Please help me simplify

A quadrilateral-shaped land property is drawn on a coordinate grid. Two adjacent sides of the property lie along the lines y = 4

Liono4ka [1.6K]

Answer:

P(-4,-8)

Step-by-step explanation:

Given two lines:

$y = 4x + 8;$ and $ \\y = -0.25x - 9$

To determine the point of intersection P of the lines, we equate the two lines and solve for (x,y).

$4x + 8= -0.25x - 9\\4x+0.25x=-9-8\\4.25x=-17\\$Divide both sides by 4.25$\\x=-4\\$Recall:$\\y = 4x + 8\\=4(-4)+8\\=-16+8\\=-8$

Therefore, the intersection point P of the lines is: P(-4,-8)

6 0

3 years ago

The backdrop is taped off into 12 equal sections for the students to paint.

cricket20 [7]

Answer:

5/12

Step-by-step explanation:

Jill 1/12

Mark 2/12

Lou 1/12

Sam 3/12

add them together

7/12

12-7=5

5/12

4 0

3 years ago

Altitute of a triange is 2 cm shorten than its base. the area is 15 cm square. find the base

jok3333 [9.3K]

The base would be 5cm

8 0

3 years ago

5 feet 4 feet What is the perimeter? 16 feet 14 feet 18 feet

brilliants [131]

perimiter means all the way around, and im assuming this is a 4-sided shape, so 5+5+4+4 = 18 ft.

8 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

4 years ago