Legal descriptions tend to prefer neat straight lines from point to point, regardless of describing a square, rectangle, triangle or even a smooth circle. When might a property boundary end up being a squiggly line?

Answer:

Step-by-step explanation:

5x(x + 3) + 6(x + 3)

The final answer is

( 5x + 6 ) ( x + 3 )

Hope this helps you

Answer:

33.777 yd = 0.030886 km

Step-by-step explanation:

==>Given:

33.777 yd

==>Required:

Convert 33.777 yd to km to 6 decimal places, using the metric and U.S systems.

==>Solution:

To convert, note that 1 km = 1093.6133 yd.

Thus,

1 km = 1093.6133 yd

x km = 33.777 yd

Cross multiply

1 × 33.777 = 1093.6133 × x

33.777 = 1093.6133x

Divide both sides by 1093.6133, to solve for x

33.777/1093.6133 = x

0.03088569 = x

x ≈ 0.030886 (to 6 decimal places)

Therefore, 33.777 yd = 0.030886 km

u/4 - 4 = -20

Add 4 to both sides:

u/4 = -16

Multiply both sides by 4:

u = -64

Answer:

u=-64

Step-by-step explanation:

u/4 -4 = -20

First add 4 to both sides.

u/4=-16

Now multiply both sides by 4

u=-64

Answer:

if m is supposed to be the equals (=) sign then x = 25

Step-by-step explanation:

45 = (2x-5)

+5         +5

50 = (2x)

÷2     ÷2

25 = x

Answer: 70

Step-by-step explanation:

Answer:

A) A = {RRR, LLL, SSS}

B) B = {LRS. LSR, RLS, RSL, SLR, SRL}

C) C = {RRL, RRS, RSR, RLR, LRR, SRR}

D) D = {RRL, RRS, RSR, RLR, LRR, SRR. LLR, LLS, LSL, LRL, RLL, SLL, SSL, SSR, SLS, SRS, LSS, RSS}

E) D' ={RRR, LLL, SSS, LRS. LSR, RLS, RSL, SLR, SRL}

F) C ∪ D = {RRL, RRS, RSR, RLR, LRR, SRR. LLR, LLS, LSL, LRL, RLL, SLL, SSL, SSR, SLS, SRS, LSS, RSS}

G) C ∩ D = {RRL, RRS, RSR, RLR, LRR, SRR}

Step-by-step explanation:

A) All vehicles must go right, left or straight ahead (three possibilities):

A = {RRR, LLL, SSS}

B) One vehicle must go right, one must go left, and the remaining one must go straight ahead (six possibilities):

B = {LRS. LSR, RLS, RSL, SLR, SRL}

C) There are three ways that exactly two vehicles go right (1 and 3, 2 and 3, 1 and 2), there are then two options for the remaining vehicle (left and straight) for a total of six possibilities:

C = {RRL, RRS, RSR, RLR, LRR, SRR}

D) Follow the same reasoning from the previous item, but multiply the number of possibilities by 3 (for each direction in which both cars can go: right, left or straight):

D = {RRL, RRS, RSR, RLR, LRR, SRR. LLR, LLS, LSL, LRL, RLL, SLL, SSL, SSR, SLS, SRS, LSS, RSS}

E) D' is the set containing all possibilities not present in set D. D' is comprised by the possibilities of all vehicles going in the same direction, or each vehicle in a different direction:

D' ={RRR, LLL, SSS, LRS. LSR, RLS, RSL, SLR, SRL}

F) The outcomes in C ∪ D is the union of elements from set C and D (neglecting repeated values), which happens to be all values in set D.

C ∪ D = {RRL, RRS, RSR, RLR, LRR, SRR. LLR, LLS, LSL, LRL, RLL, SLL, SSL, SSR, SLS, SRS, LSS, RSS}

G) The outcomes in C ∩ D is the list of values present in both sets C and D, which happens to be all values in set C:

C ∩ D = {RRL, RRS, RSR, RLR, LRR, SRR}

Answer:

a. AEB

b. BEC

c. CED

d. AED

Step-by-step explanation:

Each angle is made up of three points. All three points in order is the name of the angle.

Answer:

a. ∠1 = ∠AEB or ∠BEA

b. ∠2 = ∠BEC or ∠CEB

c. ∠3 = ∠CED or ∠DEC

d. ∠4 = ∠DEA or ∠AED

Step-by-step explanation: Penn <3

Answer:

C

Step-by-step explanation:

C is the solution

Answer:

Option C

Step-by-step explanation:

The graph is a horizontal translation 4 units left and a vertical translation 2 units down ⇒ y= |x+4|-2

Answer:

C. c = (xv - x) / (v - 1).

Step-by-step explanation:

v = (x + c) / (x - c)

(x - c) * v = x + c

vx - vc = x + c

-vc - c = x - vx

vc + c = -x + vx

c(v + 1) = -x + vx

c = (-x + vx) / (v + 1)

c = (-x + xv) / (v + 1)

c = (xv - x) / (v + 1)

So, the answer should be C. c = (xv - x) / (v + 1).

