ≈ 178

Therefore, we get;

\left (1,880-1,711  \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711  \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}

Which gives;

169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

4 0
3 years ago
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