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balu736 [363]
3 years ago
7

Hassan used the iterative process to locate StartRoot 0.15 EndRoot on the number line. A number line going from 0 to 0.9 in incr

ements of 0.1. A point is between 0.4 and 0.5. Which best describes Hassan’s estimation? Hassan is correct because StartRoot 0.15 EndRoot almost-equals 0.4 Hassan is correct because the point is on the middle of the number line. Hassan is incorrect because StartRoot 0.15 EndRoot is less than 0.4. Hassan is incorrect because the point should be located between 0.1 and 0.2.
Mathematics
2 answers:
madreJ [45]3 years ago
8 0

Answer:

Hassan is incorrect because StartRoot 0.15 EndRoot is less than 0.4

Step-by-step explanation:

√0.15 < √0.16 = √0.4² = 0.4

Correct option is:

  • Hassan is incorrect because StartRoot 0.15 EndRoot is less than 0.4
GuDViN [60]3 years ago
8 0

Answer:

Its C

Step-by-step explanation:

Credit to the guy above me

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An automobile company wants to determine the average amount of time it takes a machine to assemble a car. A sample of 40 times y
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Answer:

A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

Step-by-step explanation:

We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

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 average time = 23.92 minutes

             s = sample standard deviation = 6.72 minutes

             n = sample of times = 40

             \mu = population mean assembly time

<em> Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

<u>So, a 98% confidence interval for the population mean, </u>\mu<u> is; </u>

P(-2.426 < t_3_9 < 2.426) = 0.98  {As the critical value of z at 1%  level

                                               of significance are -2.426 & 2.426}  

P(-2.426 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.426) = 0.98

P( -2.426 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.426 \times {\frac{s}{\sqrt{n} } } ) = 0.98

P( \bar X-2.426 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.426 \times {\frac{s}{\sqrt{n} } } ) = 0.98

<u>98% confidence interval for</u> \mu = [ \bar X-2.426 \times {\frac{s}{\sqrt{n} } } , \bar X+2.426 \times {\frac{s}{\sqrt{n} } } ]

                                     = [ 23.92-2.426 \times {\frac{6.72}{\sqrt{40} } } , 23.92+2.426 \times {\frac{6.72}{\sqrt{40} } } ]  

                                    = [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

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