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Fiesta28 [93]
3 years ago
6

Among the licensed drivers in the same age group, what is the probability that

Mathematics
2 answers:
Arte-miy333 [17]3 years ago
8 0

Answer:

10.5 percent

Step-by-step explanation:

did the test

vova2212 [387]3 years ago
4 0

Answer:8%

Step-by-step explanation:

You might be interested in
How much did benji earn if he got $50 for mowing the lawn plus a 10% tip?
asambeis [7]

Answer:

$55

Step-by-step explanation:

50+(50*0.1)

50+5

55

6 0
3 years ago
Korey kept track of the number of miles he ran each week for five weeks. The median number of miles he ran during the five weeks
shepuryov [24]

Answer:

the correct answer would be A. the mean is when you add up all of them and divide it by how many numbers there are which in this case is 5. and the median is the number in the middle, in this case 20, therefore when you add up all numbers in A ,you get 105 and you divide it by 5 which gives you 21. :))

6 0
3 years ago
Iterations question two need help please :)
Contact [7]

Answer:

option b

1 , 16, 121 , 13456

Step-by-step explanation:

Given in the question a function, f(x) = (x - 5)²

initial value x_{0} = 4

First iteration

f(x0) = f(4) = (4 - 5)² = (-1)² = 1

x1 = 1

Second iteration

f(x1) = f(1) = (1 - 5)² = (-4)² = 16

x2 = 16

Third iteration

f(x2) = f(16) = ( 16 - 5)² = (11)² = 121

x3 = 121

Fourth iteration

f(x3) = f(121) = (121 - 5)² = (116)² = 13456

x4 = 13456

 

8 0
3 years ago
An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen
devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, \bar X = sample mean = \frac{\sum X}{n} = 441

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 14.34

            n = sample size = 16

            \mu = population mean

<em>Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>

<em />

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.131 < t_1_5 < 2.131) = 0.95  {As the critical value of t at 15 degrees of

                                             freedom are -2.131 & 2.131 with P = 2.5%}  

P(-2.131 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.131) = 0.95

P( -2.131 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.131 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.131 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.131 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X -2.131 \times {\frac{s}{\sqrt{n} } } , \bar X +2.131 \times {\frac{s}{\sqrt{n} } } ]

                                      = [ 441-2.131 \times {\frac{14.34}{\sqrt{16} } } , 441+2.131 \times {\frac{14.34}{\sqrt{16} } } ]

                                      = [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0
3 years ago
What is the value of the expression: 9 + 5 x (-3) – (6 – 2) ÷ 4 show and explain the work please ?
vekshin1
-7.
5x-3 = -15
-15-4\4= -16
-16 + 9 = -7
6 0
3 years ago
Read 2 more answers
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