what is 3/5 of 1800​

Answer:

x = 5, AB=5, BC = 15

Step-by-step explanation:

AC = AB + BC (Segment Addition)

AC= 20, AB =x Bc = 3x,

20= x+3x 20=4x

x=5

AB=x, AB =5

BC=3x BC= 15

The segment addition postulate states gives the value of x as 5, given

that the sum of x and 3·x is 20.

Responses:

From the given diagram, we have;

[tex]\overline{AB}[/tex] = x

[tex]\overline{BC}[/tex] = 3·x

According to segment addition postulate we have;

[tex]\overline{AB}[/tex] + [tex]\overline{BC}[/tex] = [tex]\overline{AC}[/tex] = 20 inches

Which gives;

x + 3·x = 20

Therefore;

4·x = 20

[tex]x = \dfrac{20}{4} = 5[/tex]

[tex]\mathbf{\overline{BC}}[/tex] = 3·x  

[tex]\mathbf{\overline{BC}}[/tex] = 3 × 5 = 15

Learn more about segment addition postulate here:

https://brainly.com/question/1397818

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.

The horizontal asymptote of the function f(x) = (x-2)/(x-3)² is at y = 0 which is the x-axis.

An asymptote is a line that is approached by a curve but never touches it. In other words, an asymptote is a line where the graph of a function converges.

Because a horizontal asymptote is a horizontal line, its equation is of the form y = k. The horizontal asymptote of a rational function is at y = 0, which is the x-axis if the degree of the numerator is smaller than the degree of the denominator.

Here, the function is f(x) = (x-2)/(x-3)². Here the degree of the numerator of this rational function is 1 and the degree of the denominator is 2. Since 1<2, the horizontal asymptote is at y = 0 which is the x-axis.

The horizontal asymptote of the function f(x) = (x-2)/(x-3)² is at y = 0 which is the x-axis.

Learn more about horizontal asymptotes here -

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Answer:

c

Step-by-step explanation:

Step-by-step explanation:

[tex]x = \frac{ - b \frac{ + }{ - } \sqrt{ {b}^{2} - 4ac } }{2a} [/tex]

The quadratic formula is honestly the most straightforward way of solving here.

Your other options are completing the square (which is the same thing as the quadratic formula but it's good to know that method if you have to take Integral Calculus at some point) or maybe factoring by grouping if it's appropriate. But the quadratic formula will work for you in all three equations:

1) a=1, b=0, c=8

This reduces pretty quickly into x=8i,-8i due to the negative under the radical. (Actually we didn't even really need the formula here.)

2) a=-9, b=4, c=-10

This reduces into x=(-4+i√(344))/-18, (-4-i√(344))/-18 and doesn't go any further because 344 isn't a perfect square.

3) a=1, b=8, c=17

This reduces to x=(-4+i), (-4-i)

So those are the answers for each.

Please Refer to the screenshot. Hope this helps!

Answer:

The significance level for this case would be [tex]\alpha=1-0.95=0.05[/tex] and the critical value for this case would be:

[tex] z_{\alpha/2}=1.96[/tex]

The margin of error is given by:

[tex] ME = z_{\alpha/2} \sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

And replacing we got:

[tex] ME = 1.96 \sqrt{\frac{0.23*(1-0.23)}{146}} =0.0683[/tex]

And the margin of error for this case would be [tex] ME = 0.07[/tex]

Step-by-step explanation:

For this case we have the following dataset given:

[tex] n= 146[/tex] represent the sample size

[tex] \hat p =0.23[/tex] represent the estimated proportion of interest

[tex] Conf=0.95[/tex] represent the confidence level

The significance level for this case would be [tex]\alpha=1-0.95=0.05[/tex] and the critical value for this case would be:

[tex] z_{\alpha/2}=1.96[/tex]

The margin of error is given by:

[tex] ME = z_{\alpha/2} \sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

And replacing we got:

[tex] ME = 1.96 \sqrt{\frac{0.23*(1-0.23)}{146}} =0.0683[/tex]

And the margin of error for this case would be [tex] ME = 0.07[/tex]

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

Answer:

The 99​% confidence interval estimate of the percentage of girls born is between 85.46% and 94.54%.

Usually, babies are equally as likely to be boys or girls. Here, it is desired to increase the probability of conceiving a girl, which is achieved, considering the lower bound of the confidence interval is considerably above 50%.

