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aivan3 [116]
3 years ago
9

6(t − 2) = 2(9 − 2t)

Mathematics
2 answers:
drek231 [11]3 years ago
7 0
6(t-2)=2(9-2t) 
Distribute
6t-12=18-4t
Combine like terms.
10t=30
Divide to get your answer 
30/10=3
t=3
SCORPION-xisa [38]3 years ago
4 0
Hello!

You solve this algebraically.

6(t - 2) = 2(9 - 2t)

Distribute the 6

6t - 12 = 2(9 - 2t)

Distribute the 2

6t - 12 = 18 - 4t

Add 4t to both sides

10t - 12 = 18

Add 12 to both sides

10t = 30

Divide both sides by 10

t = 3

The answer is 3

Hope this helps!
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Using the discriminant, how many real number solutions does this equation have? 3x^2 – 2 = 5x
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3 years ago
Need help homies, no steps needed
VashaNatasha [74]

Answer:

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Step-by-step explanation:

4 0
3 years ago
i bought 6 apples for0.30 dollars each and 5 oranges for 0.40 dollars each. how much did i spend ? write an equation then solve​
katovenus [111]

Answer:

3.8

Step-by-step explanation:

if the apples are 0.30 cents each then you multiply .3 X 6 and you get 1.8 because 1 apple is .30 cents you want to know how much is 6.

Next you pay .40 cents each for 5 oranges so .4 X 5 which is 2. And once again one orange is .40 cents so you multiply that by 5 to see the price of 5 oranges

Now you add the 1.8 to 2 and get 3.8

Now for the equation you just need to plug in what you did so...

(0.3 X 6) + (0.40 X 5) =  3.8

Glad i could help :)

7 0
3 years ago
A manufacturing company produces steel housings for electrical equipment. The main component part of the housing is a steel trou
Mrrafil [7]

Answer:

Check the explanation

Step-by-step explanation:

(a)

H0: Population mean = 8.46

H1: Population mean is not equal to 8.46

The test statistic (t) is given by the following expression -

{t=\frac{\overline{x}-\mu_0}{s/\sqrt{n}}}

We have calculated the sample mean (8.421) and sample standard deviation (0.0461). Therefore, the test statistic (t) will be -

t = (8.421 - 8.46)/(0.0461/sqrt(49)) = -5.92

The left-sided critical value from t-distribution (two-tailed) is -2.01. Thus the test statistic falls in the rejection region and we reject the null hypothesis at 95% confidence level and conclude that the population mean is different from 8.46 inches.

(b)

There are the following four assumptions for a single sample t-test.

The observations are in ratio scale.

The observations have been taken in such a manner that every single observation is independent and uncorrelated from the others.

There should not be any significant outliers in the sample observations.

The sample data should be approximately normal.

(c)

The first assumption is true as the data is in inches. The second assumption cannot be tested now. It will depend upon the situation and study design which we assume as correct in this case. The third assumption is checked by plotting a box plot and finding the outliers. The final assumption can be checked using the Anderson-Darling normality test.

Kindly check the graphical table in the attached images below.

(d)

The data is normal (the p-value of Anderson-Darling test in > 0.05). However, two points in the Box plot are outliers though not grossly different from the entire sample data set. So, we conclude that the assumptions are valid in this case for conducting the t-test.

7 0
3 years ago
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