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kap26 [50]
3 years ago
10

What is the scale factor from AABC to AUVW?

Mathematics
1 answer:
Hitman42 [59]3 years ago
5 0
A, 1/5 yes hehehehehehe
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A microwave cost 200 plus 20% VAT what is the total price ​
iris [78.8K]

Answer:

Total price = 240

Step-by-step explanation:

a microwave cost = 200

20% of 200 = 20/100 x 200 = 40

total price ​= 200 plus 20% VAT = 200 + 40 = 240

7 0
3 years ago
A.
Free_Kalibri [48]

Answer:

d. Yes, there appears to be a linear trend, but the fit is not exact.

Step-by-step explanation:

8 0
3 years ago
What is 2 7/9 as a improper fraction? I keep solving it wrong
Anettt [7]
25/9
9×2=18
18+7=25
Put answer over denominator
25/9
8 0
4 years ago
Read 2 more answers
Plz help me !!! Which of these statements is true for f(x) = 4 . 2X?
alexgriva [62]

Answer:

The y intercept is (0,4)

Step-by-step explanation:

Recommend using a graphing calculator for this equation

6 0
3 years ago
To get an estimate of consumer spending over a holiday season in 2009, 436 randomly sampled American adults were surveyed. Their
Tatiana [17]

Answer:

<em>-A 90% confidence interval would be narrower than the 95% confidence interval if we don't need to be as sure about our estimate. </em>

<em>-This confidence interval is not valid since the distribution of spending in the sample data is right skewed.</em>

<em>-The margin of error is $4.4.</em>

<em>-This confidence interval is valid since the sampling distribution of sample mean would be approximately normal with sample size of 436.</em>

<em>-We are 95% confident that the average spending of all American adults over this holiday season is between $80.31 and $89.11.</em>

<em />

Step-by-step explanation:

<em>A 90% confidence interval would be narrower than the 95% confidence interval if we don't need to be as sure about our estimate. </em>

TRUE. The 90% confidence is less strict in its probability of having the mean within the interval, so it is narrower than the 95% CI. It relies more in the information given by the sample.

<em />

<em>In order to decrease the margin of error of a 95% confidence interval to a third of what is is now, we would need to use a sample 3 times larger. </em>

FALSE. The margin of error is z*σ/(n^0.5). So to reduce it by two thirds, the sample size n needs to be 3^2=9 times larger.

<em>This confidence interval is not valid since the distribution of spending in the sample data is right skewed.</em>

FALSE. There is no information about the skewness in the sample.

<em>The margin of error is $4.4.</em>

TRUE. The margin of error is (89.11-80.31)/2=$4.4.

<em>We are 95% confident that the average spending of these 435 American adults over this holiday season is between $80.31 and $89.11.</em>

FALSE. The CI is related to the populations mean. We are 95% confident that the average spending of the population is between $80.31 and $89.11.

<em>This confidence interval is valid since the sampling distribution of sample mean would be approximately normal with sample size of 436.</em>

TRUE. This happens accordingly to the Central Limit Theorem.

<em>95% of random samples have a sample mean between $80.31 and $89.11.</em>

FALSE. The confidence interval refers to the population mean.

<em>We are 95% confident that the average spending of all American adults over this holiday season is between $80.31 and $89.11.</em>

TRUE. This is the conclusion that is looked for when constructing a confidence interval.

6 0
4 years ago
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