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Romashka [77]
4 years ago
8

Restaurants often slip takeout menus under Braden's apartment door. The table shows how many menus Braden has collected from dif

ferent types of restaurants. Japanese 24 Chinese 1 Thai 9 Indian 4 What is the probability that one of Braden's takeout menus, selected at random, will be from an Indian restaurant? Write your answer as a fraction or whole number.
Mathematics
1 answer:
solniwko [45]4 years ago
7 0

Answer:

2/19

Step-by-step explanation:

Total menus collected= 24+1+9+4=38

Indian menus collected= 4

Therefore= 4/38 = 2/19

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Lilit [14]
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7 0
3 years ago
The heights of 40 randomly chosen men are measured and found to follow a normal distribution. An average height of 175 cm is obt
AVprozaik [17]

Answer:

95% two-sided confidence interval for the true mean heights of men is [168.8 cm , 181.2 cm].

Step-by-step explanation:

We are given that the heights of 40 randomly chosen men are measured and found to follow a normal distribution.

An average height of 175 cm is obtained. The standard deviation of men's heights is 20 cm.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

                             P.Q. = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample average height = 175 cm

            \sigma = population standard deviation = 20 cm

            n = sample of men = 40

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

So, 95% confidence interval for the true mean, \mu is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                     level of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times }{\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times }{\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for </u>\mu = [ \bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } } ]

                                            = [ 175-1.96 \times }{\frac{20}{\sqrt{40} } } , 175+1.96 \times }{\frac{20}{\sqrt{40} } } ]

                                            = [168.8 cm , 181.2 cm]

Therefore, 95% confidence interval for the true mean height of men is [168.8 cm , 181.2 cm].

<em>The interpretation of the above interval is that we are 95% confident that the true mean height of men will be between 168.8 cm and 181.2 cm.</em>

3 0
3 years ago
A swimsuit has been marked down from an original price of $75 to $56.25. By
Xelga [282]
The answer would be 25%

75 * .25 = 18.75

75 - 18.75 = 56.25
3 0
2 years ago
How do you turn a temperature into an integer?
vichka [17]
All i found was a tempature conversion calculator online? which i could help more, sorry
5 0
4 years ago
Harry has 18 chocolates. Of the 18 chocolates, 4 of them are white chocolate. The rest are milk chocolate. Which fraction is equ
allsm [11]

Answer:

14/18

Since 4 out of the 18 chocolates are white, the rest will be 14 milk chocolates

and in total there are 18 chocolates

6 0
3 years ago
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