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Dmitry_Shevchenko [17]
3 years ago
11

Write the mixed number as a fraction. 4 1/3

Mathematics
2 answers:
Mnenie [13.5K]3 years ago
7 0
13/3 is a improper fraction
Katyanochek1 [597]3 years ago
6 0
Improper fraction: 13/3
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I don't how I would do this exactly..
Morgarella [4.7K]
<span>The sum is 10 times 11</span>
6 0
3 years ago
Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below.
garik1379 [7]

Answer:

The 95% confidence interval would be given by 0.281 \leq \mu_1 -\mu_2 \leq 0.529  

Step-by-step explanation:

Previous concepts  

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".  

\bar X_1 =9.2 represent the sample mean 1

\bar X_2 =8.8 represent the sample mean 2

n1=27 represent the sample 1 size  

n2=30 represent the sample 2 size  

\sigma_1 =0.3 population standard deviation for sample 1

\sigma_2 =0.1 population standard deviation for sample 2

\mu_1 -\mu_2 parameter of interest.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:

\bar X_1 -\bar X_2 =9.2-8.8=0.4

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that t_{\alpha/2}=1.96  

Now we have everything in order to replace into formula (1):  

0.4-1.96\sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}=0.281  

0.4+1.96\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=0.519  

So on this case the 95% confidence interval would be given by 0.281 \leq \mu_1 -\mu_2 \leq 0.529  

6 0
3 years ago
At Albemarle High School, 5/9 of the students are in band and 57% of the students are in chorus. Are more AHS students in band o
borishaifa [10]

Let find percentage of 5/9

\\ \sf\longmapsto 5/9(100)

\\ \sf\longmapsto 500/9

\\ \sf\longmapsto 55.5\%

More students are in chorus

8 0
2 years ago
Read 2 more answers
Please help! Don't answer if your not sure.
gavmur [86]

Answerthats good

Step-by-step explanation:

i will answer

5 0
2 years ago
What is the missing numerator? (4 points)
kozerog [31]

Answer:

\frac{x}{7} - \frac{3}{8} = \frac{27\\}{56}

Step-by-step explanation:

C

6 x 8 = 32

7 x 3 = 21

32 - 21 = 27

7 0
2 years ago
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