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3 years ago

Which ordered pair is a solution of the equation? y = 7 x − 3 y=7x−3

Mathematics

1 answer:

eimsori [14]3 years ago

8 0

Answer:

(1,4)

Step-by-step explanation:

Which ordered pair is a solution of the equation? y = 7 x − 3

a. (1,4) b. (-1,-4) c. both d. neither

Solution

y=7x-3

Solve by trying each ordered pair

a. (1,4)

x=1, y=4

Substituting the value of x and y into the equation

y=7x-3

4=7(1)-3

4=7-3

4=4

This is a true statement

b. (-1,-4)

x=-1, y=-4

Substitute the value into the equation

y=7x-3

-4=7(-1)-3

-4= -7-3

-4= -11

This is not a true statement

This true statement is when x=1 and y=4

So, the ordered pair (1, 4) is the solution

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The product of two consecutive odd integers is 1 less than four times their some. find the two integers.

nignag [31]

2 consecutive odd integers.....x and x + 2.....the product means multiply....the sum means add.

x(x + 2) = 4(x + x + 2) - 1
x^2 + 2x = 4(2x + 2) - 1
x^2 + 2x = 8x + 8 - 1
x^2 + 2x = 8x + 7
x^2 + 2x - 8x - 7 = 0
x^2 - 6x - 7 = 0
(x - 7)(x + 1) = 0

x - 7 = 0 x + 2 = 7 + 2 = 9
x = 7

x + 1 = 0 x + 2 = -1 + 2 = 1
x = -1

there can be 2 answers for this....
(1) 7 and 9
(2) -1 and 1

7 0

2 years ago

In a random sample of fivefive microwave ovens, the mean repair cost was $75.0075.00 and the standard deviation was $13.0013.

Radda [10]

<h2>Answer with explanation:</h2>

As per given , we have

Sample size : n= 5

Degree pf freedom = : df= 5-1=4

$\mu=\$75.00$

$\sigma=\$13.00$

Significance level for 90% confidence = $\alpha=1-0.90=0.1$

Using t-value table , t-critical value for 90% confidence:

$t_{\alpha/2, df}=t_{0.05, 4}=2.132.$

Margin of error of $\mu$ : $E=t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\(2.132)\dfrac{13}{\sqrt{5}}=12.3949720129\approx12.39$

Interpretation : The repair cost will be within $12.39 of the real population mean value $\mu$ 90% of the time.

7 0

3 years ago

35. In a simple random sample of 25 high school students, the sample mean of the SAT scores was 1450, and the sample variance wa

Tatiana [17]

Answer:

$1450-2.0\frac{30}{\sqrt{25}}=1450-12$

$1450+2.0\frac{30}{\sqrt{25}}=1450+12$

And the best option would be:

c. 1450 +/- 12

Step-by-step explanation:

Information provided

$\bar X=1450$ represent the sample mean for the SAT scores

$\mu$ population mean (variable of interest)

$s^2 = 900$ represent the sample variance given

n=25 represent the sample size

Solution

The confidence interval for the true mean is given by :

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The sample deviation would be $s=\sqrt{900}= 30$

The degrees of freedom are given by:

$df=n-1=2-25=24$

The Confidence is 0.954 or 95.4%, the value of $\alpha=0.046$ and $\alpha/2 =0.023$ , assuming that we can use the normal distribution in order to find the quantile the critical value would be $z_{\alpha/2} \approx 2.0$