Hope this helps!

Answer:

Critical value: b. 2.33

As the test statistic z=7.17 is greater than the critical value, it falls in the rejection region.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of women who think sexual harassment is a major problem in the American workplace is significantly higher than the proportion of men.

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the proportion of women who think sexual harassment is a major problem in the American workplace is significantly higher than the proportion of men.

Then, the null and alternative hypothesis are:

[tex]H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2> 0[/tex]

The significance level is 0.01.

The sample 1 (women), of size n1=150 has a proportion of p1=0.62.

The sample 2 (men), of size n2=200 has a proportion of p2=0.24.

The difference between proportions is (p1-p2)=0.38.

[tex]p_d=p_1-p_2=0.62-0.24=0.38[/tex]

The pooled proportion, needed to calculate the standard error, is:

[tex]p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{93+48}{150+200}=\dfrac{141}{350}=0.403[/tex]

The estimated standard error of the difference between means is computed using the formula:

[tex]s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.403*0.597}{150}+\dfrac{0.403*0.597}{200}}\\\\\\s_{p1-p2}=\sqrt{0.001604+0.001203}=\sqrt{0.002807}=0.053[/tex]

Then, we can calculate the z-statistic as:

[tex]z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.38-0}{0.053}=\dfrac{0.38}{0.053}=7.17[/tex]

The critical value for a right-tailed test with a signficance level of 0.01 is zc=2.33 (see picture attached).

As the test statistic z=7.17 is greater than the critical value, it falls in the rejection region.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of women who think sexual harassment is a major problem in the American workplace is significantly higher than the proportion of men.

Answer:

[tex]\frac{2}{12}[/tex] simplified to [tex]\frac{1}{6}[/tex]

Step-by-step explanation:

4 = [tex]\frac{1}{12}[/tex]

2 = [tex]\frac{1}{12}[/tex]

[tex]\frac{1}{12}[/tex] + [tex]\frac{1}{12}[/tex] = [tex]\frac{2}{12}[/tex] ÷ 2 = [tex]\frac{1}{6}[/tex]

hope it helps uh.......

Answer:

A=0.25

B=0.35

C=1.05

Step-by-step explanation:

1. A+B=0.6

2. A+C=1.3

3. C=3B

2 subtract 1:

3 substituted:

Options

Answer:

[tex](E)0.7+0.4-0.2=0.9[/tex]

Step-by-step explanation:

In probability theory

[tex]P$(A or B)=P(A)+P(B)$-$P(A and B)[/tex]

Let the event that the show had animals in it = A

P(A)=0.7

Let the event that the show aired more than 10 times =B

P(B)=0.4

P(A and B)= 0.2

[tex]P$(A or B)$=0.7+0.4-0.2=0.9[/tex]

Therefore, the equation which  shows the correct use of the addition rule to determine the probability that a randomly selected show had animals in it or aired more than 10 times is:

[tex]0.7+0.4-0.2=0.9[/tex]

The correct option is E.

Answer:

x=-52

Step-by-step explanation:

1/3x=-17 1/3

x=-52

Answer:

IT'S NOT -15 FOR SUREEE

Step-by-step explanation:

I Believe it's 15

Answer:

z = 1.476

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.86}{2} = 0.07[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.07 = 0.93[/tex], so [tex]z = 1.476[/tex]

The answer is z = 1.476

The pizza is cut into 6 slices so each slice would be 1/6 of the pizza.

He at 1/2 of a slice:

1/6 x 1/2 = 1/12 of the pizza

Answer:

1. JG (Jim gets first prize, George gets second prize)

2. JH (Jim gets first prize, Helen gets second prize)

3. JM (Jim gets first prize, Maggie gets second prize)

4. GH (George gets first prize, Helen gets second prize)

5. GJ (George gets first prize, Jim gets second prize)

6. GM (George gets first prize, Maggie gets second prize)

7. MJ (Maggie gets first prize, Jim gets second prize)

8. MG (Maggie gets first prize, George gets second prize)

9. MH (Maggie gets first prize, Helen gets second prize)

Step-by-step explanation:

The question asks for the list of outcomes in the event "Not A". We are looking for the reverse or negative of Event A.

The above given list is the list of outcomes in the event where Helen DOES NOT get first prize.

