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3 years ago

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Identify each as an example of exponential growth or decay. What is the y-intercept? What is the end behavior? 1)y=10(0.45)^x 2)

y=.75(4)^x

1 answer:

jeyben [28]3 years ago

5 0

The best answer is number 1

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Which function in Excel is used to perform a test of independence? a. T.TEST b. CHISQ.TEST c. NORM.S.DIST d. Z.TEST

denis23 [38]

Answer:

b. CHISQ.TEST

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

The statistic is given by:

$\chi^2 = \sum_{i=1}^n \frac{(O_i-E_i)^2}{E_i}$

In order to calculate the p value we need to have in count the degrees of freedom , on this case $df=(rows-1)(#cols-1)$ . And we can calculate the p value would be given by:

$p_v =P(\chi^2 >\chi^2_{calc})$

But the correct comand is given by:

b. CHISQ.TEST(Actual range, Expected range)

We just need to put in one column the Observed values and in other the expectec values.

3 0

3 years ago

Hey can u guys plz help me? its one more question

jeyben [28]

Answer:

its 7.5 do you need help

6 0

3 years ago

Read 2 more answers

PLS HELP ILL GIVE BRAINLIEST.

liberstina [14]

Answer:

Step-by-step explanation:

I think it increased by 20% I think or it’s 25% think

3 0

3 years ago

g A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of pa

adoni [48]

Complete Question

A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of parts had a mean of 1.6 millimeters with a standard deviation of 0.03 millimeters. what standard deviation will be needed to achieve a process capability index f 2.0?

Answer:

The value required is $\sigma = 0.0133$

Step-by-step explanation:

From the question we are told that

The upper specification is $USL = 1.68 \ mm$

The lower specification is $LSL = 1.52 \ mm$

The sample mean is $\mu = 1.6 \ mm$

The standard deviation is $\sigma = 0.03 \ mm$

Generally the capability index in mathematically represented as

$Cpk = min[ \frac{USL - \mu }{ 3 * \sigma } , \frac{\mu - LSL }{ 3 * \sigma } ]$

Now what min means is that the value of CPk is the minimum between the value is the bracket

substituting value given in the question

$Cpk = min[ \frac{1.68 - 1.6 }{ 3 * 0.03 } , \frac{1.60 - 1.52 }{ 3 * 0.03} ]$

=> $Cpk = min[ 0.88 , 0.88 ]$

So

$Cpk = 0.88$

Now from the question we are asked to evaluated the value of standard deviation that will produce a capability index of 2

Now let assuming that

$\frac{\mu - LSL }{ 3 * \sigma } = 2$

So

$\frac{ 1.60 - 1.52 }{ 3 * \sigma } = 2$

=> $0.08 = 6 \sigma$

=> $\sigma = 0.0133$

So

$\frac{ 1.68 - 1.60 }{ 3 * 0.0133 }$

=> $2$

Hence

$Cpk = min[ 2, 2 ]$

So

$Cpk = 2$

So $\sigma = 0.0133$ is the value of standard deviation required

3 0

3 years ago

Help me please I need to finish

IrinaK [193]

The awnser is c give me brainliest

8 0

2 years ago

Read 2 more answers