A sample is taken from all college freshman . Right-handed students are excluded.what is this an example of?

Answer:

The total number of marbles in the bag is 50.

Step-by-step explanation:

Here, we have n trials, without replacement. So the hypergeometric distribution is used.

The mean of the hypergeometric distribution is:

[tex]E(X) = \frac{n*k}{N}[/tex]

In which n is the number of items in the sample, k is the number of items in the population that are classified a success and N is the size of the population.

15 marbles are drawn:

This means that [tex]n = 15[/tex]

A bag contains some number of marbles. It is known that 20 of them are red.

This means that [tex]k = 20[/tex], since a success is drawing a red marble.

Assuming E(X)=6 red, what is the total number of marbles in the bag?

We have to find N when [tex]E(X) = 6[/tex]

So

[tex]E(X) = \frac{n*k}{N}[/tex]

[tex]6 = \frac{15*20}{N}[/tex]

[tex]6N = 300[/tex]

[tex]N = \frac{300}{6}[/tex]

[tex]N = 50[/tex]

The total number of marbles in the bag is 50.

Answer:

[tex]-x^5+9x^4-18x^3=0\\-x^3(x^2-9x+18)=0\\-x^3(x-3)(x-6)=0\\\\\\\\x=0\\x=3\\x=6[/tex]

Answer:

Hey mate , here is your answer. Hope it helps you

Step-by-step explanation:

Given Zareen has 24min to work on her math homework, and each problem is taking her 2/3 of a minute

As give one problem takes

24*(2/3) minutes = 16

Hence

2/3 divided by 24

Answer:

Step-by-step explanation:

[tex]y = f(x) =\frac{1}{\sqrt{3 \pi} } e^{-x^{2/3}}[/tex]

y = 0, x = 0 and x = 3

Consider an element of thickness dx at a distance x from the origin. By Cylindirical Shell Method, the volume of the element is given by

[tex]dV=(2\pi rdr)h=(2\pi xdx)f(x) => dV=(2\pi xdx) \frac{1}{\sqrt{3\pi}}e^{-x^{\frac{2}{3}}}[/tex]

[tex]dV=2\sqrt{\frac{\pi}{3}}xe^{-x^{\frac{2}{3}}}dx[/tex]

Integrate the above integral over the limits x=0 to x=3 which implies

[tex]\int_{0}^{V}dV=2\sqrt{\frac{\pi}{3}}\int_{0}^{3}xe^{-x^{\frac{2}{3}}}dx[/tex]

Solve by subsititution

[tex]Let,\\ -x^{\frac{2}{3}}=y => \frac{-2}{3}x^{\frac{-1}{3}}dx=dy => x^{\frac{-1}{3}}dx=\frac{-3}{2}dy[/tex]

Also, apply the new limits

[tex]At,\\\\ x=0, y=0 \ and \ At, x=3, y=-\sqrt[3]{9}[/tex]

This implies,

[tex]\int_{0}^{V}dV=2\sqrt{\frac{\pi}{3}}\int_{0}^{3}x^{\frac{4}{3}}e^{-x^{\frac{2}{3}}}x^{\frac{-1}{3}}dx=2\sqrt{\frac{\pi}{3}}\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}(\frac{-3}{2})dy[/tex]

[tex]V=-\sqrt{3\pi}\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}dy[/tex]

Let,

[tex]I=\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}dy[/tex]

Integrate by parts the above integral

[tex]u=y^2 \ and \ dv=e^ydy => du=2y \ and \ v=e^y[/tex]

Integrate by parts formula

[tex]\int udv=uv-\int vdu => y^2e^y-\int 2ye^ydy[/tex]

Again integrate by parts

[tex]u=y \ and \ dv=e^ydy => du=1 \ and \ v=e^y[/tex]

Integrate by parts formula

[tex]\int udv=uv-\int vdu => y^2e^y-2[ye^y-e^y]=e^y[y^2-2y+2][/tex]

Therefore,

[tex]I=[e^y(y^2-2y+2)]_{0}^{-\sqrt[3]{9}}\\\\=e^{-2.0802}[(2.0802)^2+2(2.0802)+2]-e^{0}[0-0+2]\\\\\frac{(4.3272+4.1604+2)}{8.0061}-2\\\\=\frac{10.4876}{8.0061}-2\\\\=1.3099-2\\\\=-0.6901[/tex]

This implies, the volume is

[tex]V=-\sqrt{3\pi}I\\\\=-\sqrt{3\times 3.142} \times (-0.6901)\\\\=3.0701 \times 0.6901\\\\=2.1186[/tex]

That is, up to three decimal places

[tex]V\approx 2.118[/tex]

Answer:

Due to the fact that there is 21% chance of getting that many girls by chance and also In conjunction to that; there is no test involved as well , we can conclude that the method does not have statistical significance.

