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kondor19780726 [428]
4 years ago
15

please do the steps solve the equation for d: 1/6d-8=5/8 2. Solve the equation for x: 3x-4+5x=10-2z 3. Solve the equation for c:

7(c-3)=14 4. Solve the equation for m: 11(m/22+3/44)=87m+m 5. Solve the equation for k: ck+5k=a
Mathematics
1 answer:
Ulleksa [173]4 years ago
8 0

Step-by-step explanation:

1. 1/6d - 8 = 5/8

1/6d -8 + 8 = 5/8 + 8

1/6d = 69/8

1/6d / 1/6 = 69/8 / 1/6

d = 51.75 or 51\frac{3}{4}

2. 3x - 4 + 5x = 10 - 2z

8x - 4 = 10 - 2z

8x - 4 + 4 = 10 - 2z + 4

8x = 14 - 2z

8x / 8 = (14 - 2z) / 8

x = \frac{14 - 2z}{8}

3. 7(c  - 3) = 14

7(c - 3) / 7 = 14 / 7

c - 3 = 2

c - 3 + 3 = 2 + 3

c = 5

4. 11(m / 22 + 3 / 44) = 87

m / 22 + 3 / 44 = 87 / 11

m / 22 + 3 / 44 = 348 / 44

m / 22 = 345 / 44

m = 172.5

5. ck + 5k = a

k(c + 5) = a

k = \frac{a}{c+5}

Franky Mason
2 years ago
your not right
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And the 90% confidence interval for the difference would be given by:(-0.022;0.0017).  

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".

p_A represent the real population proportion for brand A  

\hat p_A =\frac{80}{2000}=0.04 represent the estimated proportion for the new process

n_A=2000 is the sample size required for Brand A

p_B represent the real population proportion for brand b  

\hat p_B =\frac{75}{1500}=0.05 represent the estimated proportion for the before process

n_B=1500 is the sample size required for before process

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

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For the 90% confidence interval the value of \alpha=1-0.90=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.64  

And replacing into the confidence interval formula we got:  

 (0.04 -0.05) - 1.64 \sqrt{\frac{0.04(1-0.04)}{2000} +\frac{0.05(1-0.05)}{1500}}=-0.022  

(0.04 -0.05) + 1.64 \sqrt{\frac{0.04(1-0.04)}{2000} +\frac{0.05(1-0.05)}{1500}}=0.0017  

And the 90% confidence interval would be given (-0.022;0.0017).  

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