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qwelly [4]
3 years ago
12

Write y=2(x-1)^2-3 in stander form

Mathematics
1 answer:
KengaRu [80]3 years ago
8 0
The standard form for this equation would be y = 2x^2 - 4x -1. This can be computed by expanding first the (x - 1)^2 which is x^2 - 2x +1. Then this will be multiplied by 2 which will give us, 2x^2 - 4x +2. Lastly, we will deduct the 3 from the equation which gives us the final answer of y= 2x^2 - 4x -1.
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2330 - 0540 milatary time
Dennis_Churaev [7]

It means 11:30 PM - 5:40 AM .

3 0
3 years ago
A sample of size =n72 is drawn from a population whose standard deviation is =σ25. Part 1 of 2 (a) Find the margin of error for
kherson [118]

Answer:

a) margin of error ME = 5.77

b) Margin of error becomes smaller

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

x+/-ME

Where margin of error ME = zr/√n

a)

Given that;

Mean = x

Standard deviation r = 25

Number of samples n = 72

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

ME = 1.96(25/√72)

ME = 1.96(2.946278254943)

ME = 5.774705379690

ME = 5.77

b)

For n = 89

ME = 1.96(25/√89)

ME = 1.96(2.649994700015)

ME = 5.193989612031

ME = 5.19

5.19 is smaller than 5.77 in a) above. So,

Margin of error becomes smaller

6 0
2 years ago
Which is bigger 1/6 or 3/11?
NemiM [27]
Multiply both 1/6=11/66, 3/11=9/66. 1/6 is bigger.
8 0
3 years ago
Read 2 more answers
Airpods were marked down from $219.99 to $183.50.
lina2011 [118]

Answer:

a) 16.59%

b) $207.35

Step-by-step explanation:

you can check it

3 0
3 years ago
A group of retired admirals, generals, and other senior military leaders, recently published a report, "Too Fat to Fight". The r
weqwewe [10]

Answer:

z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735  

p_v =P(z  

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 proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .  

Step-by-step explanation:

1) Data given and notation  

n=180 represent the random sample taken  

X=125 represent the number of americans between 17 to 24 that not qualify for the military

\hat p=\frac{125}{180}=0.694 estimated proportion of americans between 17 to 24 that not qualify for the military

p_o=0.75 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that less than 75% of Americans between the ages of 17 to 24 do not qualify for the military :  

Null hypothesis: p\geq 0.75  

Alternative hypothesis:p < 0.75  

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.  

3) Calculate the statistic  

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

z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735  

4) 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  

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 proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .  

6 0
3 years ago
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