Step-by-step explanation:

Confidence interval for the proportion:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 290, \pi = \frac{261}{290} = 0.9[/tex]

99% confidence level

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.9 - 2.575\sqrt{\frac{0.9*0.1}{290}} = 0.8546[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.9 + 2.575\sqrt{\frac{0.9*0.1}{290}} = 0.9454[/tex]

Percentage:

Proportion multplied by 100.

0.8546*100 = 85.46%

0.9454*100 = 94.54%

The 99​% confidence interval estimate of the percentage of girls born is between 85.46% and 94.54%.

Based on the​ result, does the method appear to be​ effective?

Usually, babies are equally as likely to be boys or girls. Here, it is desired to increase the probability of conceiving a girl, which is achieved, considering the lower bound of the confidence interval is considerably above 50%.

The slope greater than one would be the last image, because for every step in x, you get more than one y step.

The slope between 1 and 0 would be the second image

And the slope less than 0 would be the third image

━━━━━━━☆☆━━━━━━━

▹ Answer

15.75

▹ Step-by-Step Explanation

3 ÷ 4 = .75

15 + .75 = 15.75

Hope this helps!

- CloutAnswers ❁

Brainliest is greatly appreciated!

━━━━━━━☆☆━━━━━━━

Answer:

C.

Step-by-step explanation:

When two lines are parallel, their slopes are the same.

Since the slope of line l is 2/7, the slope of its parallel line m must also be 2/7.

The answer is C.

Answer:

C. 2/7

Step-by-step explanation:

Parallel lines are lines that have the same slopes.

We know that line l is parallel to line m.

Therefore, the slope of line l is equal to the slope of line m.

[tex]m_{l} =m_{m}[/tex]

We know that line l has a slope of 2/7.

[tex]\frac{2}{7} =m_{m}[/tex]

So, line m also has a slope of 2/7. The answer is C. 2/7

Answer:

but I can do it if you want but I don't you too too y u your help and you have time can you

Step-by-step explanation:

guy who was it that you are not going to be able to make it to the meeting tonight but I can tomorrow if you have time can you come to my house

Answer:

z = -17.67 + i3.43

Step-by-step explanation:

Let us apply the formula z = r(cos Ф + i sin Ф), given 18(cos169° + isin169°) -

z = 18( cos169 + isin169 ),

z = r(cos Ф + i sin Ф)

Now we can solve this question in the form z = a + bi, in this case where a = 18 cos169, and b = 18 sin169. This is as a = r cos Ф and b = r sin Ф -

sin169 is positive, while cos169 is negative, thus -

a = -17.6692893021...,

b = 3.43456191678...

Rectangular Form, z = -17.67 + i3.43

Hope that helps!

Answer:

D

Step-by-step explanation:

Parallel lines are those that have the same slope, or coefficient of x.

Here, let's calculate the slope of the given line. Slope is the difference in the y-coordinates divided by the difference in the x-coordinates, so given the two coordinates (12, 6) and (0, -4):

slope = m = (-4 - 6) / (0 - 12) = -10 / (-12) = 10/12 = 5/6

So the slope is 5/6. That means the equation we want should also have a slope of 5/6. Already, we can eliminate A, B, and C, leaving D as our answer. But, let's check.

The equation of a line can be written as [tex]y-y_1=m(x-x_1)[/tex], where m is the slope and [tex](x_1,y_1)[/tex] is the coordinates of a given point.

Here, our slope is 5/6 and our given point is (12, -2). So plug these in:

[tex]y-y_1=m(x-x_1)[/tex]

[tex]y-(-2)=(5/6)(x-12)[/tex]

[tex]y+2=\frac{5}{6} x-10[/tex]

[tex]y=\frac{5}{6} x-12[/tex]

This matches D, so we know that it's the correct answer.

~ an aesthetics lover

The answer is D I just took the test

Answer:

1. Point estimate M (sample mean): 83.33

2. The lower bound is $73.36 and the upper bound is $93.30. One can be______% confident that the mean travel tax for all cities is between these values.

3. A. The researcher could decrease the level of confidence.

Step-by-step explanation:

A point esimate for the population mean travel tax can be done with the sample mean.

We can calculate the sample mean as:

[tex]M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{8}(67.85+78.62+70.28+84.03+79.28+87.72+101.54+97.28)\\\\\\M=\dfrac{666.6}{8}\\\\\\M=83.33\\\\\\[/tex]

2. We have to calculate a 95% confidence interval for the mean.

The sample mean is M=83.33.

