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oksian1 [2.3K]
3 years ago
11

Determine if the given ordered pair is a solution to the equation. Show work full.

Mathematics
1 answer:
sleet_krkn [62]3 years ago
5 0

Answer:

The ordered pair is not a solution of the equation

Step-by-step explanation:

we know that

If a ordered pair is a solution of a linear equation, then the ordered pair must satisfy the linear equation

we have

y=-3x-13

Substitute the value of x and the value of y of the given ordered pair in the linear equation and analyze the results

For x=-1, y=15

15=-3(-1)-13

15=3-13

15=-10 ----> is not true

so

the ordered pair not satisfy the equation

therefore

The ordered pair is not a solution of the equation

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SOMEONE PLEASE HELP ME
goldfiish [28.3K]

Answer:

First break up 72 into

36 x 2 x -1

Then take the square root of 36

6 square root 2 x -1

Finally make square root -1 = i

6i square root 2 is your answer

Step-by-step explanation:

7 0
3 years ago
If the geometric mean of a and 34 is 6 sqrt(17) find the value of a
Alborosie

Using the geometric mean concept, it is found that the value of a is 18.

----------------------

The geometric mean, of a data-set of n elements, (n_1, n_2, ..., n_n), is given by:

G = \sqrt[n]{n_1 \times n_2 \times ... \times n_n}

That is, the nth root of the multiplication of all elements.

----------------------

In this question:

  • Two elements(n = 2), a and 34.
  • G = 6\sqrt{17}

Thus:

\sqrt{34a} = 6\sqrt{17}

We find the square of each side, so:

(\sqrt{34a})^2 = (6\sqrt{17})^2

34a = 36\times17

Simplifying both sides by 17:

2a = 36

a = \frac{36}{2}

a = 18

The value of a is 18.

A similar example is given at brainly.com/question/15010240

3 0
3 years ago
The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi
eimsori [14]

Answer:

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts  and data given  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X=4.2 represent the sample mean  

s=1.4 represent the sample standard deviation  

n=16 represent the sample selected  

\alpha=0.05 significance level  

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:    

Null hypothesis:\mu = 3.6    

Alternative hypothesis:\mu \neq 3.6    

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic  

We can replace in formula (1) the info given like this:    

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with df=n-1=16-1=15 degrees of freedom. The value for \alpha=0.05 and \alpha/2=0.025 so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got t_{crit}=\pm 2.131    

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

Conclusion    

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

3 0
3 years ago
Tom gathered some mushrooms. Of those he dried 2/5 , then pickled 5/9 of the remaining mushrooms, and then fried the 28 mushroom
xxTIMURxx [149]
I am thinking 7. hope it helps

5 0
3 years ago
Read 2 more answers
there are 48 students in a school play. the ratio of boys to girls is 5:7. how many more girls than boys are in the school play?
Alex Ar [27]
We can use trial or error method here...

If we take 4 table, 4*5 = 20 and 4*7 = 28. 20+28 = 48.
20 : 28 = 5 : 7

So, there are 20 boys and 28 girls in a school play and there are 8 girls more than boys.

If you look over tables and try it out, only 4 table works for the given ratio.
3 0
3 years ago
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