Which is the best estimate of the distance between the point (–25, 10) and the origin? 8 units 15 units 23 units 27 units

Using the margin of error, it is found that he must survey 601 adults in order to be 95% confidence that his estimate is within 4 percentage points of the true population percentage.

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]\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 z-score that has a p-value of [tex]\frac{1+\alpha}{2}[/tex].

The margin of error is:

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

95% confident, hence:

[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].

We have no estimate, hence, [tex]\pi = 0.5[/tex] is used.

Within 4%, hence, it is needed to find n for which M = 0.04.

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

[tex]0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}[/tex]

[tex]0.04\sqrt{n} = 1.96(0.5)[/tex]

[tex]\sqrt{n} = \frac{1.96(0.5)}{0.04}[/tex]

[tex](\sqrt{n})^2 = \left(\frac{1.96(0.5)}{0.04}\right)^2[/tex]

[tex]n = 600.25[/tex]

Rounding up, 601 adults must be sampled.

A similar problem is given at https://brainly.com/question/15133177

Answer:

B a vertical translation 50units right

Answer:

4x-6

Step-by-step explanation:

4x-6

4 is for number of doughnuts in each package, 6 is for the taken doughnuts

Answer:

distributive property

Step-by-step explanation:

I ASSUME that you meant to lead the equation with a negative sign.

- 10x + 11x = (-10 + 11)x

Probabilities are used to determine the chances of selecting a kind of donut from the box.

The probability that Warren eats a chocolate donut, and then a custard filled donut is 0.068

The given parameters are:

[tex]\mathbf{Bars = 6}[/tex]

[tex]\mathbf{Chocolate = 3}[/tex]

[tex]\mathbf{Custard= 3}[/tex]

The total number of donuts in the box is:

[tex]\mathbf{Total= 6 + 3 + 3}[/tex]

[tex]\mathbf{Total= 12}[/tex]

The probability of eating a chocolate donut, and then a custard filled donut is calculated using:

[tex]\mathbf{Pr = \frac{Chocolate}{Total}\times \frac{Custard}{Total-1}}[/tex]

So, we have:

[tex]\mathbf{Pr = \frac{3}{12}\times \frac{3}{12-1}}[/tex]

Simplify

[tex]\mathbf{Pr = \frac{3}{12}\times \frac{3}{11}}[/tex]

Multiply

[tex]\mathbf{Pr = \frac{9}{132}}[/tex]

Divide

[tex]\mathbf{Pr = 0.068}[/tex]

Hence, the probability that Warren eats a chocolate donut, and then a custard filled donut is approximately 0.068

Read more about probabilities at:

https://brainly.com/question/9000575

Answer:

in 10 hours

Step-by-step explanation:

Answer: 4 hours

Step-by-step explanation: since the plant is already 10 centimeters, and it grows five centimeters every two hours, it only needs four hours to grow another 10 centimeters for it to be 20 centimeters.

Answer:

12 Yoga classes

Step-by-step explanation:

50 + 4x = 98

Subtract 50 on each side

4x = 48

Divide by 4 on each side

x= 12

You can attend 12 yoga classes in a year with a $98 budget.

The height of the tower is 452.66 ft

The situation will form 2 right angle triangle. One above the line of sight

and one below the line of sight.

Therefore,

Using trigonometric ratio,

tan ∅ = opposite / adjacent side

tan 30° = h(top of the tower) / 425

h(top of the tower) = tan 30 × 425 = 245.373864406  

tan 26° = h(bottom of the tower) / 425

h(bottom of the tower) = 425 × tan 26° = 207.28635014

Therefore,

Height of tower = h(top of the tower) + h(bottom of the tower)

Height of tower = 245.37 + 207.29 = 452.66 ft

learn more: https://brainly.com/question/12855949?referrer=searchResults

Step-by-step explanation:

it will be around $90.86

Answer:

No. of adults who answered 'yes'

= 20% × 1073

= 214.6

Total amount of money paid

= (18%×$77)+$77

= $13.86+$77

= $90.86

33 pupils - 17 girls = 16 total boys in the class

16 boys - 14 boys not on the team = 2 boys on the team.

Answer: 2

here are the correct answers :)

There is a 0.9968 probability that a randomly selected 50-year-old female lives through the year (based on data from the U.S. Department of Health and Human Services).

-------------------

A Fidelity life insurance company charges $226 for insuring that the female will live through the year. If she does not survive the year, the policy pays out $50,000 as a death benefit.

From the perspective of the 50-year-old female, what are the values corresponding to the two events of surviving the year and not surviving?

