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LiRa [457]
3 years ago
11

How do you solve for c in this problem? 9(c+5)+3c=11

Mathematics
2 answers:
lora16 [44]3 years ago
4 0
Distribute the 9 to what it's multiplying in the parentheses
9c+45+3c=11
combine like terms of 3c and 9c
12c+45=11
get the variable by itself on one side by subtracting the constant
12c=-34
divide by 12
c=-34/12
Keith_Richards [23]3 years ago
4 0
1. First distribute the 9(c+5) to get: 9c+45+3c=11
2. The next thing to do is combine like terms to make the equation: 12c+45=11
3. Now you should subtract 45 from 11 to get "c" by itself: 12c= -34
4. Divide -34 by 12 to finalize your answer and get: c= -34/12
5. Finally simplify the fraction to get your final answer: c= -17/6

Final Answer:  c = -17/6
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Answer:

A. 16.55

Step-by-step explanation:

What we know is the angle of 38° and the hypotenuse is 21. Before we find x (side b), we have to find side a first. TO find that multiply the sine of 38 by 21.

21 * sin(38)

<em>sin(38) calculates to 0.615661475.</em>

21 * 0.615661475

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Answer:

II. This finding is significant for a two-tailed test at .01.

III. This finding is significant for a one-tailed test at .01.

d. II and III only

Step-by-step explanation:

1) Data given and notation    

\bar X=19.2 represent the battery life sample mean    

\sigma=2.5 represent the population standard deviation    

n=25 sample size    

\mu_o =18 represent the value that we want to test    

\alpha represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)    

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:    

Null hypothesis:\mu = 18    

Alternative hypothesis:\mu \neq 18    

And for part III we have a one tailed test with the following hypothesis:

Null hypothesis:\mu \leq 18    

Alternative hypothesis:\mu > 18  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:    

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

3) Calculate the statistic    

We can replace in formula (1) the info given like this:    

z=\frac{19.2-18}{\frac{2.5}{\sqrt{25}}}=2.4    

4) P-value    

First we need to calculate the degrees of freedom given by:  

df=n-1=25-1=24  

Since is a two tailed test for parts I and II, the p value would be:    

p_v =2*P(t_{(24)}>2.4)=0.0245

And for part III since we have a one right tailed test the p value is:

p_v =P(t_{(24)}>2.4)=0.0122

5) Conclusion    

I. This finding is significant for a two-tailed test at .05.

Since the p_v. We reject the null hypothesis so we don't have a significant result. FALSE

II. This finding is significant for a two-tailed test at .01.

Since the p_v >\alpha. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

III. This finding is significant for a one-tailed test at .01.

Since the p_v >\alpha. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

So then the correct options is:

d. II and III only

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