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GuDViN [60]
4 years ago
6

In a sample of 200 books, 26 books contained at least one punctuation error. The sampling method had a margin of error of 0.03.

What is the interval estimate for the proportion that would have at least one punctuation error in the form (lower limit, upper limit) ?
Mathematics
1 answer:
exis [7]4 years ago
3 0

Answer:

confidence interval =(0.10,0.16)

Step-by-step explanation:

Given that sample proportion was

26/200 =0.13 for a sample size of 200

Margin of error is given as 0.03

We know that confidence interval is one which is calculated as sample proportion ±Margin of error

Here sample proportion=0.13

Margin of error = 0.03

Hence

confidence interval lower limit = 0.13-0.03 = 0.10

Confidence interval upper limit = 0.13+0.03 = 0.16

Hence confidence interval =(0.10,0.16)

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How many 12/4 are in 3
Anit [1.1K]

Answer:

There is one 12/4 in 3.

Step-by-step explanation:

12/4 = 3

4/4 + 4/4 + 4/4 = 12/4 = 3

8 0
3 years ago
Hey can you please help me posted picture of question :)
notka56 [123]
The given trinomial can be factored using factorization as

x²-x-12
=x²-4x+3x-12
=x(x-4)+3(x-4)
=(x-4)(x+3)

Thus x-4 and x+3 are the factors of the given trinomial. From these we can see x+3 is listed in option A

So the answer to this question is Option A
6 0
3 years ago
Marie and Whitney order eggs for $4.35, pancakes for $3.10, and 2 mugs of cocoa for $1.45 each. The tax is $0.85. How much chang
svetlana [45]

Answer:

$3.80

Step-by-step explanation:

$4.35+$3.10+$1.45+$1.45+$0.85=$11.20

$15.00-$11.20=$3.80

6 0
3 years ago
A bookkeeper bought 2,000 exercise book distributed 200 books to the student from poor economic background as donation .he sold
Andrews [41]

Books left=2000-200=1800

ATQ

\\ \sf\longmapsto 1800x+10=0.08x

\\ \sf\longmapsto 1800x-0.08x=-10

\\ \sf\longmapsto x(1800-0.08)=-10

\\ \sf\longmapsto x(1799.92)=-10

\\ \sf\longmapsto x=|\dfrac{10}{1799.92}|

\\ \sf\longmapsto x=0.005

Now

\\ \sf\longmapsto CP=0.005(2000)=10

6 0
3 years ago
Please help! I attached the question below.
kompoz [17]

Answer:

\frac{2(c+2)}{c(c-2)}

Step-by-step explanation:

\frac{c^{2}-4 }{6c^{4}+15c^{3}}=\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2}) }

Identity used:

a^{2}-b^{2}=(a-b)(a+b)

\frac{c^{2}-4c+4}{12c^{3}+30c^{2}}=\frac{(c-2)^{2}}{2(6c^{3}+15c^{2}) }

Now let us divide the modified expressions:

\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2})} ÷ \frac{(c-2)^2}{2(6c^{3}+15c^{2}) }

we get:

\frac{2(c+2)}{c(c-2)}

5 0
4 years ago
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