----

Ans: -226 ; 50,000-226 = 49774

-------------------------

If a 50-year-old female purchases the policy, what is her expected value?

WORK TRIED:

In the event she lives, the value is -$226. In the event she dies, the value is $49,774.

----

E(x) = 0.9968*(-226) + 0.0032(49774) = -$66

==================================================

Cheers,

ROR

4

the other really do not make sense

Answer:

2

Step-by-step explanation:

Answer:

Step-by-step explanation:

Let x represent the amount invested at 4% (the higher rate). Then the amount invested at 3% is (6000-x), and the total interest earned is ...

  0.04x + 0.03(6000 -x) = 216

  0.01x +180 = 216 . . . . simpilfy

  0.01x = 36 . . . . . . . . subtract 180

  x = 3600 . . . . . . . . multiply by 100

  6000-x = 2400

Katie invested $3600 at 4% and $2400 at 3%.

Answer:

3/1

Step-by-step explanation:

rise/run

9514 1404 393

Answer:

  3

Step-by-step explanation:

The line goes through the origin (0, 0) and point (1, 3), which is 3 units up and 1 to the right of the origin.

  slope = rise/run = 3/1 = 3

The slope of the line is 3.

Answer:

9 units²

Step-by-step explanation:

Parallelogram area = base × height

Base = 3 units

Height = 3 units

Area = 3 × 3

Area = 9

The area of the parallelogram is 9 units².

Here, we have,

We know that,

Parallelogram area = base × height

here, from the given figure we get,

Base = 3 units

Height = 3 units

so, we get,

Parallelogram area = base × height

Area = 3 × 3

Area = 9

Hence, The area of the parallelogram is 9 units².

To learn more on area of parallelogram click:

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#SPJ6

Answer:

You need a calculator to solve, but here: 7.7

Step-by-step explanation:

u missed a few paranthesis. I solved this instead: (3.71 + (-5.25))(4+(-9))

first subtract 3.71 by 5.25 to get -1.54 then subtract 4 by 9 to get -5. Then you multiply them to get 7.7.

Answer:

y = -1/2

Step-by-step explanation:

   (3x + 4y = 4)

+

   (2x - 4y = 6)

______________

   5x = 10

   x = 2

plug in x for any of the two equations ....

2(2) - 4y = 6

4 - 4y = 6

-4y = 2

y = -1/2

Answer:

[tex]w-0.15 = 0.25[/tex]

Step-by-step explanation:

Answer:

i dont know

Step-by-step explanation:

i really dont know

Answer:

$11,480

Step-by-step explanation:

Answer:

There is an outlier at 0, and a gap at 1,2, and 3. The peak is at 5, and the plot is more or less skewed left.

Step-by-step explanation:

Let [tex]x = \arcsin(y)[/tex], so that

[tex]\sin(x) = y[/tex]

[tex]\tan(x)=\dfrac y{\sqrt{1-y^2}}[/tex]

[tex]dx = \dfrac{dy}{\sqrt{1-y^2}}[/tex]

Then the integral transforms to

[tex]\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}[/tex]

[tex]\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy[/tex]

Integrate by parts, taking

[tex]u = \ln(y) \implies du = \dfrac{dy}y[/tex]

[tex]dv = \dfrac{y}{1-y^2} \, dy \implies v = -\dfrac12 \ln|1-y^2|[/tex]

For 0 < y < 1, we have |1 - y²| = 1 - y², so

[tex]\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy[/tex]

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

[tex]\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy[/tex]

Recall the Taylor series for ln(1 + y),

[tex]\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n y^n[/tex]

Replacing y with -y² gives the Taylor series

[tex]\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}[/tex]

and replacing ln(1 - y²) in the integral with its series representation gives

[tex]\displaystyle -\frac12 \int_0^1 \frac1y \sum_{n=1}^\infty \frac{y^{2n}}n \, dy = -\frac12 \int_0^1 \sum_{n=1}^\infty \frac{y^{2n-1}}n \, dy[/tex]

Interchanging the integral and sum (see Fubini's theorem) gives

[tex]\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy[/tex]

Compute the integral:

[tex]\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{y^{2n}}{2n^2} \bigg|_0^1 = -\frac14 \sum_{n=1}^\infty \frac1{n^2}[/tex]

and we recognize the famous sum (see Basel's problem),

[tex]\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6[/tex]

So, the value of our integral is

[tex]\displaystyle \int_0^{\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \boxed{-\frac{\pi^2}{24}}[/tex]

