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marysya [2.9K]
3 years ago
7

At the karate school, Class A has 8 students with an average age of 6 years. Class B has 12 students with an average age of 9 ye

ars. The classes combine for a group workout.
What is the average age of the combined classes?

Enter your answer as a decimal in the box.
Mathematics
2 answers:
Murrr4er [49]3 years ago
7 0

Answer: I took the test and the answer is 7.8

Good luck!

ELEN [110]3 years ago
3 0
IT IS 7.8 AND I  KNOW IT IS RIGHT
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Margo can purchase tile at a store for $0.69 per tile and rent a tile saw for $36. At another store she can borrow the tile saw
TEA [102]

Answer:

<h3>30 tiles need to buy for same cost in both store.</h3>

Step-by-step explanation:

Let x be the number of tiles to buy for same cost in both store.

18+0.69 · n = 1.29 · n,

1.29n -0.69n =18 or

0.6n =18

n =18/0.6=30

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2564.0 millimeters tall. If the boards are too long they must be trimmed, and if the boards are too short they cannot be used. A
Leokris [45]

Answer:

We conclude that the board's length is equal to 2564.0 millimeters.

Step-by-step explanation:

We are given that a sample of 26 is made, and it is found that they have a mean of 2559.5 millimeters with a standard deviation of 15.0.

Let \mu = <u><em>population mean length of the board</em></u>.

So, Null Hypothesis, H_0 : \mu = 2564.0 millimeters    {means that the board's length is equal to 2564.0 millimeters}

Alternate Hypothesis, H_A : \mu \neq 2564.0 millimeters      {means that the boards are either too long or too short}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

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

where, \bar X = sample mean length of boards = 2559.5 millimeters

            s = sample standard deviation = 15.0 millimeters

             n = sample of boards = 26

So, <em><u>the test statistics</u></em> =  \frac{2559.5-2564.0}{\frac{15.0 }{\sqrt{26}} }  ~   t_2_5

                                     =  -1.529    

The value of t-test statistics is -1.529.

Now, at a 0.05 level of significance, the t table gives a critical value of -2.06 and 2.06 at 25 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the board's length is equal to 2564.0 millimeters.

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