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Pani-rosa [81]
3 years ago
10

HELP PLEASE i cant tell wither its c or d cause im an idiot

Mathematics
2 answers:
Basile [38]3 years ago
7 0

Answer: I think it’s c

Step-by-step explanation:(don’t worry I’m a idiot too)

Tanzania [10]3 years ago
3 0

Answer:

it is b

Step-by-step explanation:

see steps below:)

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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
Please help more more!,!!,!,,
frutty [35]

Answers:

5. 100 + 400 = 500

6. 50

7. 5

8. 800-600=200

Hope this helped!!!

5 0
4 years ago
Solve the proportion for b.<br> 54<br> =<br> 812<br> b = [?]
Firlakuza [10]
B= 16

8/2 = 16/4

Multiply 2 for each sides
8 0
2 years ago
Read 2 more answers
Find the number of inches needed to represent 300 miles if 1/4 in. = 50 mi.
Kaylis [27]
1.5 inches for 300 miles. 
300/50 = 6
So 6 x 1/4 = 6/4 AKA 1 1/2 AKA 1.5
4 0
3 years ago
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If the 9th term of an A.P is 0, prove that its 29th term is twice its 19th term
Ghella [55]

Let the first term, common difference and number of terms of an AP are a, d and n respectively.

Given that, 9th term of an AP, T9 = 0 [∵ nth term of an AP, Tn = a + (n-1)d]

⇒ a + (9-1)d = 0

⇒ a + 8d = 0 ⇒ a = -8d ...(i)

Now, its 19th term , T19 = a + (19-1)d

= - 8d + 18d [from Eq.(i)]

= 10d ...(ii)

and its 29th term, T29 = a+(29-1)d

= -8d + 28d [from Eq.(i)]

= 20d = 2 × T19

Hence, its 29th term is twice its 19th term

6 0
3 years ago
Read 2 more answers
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