Change 2.56 into a percentage * 0.0256% 25.6% 256.0% none of the above

Convert 385 pounds into kg given the fact that 1 kg = 2.2 pounds.

AlladinOne [14]

Answer:

175 Kg

Step-by-step explanation:

Convert 385 pounds into kg given the fact that 1 kg = 2.2 pounds.

385 : 2.2 =

175 Kg

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6 0

2 years ago

To decrease an amount by 73% what multiplier would yo use?

Ulleksa [173]

You would use 1.73 in order to decrease an amount by 73%

5 0

2 years ago

Read 2 more answers

Given: 2x + 30 = -4(5x -2); Prove: x = -1

Lady_Fox [76]

2x + 30 = -20x + 8
2x + 20x = 8 - 30
22x = -22
x = 1

6 0

2 years ago

Use the diagram and the given information to answer the question. Gavin is in his office, deciding where to go to lunch. The dis

ivolga24 [154]

Answer:

uhh.. I can't give you an answer because I don't have enough information

Step-by-step explanation:

I need some numbers on the pizza place.. I have no information that can lead me to the answer. do you have the diagram?

4 0

1 year ago

Trials in an experiment with a polygraph include 98 results that include 24 cases of wrong results and 74 cases of correct resul

madreJ [45]

Answer:

$z=\frac{0.755 -0.8}{\sqrt{\frac{0.8(1-0.8)}{98}}}=-1.114$

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of successes is not significantly less than 0.8 or 80%

Step-by-step explanation:

Data given and notation

n=98 represent the random sample taken

X=74 represent the number of cases correct

$\hat p=\frac{74}{98}=0.755$ estimated proportion of successes

$p_o=0.8$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of successes is less than 0.8.:

Null hypothesis: $p\geq 0.8$

Alternative hypothesis: $p < 0.8$

When we conduct a proportion test we need to use the z statisitc, 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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.755 -0.8}{\sqrt{\frac{0.8(1-0.8)}{98}}}=-1.114$

Statistical decision

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.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of successes is not significantly less than 0.8 or 80%

5 0

2 years ago