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OverLord2011 [107]
3 years ago
10

The length of an English essay is 180 lines long with an average of 15 words per line If it is written with an average of 10 wor

ds per line how many lines will be needed to complete the essay?
Urgent
please answer with step by step
Mathematics
1 answer:
Ivenika [448]3 years ago
6 0

Answer:

18 lines would be needed to complete the essay.

Step-by-step explanation:

To solve this you need to divide the amount of words per line by the number of lines.

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What is 134+2355 idk the answer
Goryan [66]

Answer:

easy u can also use a calculator look up calculator online or download it... so the answer is 2489

7 0
3 years ago
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A college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of
Alla [95]

Answer:

80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

Step-by-step explanation:

We are given that a college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of hours they watch per week. The results are provided below;

<u>Hours of TV per week (X)</u>: 6, 14, 13, 6, 16, 10, 19, 4, 5, 5, 18, 8, 7, 14, 8, 8, 9, 12, 6, 5.

Firstly, the Pivotal quantity for 80% confidence interval for the true average is given by;

                                P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean number of hours of TV watched per week = \frac{\sum X}{n} = 9.65

            s = sample standard deviation = \sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }  = 4.61

            n = sample of people = 20

           \mu = true average number of hours of TV watched per week

<em>Here for constructing 80% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>

<u>So, 80% confidence interval for the true average, </u>\mu<u> is ;</u>

P(-1.33 < t_1_9 < 1.33) = 0.80  {As the critical value of t at 19 degrees of

                                               freedom are -1.33 & 1.33 with P = 10%}  

P(-1.33 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.33) = 0.80

P( -1.33 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.33 \times {\frac{s}{\sqrt{n} } } ) = 0.80

P( \bar X-1.33 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.33 \times {\frac{s}{\sqrt{n} } } ) = 0.80

<u>80% confidence interval for</u> \mu = [ \bar X-1.33 \times {\frac{s}{\sqrt{n} } } , \bar X+1.33 \times {\frac{s}{\sqrt{n} } } ]

                                         = [ 9.65-1.33 \times {\frac{4.61}{\sqrt{20} } } , 9.65+1.33 \times {\frac{4.61}{\sqrt{20} } } ]

                                         = [8.28 hours, 11.02 hours]

Therefore, 80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

7 0
3 years ago
Can anyone figure out this...55% of what is 33?
PSYCHO15rus [73]
55% of 60 is 33......
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3 years ago
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The sum of two consecutive integers is 183. What is the smaller integer?
Natali5045456 [20]
91

Explanation: Since the ones digit ends in a 3, that means the the ones digits of the two consecutive integers must be 1 and 2 or 6 and 7. However, you see 18 in the number 183. 9 + 9 = 18. When you add 6 and 7, you have to carry over one. However, when you add 1 and 2, you do not. Therefore, the two consecutive numbers are 91 and 92. 91 is the smaller integer, so that is the answer.
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3 years ago
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What's 0.312 as a fraction
VARVARA [1.3K]
312
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1000

Since it reads 0 and 312 thousandths, you would simply place 1000 as the denominator. if it reads hundredths, put 100 as the denominator. if it reads tenths, put 10 as the denominator.
5 0
4 years ago
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