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11111nata11111 [884]

3 years ago

11

My question is the picture

1 answer:

AysviL [449]3 years ago

5 0

Answer:

20 students

Step-by-step explanation:

You first have to add the number of students all up and you get 200.Then you do cross multiplication.

40/200 x s/100=

40x100= 4000=

200xs= 200s=

200s/200=

4000/200=

20=

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If ΔFGH ≅ ΔIJK, which segment is congruent to FH? PLS HELP PLS PLS PLS

Mazyrski [523]

Answer:

IK

Step-by-step explanation:

If you draw out the triangles it is easier to see

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3 years ago

3. A new process has been developed for applying pho toresist to 125-mm silicon wafers used in manufac turing integrated circu

dlinn [17]

Answer:

a

The null hypothesis is $H_o : \mu = 13.4000$

The alternative hypothesis is $H_a : \mu \ne 13.4000$

The null hypothesis is rejected

b

The 99% confidence level is $13.3930 < \mu < 13.3994$

Step-by-step explanation:

From the question we are told that

The sample size is n = 10

The population mean is $\mu = 13.4 000 \ angstroms$

The level of significance is $\alpha = 0.05$

The sample data is

13.3987, 13.3957, 13.3902, 13.4015, 13.4001, 13.3918, 13.3965, 13.3925, 13.3946, and 13.4002

Generally the sample mean is mathematically represented as

$\= x = \frac{13.3987+ 13.3957\cdots +13.4002 }{10}$

=> $\= x = 13.3962$

Generally the sample standard deviation is mathematically represented as

$\sigma = \sqrt{\frac{\sum (x_i - \= x)^2}{n} }$

=> $\sigma = \sqrt{\frac{ (13.3987 - 13.3962)^2 + (13.3987 - 13.3962)^2 + \cdots + (13.3987 -13.4002)^2 }{10} }$

=> $\sigma =0.0039$

The null hypothesis is $H_o : \mu = 13.4000$

The alternative hypothesis is $H_a : \mu \ne 13.4000$

Generally the test statistics is mathematically represented as

$z = \frac{\= x - \mu}{\frac{\sigma}{\sqrt{n} } }$

=> $z = \frac{13.3962 - 13.4000}{\frac{0.0039}{\sqrt{10} } }$

=> $z = 3.08$

Generally the p-value is mathematically represented as

$p-value = 2P( z > 3.08)$

From the z-table $P(z > 3.08)= 0.001035$

So

$p-value = 2* 0.001035$

$p-value = 0.00207$

So from the obtained value we see that

$p-value < \alpha$

Hence the null hypothesis is rejected

Consider the b question

Given that the confidence level is 99% then the level of significance is

$\alpha = (100 -99)\%$

=> $\alpha = 0.01$

Generally from the normal distribution table critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 2.58 * \frac{0.0039}{\sqrt{10} }$

=> $E = 0.00318$

Generally the 99% confidence interval is mathematically represented as

$13.3962 - 0.00318 < \mu < 13.3962 + 0.00318$

=> $13.3930 < \mu < 13.3994$

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3 years ago

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2 years ago

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nordsb [41]

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Annette [7]

Answer:

12

Step-by-step explanation:

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2 years ago