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3 years ago

Solve for x

align="absmiddle" class="latex-formula">

Mathematics

1 answer:

il63 [147K]3 years ago

3 0

S = 7x^2....divide both sides by 7
1/7s (same as s/7) = x^2...now take sqrt of both sides
(+ - ) sqrt 1/7s = x

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Please help me!!! i will gladly give brainliest :)

True [87]

Given:

ΔONP and ΔMNL.

To find:

The method and additional information that will prove ΔONP and ΔMNL similar by the AA similarity postulate?

Solution:

According to AA similarity postulate, two triangles are similar if their two corresponding angles are congruent.

In ΔONP and ΔMNL,

$\angle ONP\cong \angle MNL$ (Vertically opposite angles)

To prove ΔONP and ΔMNL similar by the AA similarity postulate, we need one more pair of corresponding congruent angles.

Using a rigid transformation, we can prove

$\angle NOP\cong \angle NML$

Since two corresponding angles are congruent in ΔONP and ΔMNL, therefore,

$\Delta ONP\sim \Delta MNL$ (AA postulate)

Therefore, the correct option is A.

8 0

2 years ago

Read 2 more answers

A sample of salary offers (in thousands of dollars) given to management majors is: 48, 51, 46, 52, 47, 48, 47, 50, 51, and 59. U

balu736 [363]

Answer:

Step-by-step explanation:

number of samples, n = 10

Mean = (48 + 51 + 46 + 52 + 47 + 48 + 47 + 50 + 51 + 59)/10 = 49.9

Standard deviation = √(summation(x - mean)/n

Summation(x - mean) = (48 - 49.9)^2 + (51 - 49.9)^2 + (46 - 49.9)^2+ (52 - 49.9)^2 + (47 - 49.9)^2 + (48 - 49.9)^2 + (47 - 49.9)^2 + (50 - 49.9)^2 + (51 - 49.9)^2 + (59- 49.9)^2 = 128.9

Standard deviation = √128.9/10 = 3.59

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation

From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 10 - 1 = 9

Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05

α/2 = 0.05/2 = 0.025

the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975

Looking at the t distribution table,

z = 2.262

Margin of error = 2.262 × 3.59/√10

= 2.57

the lower limit of this confidence interval is

49.9 - 2.57 = 47.33

the lower limit of this confidence interval is

49.9 + 2.57 = 52.47

So it is false

6 0

2 years ago

Solve the following system of equations. Write each of your answers as a fraction reduced to lowest terms. In other words, write

morpeh [17]

Answer:

x = 57/28

y = -95/84

z = 97/168

Step-by-step explanation:

Use the application in the next link: https://www.zweigmedia.com/RealWorld/tutorialsf1/scriptpivotold.html

Start with the expanded array:

$\left[\begin{array}{cccc}1&5&8&1\\3&2&2&5\\-2&-7&2&5\\\end{array}\right]$

then using the tool provided, make row operations until you find the solution:

r2 = r2-3r1

$\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\-2&-7&2&5\\\end{array}\right]$

r3 = r3+2r1

$\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\0&3&18&7\\\end{array}\right]$

r2 = r2*(-1/13)

$\left[\begin{array}{cccc}1&5&8&1\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]$

r1 = r1- r2*5

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]$

r3 = r3+ r2*-3

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&168/13&97/13\\\end{array}\right]$

r3 = r3*13/168

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&1&97/168\\\end{array}\right]$

r2 = r2- r3*22/13

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]$

r2 = r2+ r3*6/13

$\left[\begin{array}{cccc}1&0&0&57/28\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]$

Here you have a reduced array an therefore the answers to each variable are on each row:

$\left[\begin{array}{c}x\\y\\z\end{array}\right]$

8 0

2 years ago

Suppose that a bored student wants to go to a restaurant for lunch, but she only has an hour in which to go, eat, and get back i

Blababa [14]

Missing information

Take local speed ad 30 mph and express your answer in miles

Answer:

7.5 miles

Step-by-step explanation:

Time for lunch is 1 hour but once will wait for 30 minutes before served and eat.

Out of 1 hour, 30 mins is for service and eating to mean time to get from work to restaurant, to and fro is the other 30 mins

Since 30 mins is for to and fro, then one way direction takes 15 mins.

Speed, s=d/t and distance d=speed*time

Given local speed of 30 mph and time as 15 mins,

To convert minutes to hours, we divide it by 60 hence the distance is

30*15/60=7.5 miles

Therefore, the student can go to a restaurant that is 7.5 miles from the office.

4 0

2 years ago

Assume that a simple random sample has been selected from a normally distributed population. State the hypotheses, find the test

BaLLatris [955]

Solution

Hypotheses:

- The population mean is 132. In order to test the claim that the mean is 132, we should check for if the mean is not 132.

- Thus, the Hypotheses are:

$\begin{gathered} H_0:\mu=132 \\ H_1:\mu\ne132 \end{gathered}$

Test statistic:

- The test statistic has to be a t-statistic because the sample size (n) is less than 30.

- The formula for finding the t-statistic is:

$\begin{gathered} t=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}} \\ \\ where, \\ \bar{X}=\text{ Sample mean} \\ \mu=\text{ Population mean} \\ s=\text{ Standard deviation} \\ n=\text{ Sample size} \end{gathered}$

- Applying the formula, we have:

$\begin{gathered} t=\frac{137-132}{\frac{14.2}{\sqrt{20}}} \\ \\ t=\frac{5}{3.1752} \\ \\ t\approx1.5747 \end{gathered}$

Critical value:

- The critical value t-critical, is gotten by reading off the t-distribution table.

- For this, we need the degrees of freedom (df) which is gotten by the formula:

$\begin{gathered} df=n-1 \\ df=20-1=19 \end{gathered}$

- And then we also use the significance level of 0.1 and the fact that it is a two-tailed test to trace out the t-critical. (Note: significance level of 0.1 implies 10% significance level)

- This is done below:

- The critical value is 1.729

P-value:

- To find the p-value, we simply check the table for where the t-statistic falls.

- The t-statistic given is 1.5747. We simply check which values this falls between in the t-distribution table. It falls between 1.328 and 1.729. We can simply choose a value between 0.1 and 0.05 and multiply the result by 2 since it is a two-tailed test.

- However, we can also use a t-distribution calculator, we have:

- Thus, the p-value is 0.13183

Final Conclusion:

- The p-value is 0.13183, and comparing this to the significance level of 0.1, we can see that 0.13183 is outside the rejection region.

- Thus, the result is not significant and we fail to reject the null hypothesis

7 0

1 year ago