The sample size is N=8.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

We calculate the sample standard deviation as:

[tex]s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{7}((67.85-83.33)^2+(78.62-83.33)^2+(70.28-83.33)^2+. . . +(97.28-83.33)^2)}\\\\\\s=\sqrt{\dfrac{994.49}{7}}\\\\\\s=\sqrt{142.07}=11.92\\\\\\[/tex]

The standard error is:

[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{11.92}{\sqrt{8}}=\dfrac{11.92}{2.828}=4.214[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=8-1=7[/tex]

The t-value for a 95% confidence interval and 7 degrees of freedom is t=2.36.

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

[tex]MOE=t\cdot s_M=2.36 \cdot 4.214=9.97[/tex]

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

[tex]LL=M-t \cdot s_M = 83.33-9.97=73.36\\\\UL=M+t \cdot s_M = 83.33+9.97=93.30[/tex]

The 95% confidence interval for the mean travel tax is (73.36, 93.30).

We can be 95% confident that the true mean travel tax is within this interval.

3.. If we have no access to additional data, we can not decrease the standard deviation or increase the sample size.

The only way to have a narrower confidence interval is decreasing its level of confidence. With the same sample information, the lower the confidence, the narrower is the interval.

Answer:

Minimum 8 at x=0, Maximum value: 24 at x=4

Step-by-step explanation:

Retrieving data from the original question:

[tex]f(x)=x^{2}+8\:over\:[-1,4][/tex]

1) Calculating the first derivative

[tex]f'(x)=2x[/tex]

2) Now, let's work to find the critical points

Set this

[tex]2x=0\\x=0[/tex]    

0, belongs to the interval. Plug it in the original function

[tex]f(0)=(0)^2+8\\f(0)=8[/tex]

3)  Making a table x, f(x) then compare

x|  f(x)

-1 | f(-1)=9  

0 | f(0)=8   Minimum

4 | f(4)=24 Maximum

4) The absolute maximum value is 24 at x=4 and the absolute minimum value is 8 at x=0.    

Answer:

x = 5 and y = -4

Step-by-step explanation:

y = - 3x + 11 ______(1)

5x + y = 21______(2)

Substitute (1) into (2).

5x + (-3x + 11) = 21

5x - 3x + 11 = 21

2x = 21-11

2x = 10

x = 10/2

x = 5

Now substitute x = 5 into (1).

y = -3(5) + 11

y = -15 + 11

y = -4

Hence, x = 5 and y = -4

Answer:

probability of chosing a student that has a cat and a dog is 9/25

Step-by-step explanation:

And yes the Venn diagram is wrong because you forgot to subtract 9 from 15 and 16

This makes it

[ 3 ( 6 ( 9 ) 7 ) ]

3 + 6 + 9 + 7 = 25

Answer:

Fail to reject the null hypothesis because the p-value =0.10 is more than the significance level α=0.05.

Step-by-step explanation:

The significant level for this case study is 5% - 0.05. If the results gotten is less than the significance level, we reject the null hypothesis, but if greater, we fail to reject the null hypothesis.

In this case study, the p-value is 0.10 which is a lot higher than 0.05, thus we will fail to reject the null hypothesis because the p-value =0.10 is more than the significance level α=0.05.

Answer:

3

Step-by-step explanation:

minus the variable, 4-1 is 3.

Answer:

[tex] P(X\geq 20)= P(X=20)+P(X=21)+P(X=22)+P(X=23)+P(X=24)+P(X=25)[/tex]

And replacing using the mass function we got:

[tex]P(X=20)=(25C20)(0.85)^{20} (1-0.85)^{25-20}=0.156[/tex]  

[tex]P(X=21)=(25C21)(0.85)^{21} (1-0.85)^{25-21}=0.211[/tex]  

[tex]P(X=22)=(25C22)(0.85)^{22} (1-0.85)^{25-22}=0.217[/tex]  

[tex]P(X=23)=(25C23)(0.85)^{23} (1-0.85)^{25-23}=0.161[/tex]  

[tex]P(X=24)=(25C24)(0.85)^{24} (1-0.85)^{25-24}=0.0759[/tex]  

[tex]P(X=25)=(25C25)(0.85)^{25} (1-0.85)^{25-25}=0.0172[/tex]  

And adding the values we got:

[tex] P(X\geq 20) = 0.8381[/tex]

Step-by-step explanation:

Let X the random variable of interest, on this case we now that:  

[tex]X \sim Binom(n=25, p=0.85)[/tex]  

The probability mass function for the Binomial distribution is given as:  

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]  

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

We want to find the following probability:

[tex] P(X\geq 20)= P(X=20)+P(X=21)+P(X=22)+P(X=23)+P(X=24)+P(X=25)[/tex]

