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lutik1710 [3]
3 years ago
9

When you divide a decimal by a decimal why do you multiply the dividend and the divisor by a power of 10

Mathematics
1 answer:
Xelga [282]3 years ago
7 0
To make both sides of the equations equal and balanced. 
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Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf
kobusy [5.1K]

Answer:

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

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097

And the confidence interval for the difference of means would be given by:

0.2903 \leq \mu_1 -\mu_2 \leq 0.5097

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".  

We have the following data given:

\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by \alpha=1-0.93=0.07 and \alpha/2 =0.035. And the critical value would be given by:  

z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811  

The confidence interval is given by:

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

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097

And the confidence interval for the difference of means would be given by:

0.2903 \leq \mu_1 -\mu_2 \leq 0.5097

4 0
3 years ago
Astronomers measure large distances in light-years. One light-year is the distance that light can travel in one year, or approxi
sashaice [31]
7.23 x 10^14
your want you answer in miles ... therefore you multiply 1.23 x 10^2 by 5.88 x 10^12
4 0
3 years ago
Type the missing number to complete the proportion.
zysi [14]

Step-by-step explanation:

if 39 passengers in 13 cars then 39 multiplied by 2 = 78 so 13 by 2 = 26 so the cars will be 26

5 0
2 years ago
The quality control manager at a computer manufacturing company believes that the mean life of a computer is 120 months, with a
IrinaK [193]

Answer:

P(X<118.81)=0.0803

Step-by-step explanation:

Assuming the distribution for the mean life is approximately normal, with mean 120 months and variance 64 months^2, we can calculate the parameters for a sampling distribution with sample size = 89 computers.

The sampling distribution mean will be equal to the mean for a single computer:

\mu_{89}=\mu=120

The standard deviation will be adjusted by the sample size as:

\sigma_{89}=\sqrt{\sigma^2/n}=\sqrt{64/89}=\sqrt{0.719}=0.848

With these parameters, we can calculate the z-score for X=118.81.

z=\dfrac{X-\mu}{\sigma}=\dfrac{118.81-120}{0.848}=\dfrac{-1.19}{0.848}=-1.403

Then, the probability that the mean of a sample of 89 computers is less than 118.81 months is:

P(X

7 0
3 years ago
Combine the like terms to create an equivalent expression.
iragen [17]
4r+14-r-6
Combine like terms
4r and -r are like terms and 14 and -6 are like terms.
(4r-r)+(14-6)
3r+8
Hope I helped!
4 0
3 years ago
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