The result does not appear to have a practical significance.

Step-by-step explanation:

Given that:

In  a random selection 1936 users, we observed that the method gave birth to  950 boys and 986 girls

There is about a 21​% chance of getting that many girls if the method had no effect.

Due to the fact that there is 21% chance of getting that many girls by chance and also In conjunction to that; there is no test involved as well , we can conclude that the method does not have statistical significance.

Given that:

The number of girls = 986

Number of boys = 950

Number of babies born = 1936

The percentage of girls = number of girls born/ number of babies born

The percentage of girls = 986 /1936

The percentage of girls = 0.5093

The percentage of girls =  50.93%

We can infer that this method does not have a practical significance  because most couples would not prefer to use a method that raise the likelihood of a girl from the approximately 50% rate expected by chance to the 50.93% .

Answer:

169.5 yd²

Step-by-step explanation:

See attachment.

In situations like this, ALWAYS (!) make as sketch.

Divide the areas by adding dotted lines on sensible places, and write in the missing numbers for the correct distances.

The only possible difficult one is the triangle, which is the half of a rectangle. So you are dealing with a series of areas of rectangles. That is really easy if you understand what you are doing.

Total area =

area 1 + area 2 + area 3 + area 4

10*4 + 4*7 + 3*13 + (0.5 * 7*17)

40 + 28 + 42 + 59.5

Total area = 169.5 yd²

Answer:

Step-by-step explanation:

There are no factors outside the parentheses that need to be distributed, so the parentheses can be simply dropped:

  6 -2x +15 -3x

The terms can be rearranged to put like terms next to each other:

  -2x -3x +6 +15

and the like terms can be combined.

  -5x +21 . . . . simplified expression

__

Put the value of x where x is, then do the arithmetic.

  -5(-0.2) +21 = 1 +21 = 22

Answer:

[tex] T_1 = 10*3.5 = 35[/tex]

[tex] T_2 = 20*2.9 = 58[/tex]

[tex] T_3 = 25*3.2 = 80[/tex]

[tex] T_4 = 15*3.4 = 51[/tex]

[tex] \bar X= \frac{T_1 +T_2 +T_3 +T_4}{10+20+25+15}= \frac{35+58+80+51}{10+20+25+15} = 3.2[/tex]

Step-by-step explanation:

For this case we have the following info given:

[tex] n_1= 10 , \bar X_1 = 3.5[/tex] for freshmen

[tex] n_2= 20 , \bar X_2 = 2.9[/tex] for sophomores

[tex] n_3= 25 , \bar X_3 = 3.2[/tex] for juniors

[tex] n_4= 15 , \bar X_4 = 3.4[/tex] for seniors

For this case we can use the formula for the sample mean in order to find the total of each group:

[tex] \bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex]T= \sum_{i=1}^n X_i = n *\bar X[/tex]

And replacing we got:

[tex] T_1 = 10*3.5 = 35[/tex]

[tex] T_2 = 20*2.9 = 58[/tex]

[tex] T_3 = 25*3.2 = 80[/tex]

[tex] T_4 = 15*3.4 = 51[/tex]

And the grand mean would be given by:

[tex] \bar X= \frac{T_1 +T_2 +T_3 +T_4}{10+20+25+15}= \frac{35+58+80+51}{10+20+25+15} = 3.2[/tex]

Answer:

me marca como melhor resposta ☺️❤️

Step-by-step explanation:

.....

Answer:

400

brigadaaaaaaaaaaaaaa

Answer:

yes it does

17 is the longest side.

Iff  Iff 17%5E2+=+8%5E2+%2B+15%5E2 it's a right triangle.  it's a right triangle.

Ps "iff" means if and only if

hope it helps

if so please mark me  as brainliest

Answer:

Step-by-step explanation:

Let us assume that x is normally distributed. The sample size is greater than 30. Since the the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = standard deviation

n = number of samples

From the information given,

µ = 50

σ = 7

n = 36

If x is within 0.5 of the population mean, it means that x is between (50 - 0.5) and (50 + 0.5)

the probability is expressed as

P(49.5 ≤ x ≤ 50.5)

For x = 49.5

z = (49.5 - 50)/(7/√36) = - 0.43

Looking at the normal distribution table, the probability corresponding to the z score is 0.334

For x = 49.5

z = (50.5 - 50)/(7/√36) = 0.43

Looking at the normal distribution table, the probability corresponding to the z score is 0.666

Therefore,

P(49.5 ≤ x ≤ 50.5) = 0.666 - 0.334 = 0.332

Answer:

D

Step-by-step explanation:

The volume of pyramid = 1/3 wlh

Where w = width, l = length and h = height

While,

The volume of rectangular prism = wlh

So,

The volume of pyramid = 1/3(the volume of prism)

Answer:

Step-by-step explanation:

Hello!

