All categories

3 years ago

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

the ordered pair not satisfy the equation

therefore

The ordered pair is not a solution of the equation

You might be interested in

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