And replacing using the mass function we got:

[tex]P(X=20)=(25C20)(0.85)^{20} (1-0.85)^{25-20}=0.156[/tex]  

[tex]P(X=21)=(25C21)(0.85)^{21} (1-0.85)^{25-21}=0.211[/tex]  

[tex]P(X=22)=(25C22)(0.85)^{22} (1-0.85)^{25-22}=0.217[/tex]  

[tex]P(X=23)=(25C23)(0.85)^{23} (1-0.85)^{25-23}=0.161[/tex]  

[tex]P(X=24)=(25C24)(0.85)^{24} (1-0.85)^{25-24}=0.0759[/tex]  

[tex]P(X=25)=(25C25)(0.85)^{25} (1-0.85)^{25-25}=0.0172[/tex]  

And adding the values we got:

[tex] P(X\geq 20) = 0.8381[/tex]

Answer: 52.8

Step-by-step explanation: it’s on khan ,

Answer:

The general term is

Sn = -(-2)ⁿ.3¹⁻ⁿ

step by step Explanation:

we were told to find a general term of the above sequence, what should come to mind is that the terms will follow an order....

CHECK THE ATTACHMENT FOR DETAILED EXPLANATION

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:

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:

[tex]P(X<0.4375)=P(\frac{X-\mu}{\sigma}<\frac{0.4375-\mu}{\sigma})=P(Z<\frac{0.4375-0.635}{0.082})=P(z<-2.41)[/tex]

And we can find this probability using the z table and we got:

[tex]P(z<-2.41)=0.0080[/tex]

Step-by-step explanation:

Let X the random variable that represent the thickness of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(0.635,0.082)[/tex]  

Where [tex]\mu=0.635[/tex] and [tex]\sigma=0.032[/tex]

We are interested on this probability

[tex]P(X<0.4375)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X<0.4375)=P(\frac{X-\mu}{\sigma}<\frac{0.4375-\mu}{\sigma})=P(Z<\frac{0.4375-0.635}{0.082})=P(z<-2.41)[/tex]

And we can find this probability using the z table and we got:

[tex]P(z<-2.41)=0.0080[/tex]

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 = 4 /b

or ab = 4

Step-by-step explanation:

An inverse relation is given by

a = k/b where k is the constant

Rewriting

ab = k

12 * 1/3 = k

4 = k

a = 4 /b

or ab = 4

Answer:

B

Step-by-step explanation:

area of top on triangle=1/2×4×3=6m²

area of four top triangles=6×4=24 m²

area of bottom square=4×4=16 m²

area of four side rectangles=4×(4×5)=80 m²

Total area= 24+16+80=120m²

Answer:

p > α

0.7038 > 0.05

Also since F < F critical

0.475 < 3.238

We failed to reject H₀

We do not have significant evidence at the given significance level to show that there is a difference among the four chemical agents on the strength of a particular type of  cloth.

Step-by-step explanation:

We are given that a chemist wishes to test the effect of four chemical agents on the strength of a particular type of  cloth.

Since we are given data for four independent chemical agents to determine the effect on the strength of a particular type of cloth, therefore, a one-way analysis of variance may be used for the given problem.

ANOVA:

The one-way analysis of variance (ANOVA) may be used to find out whether there is any significant difference between the means of two or more independent categories of data.

Set up hypotheses:

Null hypotheses = H₀: μ₁ = μ₂ = μ₃ = μ₄

Alternate hypotheses = H₁: μ₁ ≠ μ₂ ≠ μ₃ ≠ μ₄

Set up decision rule:

We Reject H₀ if p ≤ α

OR

We Reject H₀ if F > F critical

ANOVA in Excel:

Step 1:

In the data tab, select data analysis

Step 2:

Select "Anova single factor" from the analysis tools

Step 3:

Select the destination of input data in the "input range"

Step 4:

Select "rows" for the option "Group By"

Step 5:

Tick the option "labels in first row"

Step 6:

Set alpha = 0.05

Step 7:

Select the destination of output data in the "output options"

Conclusion:

Please refer to the attached results.

The p-value is found to be

p = 0.7038

The F value is found to be

F = 0.475

The F critical value is found to be

F critical = 3.238

Since p > α

0.7038 > 0.05

We failed to reject H₀

Also since F < F critical

0.475 < 3.238

We failed to reject H₀

We do not have significant evidence at the given significance level to show that there is a difference among the four chemical agents on the strength of a particular type of cloth.