Given the variables

X: daily hotel room rate

Y: amount spent on the entertainment

See second attachment for scatter plot.

The population regression equation is E(Yi)= α + βXi

To estimate the y-intercept and the slope of the regression equation you have to apply the following formulas:

[tex]b= \frac{sum XY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }[/tex]

a= Y[bar]-bX[bar]

n= 9; ∑X= 945; ∑X²= 103325; ∑Y= 1134 ∑Y²= 148804; ∑XY= 123307

X[bar]= ∑X/n= 945/9= 105

Y[bar]= ∑Y/n= 1134/9= 126

[tex]b= \frac{123307-\frac{945*1134}{9} }{103325-\frac{(945)^2}{9} }= 1.03[/tex]

a= 126 - 1.03*105= 17.49

^Y= 17.49 + 1.03Xi

Slope interpretation: The estimated average amount spent on entertainment increases 1.03 every time the daily hotel room rate increases one unit.

If the room rate for Chicago is $128 (X), to predict the mount spent in entertainment (Y) you have replace it in the estimated regression line:

^Y= 17.49 + 1.03Xi= 17.49 + 1.03*128= 149.33

The expected amount spent on entertainment for Chicago is $149.33

I hope this helps!

Answer:

0.35 cups/hour

Step-by-step explanation:

To be able to determine the rate at which the water is leaking from the pipe with the information given, you have to divide the number of cups by the number of hours in which they were collected:

12 cups/34 hours= 0.35 cups/hour

According to this, the answer is that the rate at which the water is leaking from the pipe is 0.35 cups/hour.

Answer:

False

Step-by-step explanation:

AAS is one of the POSTULATE to prove triangles' congruency.

Answer:n

Step-by-step explanation:

Answer:

y= 2x+1

Step-by-step explanation:

Points:

Form of the line:

Finding the slope:

Line is now:

Using one of the given points to find out the value of b:

So the equation for the line is:

Answer:

Read below

Step-by-step explanation:

To graph the inequality, place an open circle on -2.5 because there is no line under the > sign. Draw the arrow pointing to the right because the inequality reads "x is greater than -2.5.

As for the check box questions, only B and C should be checked. The arrow points right, and the circle is open.

In polar coordinates, the inequality changes to

[tex]x^2+y^2\le4x\implies r^2\le4r\cos\theta\implies r\le4\cos\theta[/tex]

which is a circle of radius 2 and centered at (2, 0). The set D is then

[tex]D=\left\{(r,\theta)\mid0\le r\le4\cos\theta\land0\le\theta\le\pi\right\}[/tex]

The integral is then

[tex]\displaystyle\iint_D\frac{y^2}{x^2+y^2}\,\mathrm dx\,\mathrm dy=\int_0^\pi\int_0^{4\cos\theta}\frac{r^2\sin^2\theta}{r^2}r\,\mathrm dr\,\mathrm d\theta[/tex]

[tex]=\displaystyle\int_0^\pi\int_0^{4\cos\theta}r\sin^2\theta\,\mathrm dr\,\mathrm d\theta[/tex]

[tex]=\displaystyle\frac12\int_0^\pi((4\cos\theta)^2-0^2)\sin^2\theta\,\mathrm d\theta[/tex]

[tex]=\displaystyle8\int_0^\pi\cos^2\theta\sin^2\theta\,\mathrm d\theta[/tex]

There are several ways to compute the remaining integral; one would be to invoke the double-angle formula,

[tex]\sin(2\theta)=2\sin\theta\cos\theta[/tex]

so that the integral is

[tex]=\displaystyle8\int_0^\pi\frac{\sin^2(2\theta)}4\,\mathrm d\theta[/tex]

[tex]=\displaystyle2\int_0^\pi\sin^2(2\theta)\,\mathrm d\theta[/tex]

Then invoke another double-angle formula,

[tex]\sin^2\theta=\dfrac{1-\cos(2\theta)}2[/tex]

to change the integral to

[tex]=\displaystyle\int_0^\pi1-\cos(4\theta)\,\mathrm d\theta[/tex]

[tex]=(\pi-\cos(4\pi))-(0-\cos0)=\boxed{\pi}[/tex]

Answer:

numerators:      11  7. 12. 0

                          _  _   _.  _

denominators.  12 512. 10. 78

Step-by-step explanation:

Answer:

11/12 n:11 d:12

7/512 n:7 d:512

12/10 n:12 d:10

0/78 n:0 d:78

Step-by-step explanation:

n=numerator

d=denominator

Answer:

x > 1

Step-by-step explanation:

Add 7 to both sides

x > 1

Answer:

a)Randomization condition: Satisfied, as the subjects were randomly selected.

