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BabaBlast [244]
3 years ago
6

What is the mean of the data set? 17, 24, 26, 13 4 13 20 80

Mathematics
2 answers:
dangina [55]3 years ago
8 0

Answer:

20

Step-by-step explanation:

17+24+26+13= 80

80/4= 20

vekshin13 years ago
4 0

Answer:

17 + 24 + 26 + 13 = 80

Divide by # of numbers in data set

80/4 = 20

<em>20 is the answer.</em>

<em>Hope this helps</em>

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Nastasia [14]
3miles/ 28 minutes times 42 minutes is equal to 4.5 miles.

Therefore, Eric run 4.5 miles in 42 minutes.

Hopefully this helps!
7 0
3 years ago
Explain:<br><br><br><br><br><br><br><br><br><br> 10 to the 3rd power
irina [24]
10x10x10 which is 1000 have a great day
7 0
3 years ago
Read 2 more answers
Solve each equation
Darya [45]

Answer:

r= 0

r= 1

b= -3

x= -3

Step-by-step explanation:

4r-8r=0

-4r=0

r=0

-15=-8r-7r

-15= -15r

r=1

3b+7-1=-3

3b= -3-7+1

3b= -9

b= -9/3

b=-3

3x+1-6x=7

-3x= 7-1

-3x= 6

x= -6/3

x= -3

7 0
3 years ago
Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa
Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation  

n=580 represent the random sample taken    

X represent the seafood sold in the country that is mislabeled or misidentified by the people

\hat p=0.25 estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

\alpha=0.01 represent the significance level (no given, but is assumed)    

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

The confidence interval would be given by this formula

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

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204

0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

Part c

A government spokesperson claimed that the sample size was too​ small, relative to the billions of pieces of seafood sold each​ year, to generalize. Is this criticism​ valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

4 0
3 years ago
he XO Group Inc. conducted a survey of brides and grooms married in the United States and found that the average cost of a weddi
ahrayia [7]

Answer:

A. P(X<20,000) = 0.0392

B. P(20,000 < x < 30,000) = 0.488

C. Amount = $39,070

Step-by-step explanation:

From XO group website

Cost of a wedding = $29,858

Mean, μ = $29,858

Standard Deviation, σ =$5,600

a. Calculating the probability that a wedding costs less than $20,000

P(X<20,000)?

First, the z value needs to be calculated.

z = (x - μ)/σ

x = 20,000

z = (20,000 - 29,858)/5600

z = -1.76

So, P(X<20,000) = P(Z<-1.76)

From the z table,

P(Z<-1.76) = 0.0392

P(X<20,000) = 0.0392

b. Calculating the probability that a wedding costs between $20,000 and $30,000

P(20,000 < x < 30,000)

First, the z value needs to be calculated.

z = (x - μ)/σ

When x = 20,000

z = (20,000 - 29,858)/5600

z = -1.76

When x = 30,000

z = (30,000 - 29,858)/5600

z = 0.02

P(20,000 < x < 30,000) = P(-1.76 < z <0.02) --- using the z table

P(-1.76 < z <0.02) = 0.508 - 0.0392

P(-1.76 < z <0.02) = 0.4688

P(20,000 < x < 30,000) = 0.488

c. Using the following formula, we'll get the amount it'll a wedding to be among the 5% most expensive

z = (x - μ)/σ where x = amount

Make x the subject of formula

x = σz + μ

Fist we need to get the z value of 5%

z0.05 = 1.645

x = σz + μ becomes

x = 5600 * 1.645 + 29,858

x = $39,070

Amount = $39,070

5 0
3 years ago
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