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zepelin [54]
4 years ago
9

who would win? Justice leage or the Teen titans, the Thundercats and the Teenage mutant ninja turtles

Mathematics
2 answers:
OverLord2011 [107]4 years ago
4 0

Answer:

JUSTICE LEAGUE

Step-by-step explanation:

Marat540 [252]4 years ago
3 0
Justice league by murder
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How do you solve<br> -2=u-93÷-3
viktelen [127]

Answer:

u=91

Step-by-step explanation:

-2=u-93÷-3

multiply both sides by 3

3(-2)=(u)3+(-93÷-3)3

-6=3u-279

add 279 to both sides

-6=3u-279

+279. +279

273=3u

divide by 3

273÷3=3u÷3

91=u

4 0
3 years ago
Read 2 more answers
Write the expression shown below with the least amount of terms. <br>8y + 13 - 6y + 3y + 2​
Helga [31]

Answer:

5y + 15

Step-by-step explanation:

8y - 6y = 2y + 3y = 5y

13 + 2 = 15

4 0
3 years ago
$2.01 divided by 2.5 round to the nearest cent?
maw [93]

Answer: $0.80

Step-by-step explanation:

2.01/2.5=

0.804=

<u>0.80</u>4=$0.80

4 0
1 year ago
Which measurement would not be used to calculate the area of this triangle?
bonufazy [111]

Answer:

Im not sure but i think 2 is the answer

Step-by-step explanation:

5 0
3 years ago
Students in a representative sample of 69 second-year students selected from a large university in England participated in a stu
Serhud [2]

Answer:

95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

Step-by-step explanation:

We are given that for the 69 second-year students in the study at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.89.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

                         P.Q. = \frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean procrastination score = 41

             s = sample standard deviation = 6.89

            n = sample of students = 69

            \mu =  population mean estimate

<em>Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.</em>

So, 95% confidence interval for the true mean, \mu is ;

P(-1.9973 < t_6_8 < 1.9973) = 0.95  {As the critical value of t at 68 degree

                                        of freedom are -1.9973 & 1.9973 with P = 2.5%}  

P(-1.9973 < \frac{\bar X -\mu}{\frac{s}{\sqrt{n} } } < 1.9973) = 0.95

P( -1.9973 \times{\frac{s}{\sqrt{n} } } < {\bar X -\mu} < 1.9973 \times{\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-1.9973 \times{\frac{s}{\sqrt{n} } } < \mu < \bar X+1.9973 \times{\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for </u>\mu =[\bar X-1.9973 \times{\frac{s}{\sqrt{n} } } , \bar X+1.9973 \times{\frac{s}{\sqrt{n} } }]

                              = [ 41-1.9973 \times{\frac{6.89}{\sqrt{69} } } , 41+1.9973 \times{\frac{6.89}{\sqrt{69} } } ]

                              = [39.34 , 42.66]

Therefore, 95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

5 0
3 years ago
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