10% condition: Satisfied, as the sample size is less than 10% of the population (U.S. adults).

Sample size condition: Satisfied, as the product between the smaller proportion and the sample size is bigger than 10.

b) The 90% confidence interval for the population proportion is (0.68, 0.70).

Step-by-step explanation:

a) Evaluating the necessary conditions:

Randomization condition: Satisfied, as the subjects were randomly selected.

10% condition: Satisfied, as the sample size is less than 10% of the population (U.S. adults).

Sample size condition: Satisfied, as the product between the smaller proportion and the sample size is bigger than 10.

[tex]n(1-p)=4,726\cdot (1-0.69)=4,726\cdot 0.31=1,465>10[/tex]

b) We have to calculate a 90% confidence interval for the proportion.

The sample proportion is p=0.69.

The standard error of the proportion is:

[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.69*0.31}{4726}}\\\\\\ \sigma_p=\sqrt{0.000045}=0.007[/tex]

The critical z-value for a 90% confidence interval is z=1.645.

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

[tex]MOE=z\cdot \sigma_p=1.645 \cdot 0.007=0.01[/tex]

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

[tex]LL=p-z \cdot \sigma_p = 0.69-0.01=0.68\\\\UL=p+z \cdot \sigma_p = 0.69+0.011=0.70[/tex]

The 90% confidence interval for the population proportion is (0.68, 0.70).

Answer:

1

Step-by-step explanation:

Set the height of the bar to 1 because there is only 1 number between 40-49 i.e. 49

Answer:

72°

Step-by-step explanation:

This is called an isosceles triangle. This means that the 2 angles related to the equal sides, are also equal. Hence, the answer is 72°

Answer:

A

Step-by-step explanation:

Since it is isosceles triangle (two equal sides) therefore, there are 2 equal angles too which at the base (72°)

The total angle of triange is 180°

So 180-72-72=36°

Answer:

As per ASA postulate, the two triangles are congruent.

Step-by-step explanation:

We are given two triangles:

[tex]\triangle ABC[/tex] and [tex]\triangle DEC[/tex].

AD bisects BE.

AB || DE.

Let us have a look at two properties.

1. When two lines are parallel and a line intersects both of them, then alternate angles are equal.

i.e. AB || ED and [tex]\angle B[/tex] and [tex]\angle E[/tex] are alternate angles [tex]\Rightarrow[/tex] [tex]\angle B = \angle E[/tex].

2. When two lines are cutting each other, angles formed at the crossing of two, are known as Vertically opposite angles and they are are equal.

[tex]\Rightarrow \angle ACB = \angle DCE[/tex]

Also, it is given that AD bisects BE.

i.e. EC = CB

1. [tex]\angle B = \angle E[/tex]

2. EC = CB

3. [tex]\angle ACB = \angle DCE[/tex]

So, we can in see that in [tex]\triangle ABC[/tex] and [tex]\triangle DEC[/tex], two angles are equal and side between them is also equal to each other.

Hence, proved that [tex]\triangle ABC[/tex] [tex]\cong[/tex] [tex]\triangle DEC[/tex].

Answer:

1/x = 1/2÷4/5

1/x=1/2 x 5/4

1/x=5/8

5x=8

x=1.6

Answer:

x = 5/v

Step-by-step explanation:

Solve for x:

2 v x - 5 = 5

Add 5 to both sides:

2 v x = 10

Divide both sides by 2 v:

Answer: x = 5/v

Answer:

I'd say start with "Add 5 to both sides"

Step-by-step explanation:

VX +VX-5 = 5

Add 5 to both sides

2VX=10

Divide both sides by 2

VX=5

Divide both sides by V

X=[tex]\frac{5}{V}[/tex]

Answer:

D.

Step-by-step explanation:

Since a quadrilateral inscribed in a circle has its opposite angles adding up to 180°

So,

<MNO + <OLM = 180

82 + <OLM = 180

<OLM = 180-82

<OLM = 98°

Answer: The answer is (x-5)^2

Answer:

63.25 not an integer

Step-by-step explanation:

HCF(a,b)*LCM(a,b)=ab

11*368=64*x

x=11*368/64

x=63.25 not an integer, one of the given numbers must be incorrect

but you may use this method to find it yourself