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Oliga [24]
3 years ago
9

I need help. Please give me an example

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
8 0

Answer:

9

Step-by-step explanation:

3/729= 9

729=9^3

=3/9^3= 9


Hope this helps! Please give brainiest!

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What is the slope of the linear function represented in the table​
Yanka [14]

Answer: m (the slope) = 1/7

Step-by-step explanation:

The slope formula is attached. When you plug in the numbers, you get 1/7 -

0 - 1 / -7 - 0

8 0
3 years ago
Eric is adding water to a 60
Archy [21]

Answer:

After 2.5 minutes the pool will have 27 gallons of water.

Step-by-step explanation:

The pool has already 12 gallons of water and Eric wants to fill it to at least 27 gallons.

The water is flowing at a rate of 6 gallons per minute.

Let, after t minutes the pool will have at least 27 gallons of water.

Therefore, we can write the equation as  

27 = 12 + 6t

⇒ 6t = 15

⇒ t = 2.5 minutes.

Therefore, after 2.5 minutes the pool will have 27 gallons of water. (Answer)

8 0
3 years ago
A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.
TEA [102]

Answer:

(a) Null Hypothesis, H_0 : \mu = $1,150  

    Alternate Hypothesis, H_A : \mu \neq $1,150

(b) The test statistic is 3.571.

(c) We conclude that the mean of all account balances is significantly different from $1,150.

Step-by-step explanation:

We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.

<u><em>Let </em></u>\mu<u><em> = mean of all account balances</em></u>

(a)So, Null Hypothesis, H_0 : \mu = $1,150     {means that the mean of all account balances is equal to $1,150}

Alternate Hypothesis, H_A : \mu \neq $1,150    {means that the mean of all account balances is significantly different from $1,150}

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

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

where, \bar X = sample average balance = $1,200

             s = sample standard deviation = $126

             n = sample of account balances = 81

(b) So, <u>test statistics</u>  =  \frac{1,200-1,150}{\frac{126}{\sqrt{81} } }  ~ t_8_0

                               =  3.571

The value of the sample test statistics is 3.571.

(c) <u>Now, P-value of the test statistics is given by the following formula;</u>

          P-value = P( t_8_0 > 3.571) = <u>Less than 0.05%</u>

Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.

Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean of all account balances is significantly different from $1,150.

8 0
4 years ago
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Goshia [24]

Answer:

I think the awsner is 180 deegrees

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Answer:

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5 0
3 years ago
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