[tex]\displaystyle \int \limits_{0}^{ \frac{ \pi}{2} } \tt\tan (x) \ln ( \sin (x))\\ \\\displaystyle \sf{ \implies \: I = \int^{ \frac{\pi}{2} }_{0} \: \ln \left( sin \left( \dfrac{\pi}{2} - x \right) \right) \: dx } \\ \\\displaystyle \sf{ \implies \: I = \int^{ \frac{\pi}{2} }_{0} \: \ln \left( cos(x) \right) \: dx } \\ \\\displaystyle \sf{ \implies \:2I =\int^{ \frac{\pi}{2} }_{0} \:\ln \left( sin(x) \right) \: dx + \int^{ \frac{\pi}{2} }_{0} \: \ln \left( cos(x) \right) \: dx } \\ \\\displaystyle \sf{ \implies \: 2I =\int^{ \frac{\pi}{2} }_{0} \: \ln \left( sin(x) \right) +\ln\left( cos(x) \right) \: dx }[/tex]

[tex]\displaystyle \sf{ \implies \: 2I =\int^{ \frac{\pi}{2} }_{0} \: \ln \left( sin(x) \: cos(x) \right) \: dx } \\ \\\displaystyle\sf{ \implies \: 2I =\int^{ \frac{\pi}{2} }_{0} \: \ln \left( \dfrac{sin(2x)}{2}\right) \: dx } \\ \\\displaystyle \sf{ \implies \: 2I =\int^{ \frac{\pi}{2} }_{0} \:\ln \left( sin(2x) \right)\:dx-\ln(2) \int^{ \frac{\pi}{2} }_{0} \: dx}[/tex]

[tex]\displaystyle\sf{ \implies \: 2I = \dfrac{1}{2} \int^{ \pi}_{0} \: \ln \left( sin(t) \right) \:dt - \ln(2) \int^{ \frac{\pi}{2} }_{0} \: dx} \\ \\\displaystyle\tt{ \implies \: 2I = \dfrac{1}{2} \cdot2 \int^{ \frac{\pi}{2}}_{0} \: \ln \left( sin(t) \right) \: dt - \ln(2) \int^{ \frac{\pi}{2} }_{0} \: dx}[/tex]

[tex]\displaystyle \tt{ \implies \: 2I =\int^{ \frac{\pi}{2}}_{0} \: \ln \left( sin(t) \right) \: dt - \ln(2)\int^{ \frac{\pi}{2} }_{0} \: dx} \\ \\ \displaystyle \tt{ \implies \: 2I = I - \ln(2) \left[x \right] ^{ \frac{\pi}{2} }_{0} } \\ \\ \displaystyle \tt{ \implies \: I = -\ln(2)\left[ \dfrac{\pi}{2} - 0 \right] }[/tex]

[tex]\displaystyle \sf{ \implies \: I = - \dfrac{\pi}{2} \ln(2) }[/tex]

☞︎︎︎Apologies,if incorrect.

Answer:

The domain of f–1(x) is all real numbers and the range is all real numbers greater than 0.

Step-by-step explanation:

Answer:

6+x=15 (x=9)

Step-by-step explanation:

6 cookies + the newly baked ones = 15 cookies

we know there was 6 cookies and we knew that the result is 15, which means that we "don't know" how many cookies are baked. So the answer is 6+x=15

Answer:

m∠O = 41°

Step-by-step explanation:

∠NOM=∠NMO=(4y-15)° (base angles of isos triangle)

7y+2(4y-15)=180 (angle sum of triangle)

  7y+8y-30=180

       15y-30=180

             15y=180+30

                  =210

                y=210÷15

                  =14

Hence, m∠O = (4y-15)°

                      = [4(14)-15]°

                      = (56-15)°

                      = 41°

Answer:

[tex]x\geq 2[/tex]

Step-by-step explanation:

Write out the problem

[tex]\frac{x}{-5+6} \geq 2[/tex]

Simplifiy -5 + 6 = 1

[tex]\frac{x}{1} \geq 2[/tex]

Remember that any fraction with a denominator of 1 is equilvalent to the numerator, so

[tex]\frac{x}{1} =x[/tex]

[tex]x\geq 2[/tex]

Answer:

B=24

A= I don't know what that means

Step-by-step explanation:

Answer:

1.25 miles

Step-by-step explanation:

0.25 miles = 8 seconds

x miles = 40 seconds

0.25/8 = x/40

0.25 * 40 = 8x

10 = 8x

x = 1.25

-Chetan K