Which one is the answer???

If 1 inch = 50 miles how many miles is 8.5 in

GenaCL600 [577]

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Hello!

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❖ 8.5 in = 425 miles

If 1 mi = 50 inches, multiply by 8.5:

50 x 8.5 = 425

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7 0

2 years ago

Clare estimates that her brother is 4 feet tall. When they get measured at the doctor’s office, her brother’s actual height is 4

chubhunter [2.5K]

Problem 1

The error is 2 inches since her estimate is 2 inches off the true value.

We can think of it like this

4 feet = 4*12 = 48 inches

4 feet, 2 inches = 4 ft + 2 in = 48 in + 2 in = 50 inches

So she guesses he is 48 inches, but he's really 50 inches, so 50-48 = 2 inches is her error.

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Problem 2

Divide the error (2 inches) over the actual height (50 inches) to get

2/50 = 4/100 = 4%

The percentage error is 4%

This means she is 4% off the target.

Note how 4% of 50 = 0.04*50 = 2 which was the error we found back in problem 1.

8 0

3 years ago

Solve for x in the equation x2+14 x+17=-98

Softa [21]

Answer:

x=−7+√130 or x=−7−√130

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the measure of angle y. Round your answer to the nearest hundredth. (Numerical answer only)

tino4ka555 [31]

Check the picture below.

3 0

3 years ago

Read 2 more answers

In a recent year, a poll asked 2362 random adult citizens of a large country how they rated economic conditions. In the poll,

Harman [31]

Answer:

a) The 99% confidence interval is given by (0.198;0.242).

b) Based on the p value obtained and using the significance level assumed $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

c) $\alpha=0.01$

Step-by-step explanation:

Data given and notation

n=2362 represent the random sample taken

X represent the people who says that they would watch one of the television shows.

$\hat p=\frac{X}{n}=0.22$ estimated proportion of people rated as Excellent/Good economic conditions.

$p_o=0.24$ 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that 24% of people are rated with good economic conditions:

Null hypothesis: $p=0.24$

Alternative hypothesis: $p \neq 0.24$

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 $p_o$ .

Part a: Test the hypothesis

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

np = 2362x0.22=519.64>10 and n(1-p)=2364*(1-0.22)=1843.92>10

Condition satisfied.

Calculate the statistic

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

$z=\frac{0.22 -0.24}{\sqrt{\frac{0.24(1-0.24)}{2362}}}=-2.28$

The confidence interval would be given by:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

The critical value using $\alpha=0.01$ and $\alpha/2 =0.005$ would be $z_{\alpha/2}=2.58$ . Replacing the values given we have:

$0.22 - (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.198$

$0.22 + (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.242$

So the 99% confidence interval is given by (0.198;0.242).

Part b

Statistical decision 

P value method or p value approach . "This method consists on 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 is $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z$

So based on the p value obtained and using the significance level assumed $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

Part c

The confidence level assumed was 99%, so then the signficance is given by $\alpha=1-confidence=1-0.99=0.01$

6 0

2 years ago