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vovikov84 [41]
3 years ago
13

How does 1.848 look as a fraction

Mathematics
1 answer:
Alex3 years ago
7 0
Well 1 is in the ones place so its 1
.848 is 848 thousandths so it would be 1 \frac{848}{1000}
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Answer:Its O.x^2-4x-21

Step-by-step explanation:

(x-7)(x+3)we multiplie one value with anothet in different brackets and we get x^2+3x-7x-21,add same constant with x and u get .O

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The thickness of metal wires used in the manufacture of silicon wafers is assumed to be normallydistributed with meanμ. To monit
yanalaym [24]

Answer:

(a) Null Hypothesis, H_0 : \mu = 10

(b) Alternate Hypothesis, H_A : \mu\neq 10

(c) One-sample t test statistics distribution :  T.S. = \frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

(d) The value of the test statistic is 1.054.

(e) The p-value = 0.1493

(f) We conclude that the mean equals the target value of 10 which means that the output can be considered acceptable as it doesn't differs from the target value of 10.

Step-by-step explanation:

We are given that the thickness of metal wires used in the manufacture of silicon wafers is assumed to be normally distributed with mean μ. To monitor the production process, the thickness of 40 wires is taken.

The output is considered unacceptable if the mean differs from the target value of 10. The 40 measurements yield a sample mean of 10.2 and sample standard deviation of 1.2.

<u><em>Let </em></u>\mu<u><em> = mean thickness of metal wires used in the manufacture of silicon wafers.</em></u>

(a) Null Hypothesis, H_0 : \mu = 10   {means that the mean equals the target value of 10}

(b) Alternate Hypothesis, H_A : \mu\neq 10   {means that the mean differs from the target value of 10}

(c) The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

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

where, \mu = sample mean = 10.2

            s = sample standard deviation = 1.2

            n = sample of yields = 40

(d) So, <u><em>test statistics</em></u>  =  \frac{10.2-10}{\frac{1.2}{\sqrt{40} } }  ~ t_3_9     

                               =  1.054

The value of the test statistic is 1.054.

(e) <u>Now, P-value of the test statistics is given by;</u>

        P-value = P(t_3_9 > 1.054) = 0.1493

<em>Now at 0.05 significance level, the t table gives critical values between -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test. Since our test statistics lies within the critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.</em>

(e) Therefore, we conclude that the mean equals the target value of 10 which means that the output can be considered acceptable as it doesn't differs from the target value of 10.

7 0
3 years ago
4000 is blank times as much as 4
Goryan [66]
4000 is 1000 times as much as 4.
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Answer:

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Step-by-step explanation:

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