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A ____ is written as an if-then statement A.conjecture B.counterexample C.conditional D.negation

Kaylis [27]

Answer:

D

Step-by-step explanation:

I think D is the answer.

5 0

3 years ago

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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

stealth61 [152]

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.

5 0

3 years ago

Sarah would like to make a 6 lb nut mixture that is 60% peanuts and 40% almonds. She has several pounds of a mixture that is 80%

ivolga24 [154]

Answer:

a. The system of equations that models the situation is....

$0.80x+0.50y=3.60\\\\ 0.20x+0.50y=2.40$

b. The solution to the system: x = 2 and y = 4

The amount of 80/20 mixture is 2 pounds and the amount of 50/50 mixture is 4 pounds.

Step-by-step explanation:

Suppose, the amount of 80/20 mixture is $x$ pounds and the amount of 50/50 mixture is $y$ pounds.

So, the amount of peanuts in 80/20 mixture $= 0.80x$ pound and the amount of almonds in 80/20 mixture $=0.20x$ pound.

And the amount of peanuts in 50/50 mixture $=0.50y$ pound and the amount of almonds in 50/50 mixture $=0.50y$ pound.

Now, Sarah would like to make a 6 pounds nut mixture that is 60% peanuts and 40% almonds.

So, the amount of peanuts in that mixture $=(6\times 0.60)=3.60$ pounds

and the amount of almonds in that mixture $=(6 \times 0.40)= 2.40$ pounds.

So, the system of equations will be.........

$0.80x+0.50y=3.60 ...................(1)\\\\ 0.20x+0.50y=2.40...................(2)$

Subtracting equation (2) from equation (1), we will get.....

$(0.80x+0.50y)-(0.20x+0.50y)=3.60-2.40\\ \\ 0.60x=1.20\\ \\ x= \frac{1.20}{0.60}=2$

Now, plugging this $x=2$ into equation (1), we will get......

$0.80(2)+0.50y=3.60\\ \\ 1.60+0.50y=3.60\\ \\ 0.50y=3.60-1.60=2\\ \\ y=\frac{2}{0.50}=4$

So, the amount of 80/20 mixture is 2 pounds and the amount of 50/50 mixture is 4 pounds.

7 0

3 years ago

Am I right?

Yuliya22 [10]

I think this MIGHT be right I’m not sure but yeah

5 0

3 years ago

Read 2 more answers

Please help me!! -6+x/4=-5

damaskus [11]

y= 4 hope this helps :)

8 0

3 years ago