Answer:

[tex]\frac{7\sqrt{x} }{x}[/tex]

Step-by-step explanation:

To simplify 7/√x, we need to rationalize:

[tex]\frac{7}{\sqrt{x} } (\frac{\sqrt{x} }{\sqrt{x} } )[/tex]

When we multiply the 2, we should get our answer:

[tex]\frac{7\sqrt{x} }{x}[/tex]

Answer:

[tex]\frac{7\sqrt{x} }{x}[/tex]

Step-by-step explanation:

[tex]\frac{7}{\sqrt{x} } \\\\\frac{7}{\sqrt{x} } * \frac{\sqrt{x} }{\sqrt{x} } \\\\\frac{7\sqrt{x} }{\sqrt{x\sqrt{x} } } \\[/tex]

[tex]\frac{7\sqrt{x} }{x}[/tex]

Hope this helps! :)

Answer:

16 km due west

Step-by-step explanation:

The bearing of the school p from school q is 16 km due west.

To find the bearing of school q from school p, we have to find the direction that the school q is with respect to school p.

Since p is directly west of q, then it implies that q must be directly east of p.

We now know the direction.

Since the distance from q to p is exactly the same as the distance from p to q, then, the distance from p to q is 16 km.

Hence, the bearing of q from p is 16 km due west.

Answer:

The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is (-11.937, -6.863).

Step-by-step explanation:

We have to calculate a 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams.

The sample 1 (Fiberglass), of size n1=26 has a mean of 33.4 and a standard deviation of 2.2.

The sample 2 (Carbon), of size n2=26 has a mean of 42.8 and a standard deviation of 4.3.

The difference between sample means is Md=-9.4.

[tex]M_d=M_1-M_2=33.4-42.8=-9.4[/tex]

The estimated standard error of the difference between means is computed using the formula:

[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{2.2^2}{26}+\dfrac{4.3^2}{26}}\\\\\\s_{M_d}=\sqrt{0.186+0.711}=\sqrt{0.897}=0.9473[/tex]

The critical t-value for a 99% confidence interval is t=2.678.

The margin of error (MOE) can be calculated as:

[tex]MOE=t\cdot s_{M_d}=2.678 \cdot 0.9473=2.537[/tex]

Then, the lower and upper bounds of the confidence interval are:

[tex]LL=M_d-t \cdot s_{M_d} = -9.4-2.537=-11.937\\\\UL=M_d+t \cdot s_{M_d} = -9.4+2.537=-6.863[/tex]

The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is (-11.937, -6.863).

In this way, we can calculate the individual duration of each one and the duration time, knowing that the sample means:

The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is -11.937 and -6.863.

We have to calculate a 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams. The sample 1 (Fiberglass), of size n1=26 has a mean of 33.4 and a standard deviation of 2.2. The sample 2 (Carbon), of size n2=26 has a mean of 42.8 and a standard deviation of 4.3. The difference between sample means is Md=-9.4.

[tex]Sm_d= \sqrt{\frac{\sigma^2_1}{n_1} +\frac{\sigma^2_2}{n_2}} = \sqrt{(0.186)+(0.711) }= 0.9473[/tex]

The critical t-value for a 99% confidednce interval is t=2.678. The margin of error (MOE) can be calculated as:

[tex]MOE=t*8M_d = (2.678)(0.9473)= 2.537[/tex]

Then, the lower and upper bounds of the confidence interval are:

[tex]LL= M_d-t*SM_d = -9.4-2.537= -11.937\\UL= M_d+t*SM_d= -9.4+2.537= -6.863[/tex]

The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is (-11.937, -6.863).

See more about statistics at brainly.com/question/2289255

Answer:

A 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].

Step-by-step explanation:

We are given that statistics students at the Air Force Academy (no kidding) purchased some randomly selected bags of cookies and counted the chocolate chips.

Some of their data are given below; 1219, 1214, 1087, 1200, 1419, 1121, 1325, 1345, 1244, 1258, 1356, 1132, 1191, 1270, 1295, 1135.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                          P.Q.  =  [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean number of chocolate chips = [tex]\frac{\sum X}{n}[/tex] = 1238.2

            s = sample standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }[/tex]  = 94.3

            n = sample of car drivers = 16

            [tex]\mu[/tex] = population mean number of chips in a bag

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;

P(-2.131 < [tex]t_1_5[/tex] < 2.131) = 0.95  {As the critical value of t at 15 degrees of

                                             freedom are -2.131 & 2.131 with P = 2.5%}  

P(-2.131 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.131) = 0.95

P( [tex]-2.131 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.131 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

P( [tex]\bar X-2.131 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.131 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.131 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.131 \times {\frac{s}{\sqrt{n} } }[/tex] ]

                                   = [ [tex]1238.2-2.131 \times {\frac{94.3}{\sqrt{16} } }[/tex] , [tex]1238.2+2.131 \times {\frac{94.3}{\sqrt{16} } }[/tex] ]

                                   = [1187.96, 1288.44]

Therefore, a 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].

