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MA_775_DIABLO [31]
3 years ago
6

A survey of Ghasi's customers showed that 1,200 of the customers buying yarn would buy bamboo yarn. In the first month, 725 cust

omers actually bought bamboo yarn. What was the percent error in the survey estimate of the number of people buying bamboo yarn? Round to the nearest tenth.
Mathematics
2 answers:
Arturiano [62]3 years ago
5 0

Answer:

<h2>40%</h2>

Step-by-step explanation:

Step one:

given data

expected customers= 1200

actual customers= 725

the percent error= (actual-expected)/expected

substituting the given values we have'

%error=725-1200/1200

%error=475/1200

%error=0.395

To the nearest tenth we have 0.4

%error=0.4*100

%error=40%

Hence the percent error is 40%

hjlf3 years ago
4 0

Answer:

40%

Step-by-step explanation:

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Step-by-step explanation: subtract 160 and 125 and you’ll get 35

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The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims
kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is \mu=600, we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it \sigma) is

\sigma=48

Next they draw a random sample of n=70 students, and they got a mean score (denoted by \bar x) of \bar x=613

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis H_0:\bar x \geq \mu

- The alternative would be then the opposite H_0:\bar x < \mu

The test statistic for this type of test takes the form

t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\

<h3>since 2.266>1.645 we  can reject the null hypothesis.</h3>
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4 years ago
Read 2 more answers
10. Explain why log328 is between 3 and 4.
maw [93]

Answer:

See explanation.

(Before continuing reading, I took the base to be 3. Please tell me if you didn't want the base to be 3.)

Step-by-step explanation:

I assume 3 is suppose to be the base. Let's list some values that can be written as 3 to some integer.

3^0=1

3^1=3

3^2=9

3^3=27

3^4=81

3^5=243

......

I could have also did negative integer powers, but this is all I really need to convince you that log_3(28) is between 3 and 4.

log_3(28) means the value x such that 3^x=28.

Since 28 is between 27 and 81 in my list above, that means 3^x is between 3^3 and 3^4. This means that x is a value between 3 and 4.

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