Please answer this math

Help asap!!!!!!!!!!!

svlad2 [7]

Answer:

The answer to your question is:

Step-by-step explanation:

12.-

x² - 6x = 5

x² - 6x + (3)² = 5 + (3)²

x² - 6x + 9 = 5 + 9

(x - 3)² = 14

13.- x² - 6x = 12

x² - 6x + (3)² = 12 + (3)²

x² - 6x + 9 = 12 + 9

(x - 3)² = 21

x - 3 = ±√21

x1 = √21 + 3 x2 = -√21 + 3

x1 = 7.58 x2 = -1.58

14.- 2x² - 24x = - 20 Divide by 2

x² - 12 x = -10

x² - 12x + (6)² = - 10 + (6)²

x² - 12x + 36 = - 10 + 36

( x - 6) ² = 26

x - 6 = ±√26

x1 = √26 + 6 x2 = -√26 + 6

x1 = 11.1 x2 = 0.9

3 0

3 years ago

Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8

Delvig [45]

We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51$

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98$

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34$
Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86$

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.

3 0

3 years ago

A.16cm<br> b. 16 root 2cm<br> c. 8 root 2cm<br> d. 8cm

Stolb23 [73]

Answer:

C

Step-by-step explanation:

Mark as Brainliest :)

7 0

3 years ago

The sum of two consecutive integers is at least 185. what is the smaller of the two integers? enter your answer in the box.

ss7ja [257]

Let us assume the first integer to be = x
Then
The second integer will be = x + 1
The sum of the two consecutive integers =185
Then
x + x + 1 = 185
2x + 1 = 185
2x = 184
x = 92
From the above deduction, we can easily conclude that the smallest of the two integers is 92. I hope that the answer has helped you.

4 0

3 years ago

Read 2 more answers

7. When dy/dx = (x -6)1/2, what does y equal?

Sergio039 [100]

The answer is below
$\frac{dy}{dx} = \frac{1}{2} (x - 6)$
multiply both sides by dx

$dy = \frac{1}{2} (x - 6)dx$
integrate both sides of the diffrential

$integral \: dy = integral \: \frac{1}{2} (x - 6)dx$
simplify and integrate to get y alone don't forget to tack on c with your indefinite integral

$y = \frac{x {}^{2} }{4} - 3x + c$

3 0

3 years ago