━━━━━━━☆☆━━━━━━━

▹ Answer

Area = 9

▹ Step-by-Step Explanation

A = b * h ÷ 2

A = 9 * 2 ÷ 2

A = 9

Hope this helps!

- CloutAnswers ❁

Brainliest is greatly appreciated!

━━━━━━━☆☆━━━━━━━

Answer:

48.3 cm²

Step-by-step explanation:

Let A be the area of the yellow region

A= the area of the square - the area of the quarter square

A= 15²-(15²*π)/4= 48.28≈ 48.3 cm²

The greatest number is 6.23 times 10 superscript 12.

The number is written in the form [tex]a \times 10^b[/tex] where we have [tex]1 \leq a < 10[/tex]

The number b shows the order, which is the most important figure for which scientific notation is used. It tells us how much order large or small a value is in powers of 10. We can for a time, ignore the value of 'a' for two comparable quantities and only compare their orders(this type of comparison is useful when difference is too big, like size of human to size of a star etc sort of comparisons).

We are given that the number so;

A.6.23 x 10^12 is equivalent to rolling the decimal 12 times to the right.

B.6.23 x 10^8 is equivalent to rolling the decimal 8 times to the right.

C.6.23 x 10^-6 is equivalent to rolling the decimal 6 times to the left.

D.6.23 x 10^3 is equivalent to rolling the decimal 3 times to the right.

This shows the 10 has been multiplied by itself thrice.

Learn more about scientific notation here:

https://brainly.com/question/3112062

#SPJ6

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: time it takes a customer from Andrew's computer store to pay his invoices.

You have the information of a sample of n= 17 customers

19, 15, 43, 39, 35, 31, 27, 34, 34, 30, 30, 26, 26, 26, 21, 21, 17

To determine the class width of the intervals for the divide the difference between the ending and initial class boundaries by the number of intervals that you want to determine:

Class width: (49.5-14.5)/5= 7

Then, starting from the initial class boundary, you have to add the class width to determine the next boundary, and so on until the ending class boundary:

Initial class boundary: 14.5

14.5 + 5.6= 20.1

1st interval: [14.5; 21.5]

and so on:

[21.5; 28.5]

[28.5; 35.5]

[35.5; 42.5]

[42.5; 49.5]

Once you determined all class intervals, you have to order the values of the data set from least to greatest and then count how many observations correspond to each interval and arrange it in a frequency table.

15, 17, 19, 21, 21, 26, 26, 26, 27, 30, 30, 31, 34, 34, 35, 39, 43

[14.5; 21.5] ⇒ 5

[21.5; 28.5] ⇒ 4

[28.5; 35.5] ⇒ 6

[35.5; 42.5] ⇒ 1

[42.5; 49.5] ⇒ 1

Once you have the data set organized in the table, you can proceed to draw the histogram.

(See attachment)

I hope this helps!

Answer:

d = 3 in

Step-by-step explanation:

Since we are trying to find the diameter, and the diameter is given to us as 3 in, our diameter is 3 in.

Answer:

In terms of construction, copying a line segment and an angle requires a fixed compass width as a basic tool

Step-by-step explanation:

The basic similarity is in both constructions, or copies is that we are going to use the same compass width in each case as the basic tool to copy a line segment or an angle.

hope this helpes

be sure to give brainliest

Answer:

An angle is form by two rays and the two line segment share a common points and we utilize a straightedge for drawing the comparative figure on paper.

At that point, utilize the straightedge and the compass used to copy this type of figure precisely. To duplicate the given figure, we should copy line as well as angle.

The line of segment are basically formed by adjusting the compass and makes it equal to the line segment length and then copy each point in the figure.

Answer:

true

Step-by-step explanation:

doesn't over lap each other

Answer:

c. 0.10

Step-by-step explanation:

Hello!

To accept a batch of components, the proportion of defective components is at most 0.10.

X: Number of defective components in a sample of 10.

This variable has a binomial distribution with parameters n=10 and p= 0.10 (for this binomial experiment, the "success" is finding a defective component)

The distributor will accept the batch if at most two components are defective, symbolically:

P(X≤2)

Using the tables for the binomial distribution you can find the accumulated probability for a sample of n=10 with probability of success of p= 0.10 and number of successes x= 2

P(X≤2)= 0.9298

I hope this helps!