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Ugo [173]
3 years ago
7

L.c.m of 84 ,27,35,45 please help​

Mathematics
1 answer:
den301095 [7]3 years ago
8 0

Answer:

3780

Step-by-step explanation:

84 = 2^2 * 3 * 7

27 = 3^3

35 = 5 * 7

45 = 3^2 * 5

lcm = 2^2 * 3^3 * 5 * 7 = 3780

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A university interested in tracking its honors program believes that the proportion of graduates with a GPA of 3.00 or below is
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Answer:

a) On this case we are interested on the population proportion of students that have a GPA of 3.00 or below.

b) z=\frac{0.15 -0.2}{\sqrt{\frac{0.2(1-0.2)}{200}}}=-1.77  

p_v =P(z  

c) Null hypothesis:p\geq 0.2  

Alternative hypothesis:p  

d) The significance level provided is \alpha=0.05. If we compare the  p value obtained and the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the population proportion of students that have a GPA of 3.00 or below is significantly lower than 0.2.  

Step-by-step explanation:

Data given and notation n  

n=200 represent the random sample taken

X=30 represent the  students with a GPA of 3.00 or below.

\hat p=\frac{30}{200}=0.15 estimated proportion of  students with a GPA of 3.00 or below.  

p_o=0.2 is the value that we want to test

\alpha represent the significance level  

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

a. In testing the university's belief, how does on define the population parameter of interest?

On this case we are interested on the population proportion of students that have a GPA of 3.00 or below.

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of graduates with GPA of 3.00 or below is less than 0.2.:  

Null hypothesis:p\geq 0.2  

Alternative hypothesis:p  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

b. The value of the test statistics and its associated p-value are?

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.15 -0.2}{\sqrt{\frac{0.2(1-0.2)}{200}}}=-1.77  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

Since is a one left tailed test the p value would be:  

p_v =P(z  

c. In testing the university's belief, the appropriate hypothesis are?

Null hypothesis:p\geq 0.2  

Alternative hypothesis:p  

d. At a 5% significance level, the decision is to?

The significance level provided is \alpha=0.05. If we compare the  p value obtained and the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the population proportion of students that have a GPA of 3.00 or below is significantly lower than 0.2.  

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3 years ago
PLs Help...................
Grace [21]

Answer:

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Step-by-step explanation:

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2 years ago
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andrew11 [14]

Answer:

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Step-by-step explanation:

We are told that:

The ratio of Val's savings to Dan is 2:3 at first.

Now, Dan spent P3000.00 and Val's saving became 1⅓ more of that Dan's remaining savings.

Let's say the total amount of savings they had at first was x.

Thus;

Val had: 2x/5

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Now, Dan spent P3000.00.

So amount Dan has left = (3x/5) - 3000

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2x/5 = 12x/15 - 4000

Multiply through by 15 to get;

6x = 12x - (3000 × 15)

6x = 12x - 45000

Rearranging, we have;

12x - 6x = 45000

6x = 45000

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x = P.7500

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