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lilavasa [31]
3 years ago
8

I don't understand this one help me please

Mathematics
1 answer:
Nadya [2.5K]3 years ago
8 0
Well you need to find the value of x. So this is what you do. You have to isolate the variable in other words it needs to be on one side of the equation. You would need to add 11 to both sides and you would get 13x= 9x -2. Then you do the same with 9x and subtract it from boths sides and end up with 4x= -2. You divide -2 by 4 and get -1/2 as your answer. So x= -1/2. I hope this helps love! :)

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Need help with b, c and d for 15 points
stich3 [128]
B is miles he went, for c at the rate he's going at he'll probably be there after five minutes that is all I know 
4 0
3 years ago
Evaluate 1+(-2/3)-(-m)where m = 9/2
slava [35]

Answer:

29/6

Step-by-step explanation:

The expression that we have to evaluate in this problem is

1+(-\frac{2}{3}) -(-m)

for m=\frac{9}{2}.

First of all, we re-arrange the expression. We know that:

- The product of a positive number (+) and a negative number (-) is negative (-)

- The product of a negative number (-) and a negative number (-) is positive (+)

So we can rewrite the expression as

1-\frac{2}{3}+m

Also, we can combine the two numerical terms:

1-\frac{2}{3}=\frac{3}{3}-\frac{2}{3}=\frac{1}{3}

Therefore the expression becomes

\frac{1}{3}+m

Now we can substitute m=\frac{9}{2}, and we find:

\frac{1}{3}+\frac{9}{2}=\\\frac{2}{6}+\frac{27}{6}=\\\frac{2+27}{6}=\frac{29}{6}

8 0
4 years ago
g what is the 95% confidence interval a researcher wishes to compare the average amount of time spent in extracurricular
Serggg [28]

Complete question is;

A researcher wished to compare the average amount of time spent in extracurricular activities by high school students in a suburban school district with that in a school district of a large city. The researcher obtained a simple random sample of 60 high school students in a large suburban school district and found the mean time spent in extracurricular activities per week to be six hours with a standard deviation of three hours. The researcher also obtained an independent simple random sample of 40 high school students in a large city school district and found the mean time spent in extracurricular activities per week to be four hours with a standard deviation of two hours. Let x¯1 and x¯2 represent the mean amount of time spent in extracurricular activities per week by the populations of all high school students in the suburban and city school districts, respectively. Assume two-sample t procedures are safe to use?

what is the 95% confidence interval a researcher wishes to compare the average amount of time spent in extracurricular?

Answer:

CI = (0.755, 3.245)

Step-by-step explanation:

For SRS of 60;

Mean: x1¯ = 6

Standard deviation: s1 = 3

For SRS of 40;

Mean: x2¯ = 4

Standard deviation; s2 = 2

Critical value for the confidence interval of 95% is: t = 1.96

Formula for the CI is;

CI = (x¯1 - x¯2) ± t√[(s1²/n1) + ((s2)²/n1)]

Plugging in the relevant values, we have:

CI = (6 - 4) ± 1.96√[(3²/60) + ((4)²/40)]

CI = 2 ± 1.96√[(3²/60) + ((4)²/40)]

CI = 2 ± 1.96√0.55

CI = 2 ± 1.245

CI = [(2 - 1.245), (2 + 1.245)]

CI = (0.755, 3.245)

5 0
3 years ago
Can someone please help me !
Schach [20]
2)~3x\to\text{Linear Monomial, because it's of 1st degree.}\to B\\\\
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8 0
3 years ago
If ABCD~PQRS, write a proportion and solve for the value of x. Show your work.
sergeinik [125]
X=34

34=34
45=45
But because OF this equation the answer is 34
6 0
3 years ago
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