PQR is an isosceles triangle in which PQ = PR

Mand N are points on PQ and PR such that angle MRQ = angle NQR.

Prove that triangles QNR and RMQ are congruent.

Answer:

9.375 in^2

Step-by-step explanation:

Answer:

-30000/100

300 0/1

Step-by-step explanation:

We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.

16 2/3 = 16.67

We do the multiplication:

−18⋅16 2/3 = -300

We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:

-30000/100

How mixed number would it be:

300 0/1

Answer:

Step-by-step explanation:

Let the money earned during fundraising activities =x

Since the World Issues club has decided to donate 60% of all their fundraising activities this year to Stephen Lewis Foundation.

The amount of money the foundation will receive

=60% of x

= 0.6x

In the bake sale, the club raised $72.

Therefore, the amount the foundation will receive =0.6*72=$43.20

At the end of the year, the World Issues Club mailed a cheque to the foundation for $850.

Therefore:

0.6x=850

x=850/0.6

x=$1416.67

The total amount of money the club raised is $1416.67.

Answer:

9.42 in²

Step-by-step explanation:

The area of whole circle S=pi*R²    , where pi is appr. 3.14,  R= 6 in

S= 3.14*6² =113.04 in²

The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²

The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.

To find the area of a sector, you can use the formula:

Area of sector = (θ/360) × π × r²

where θ is the central angle in degrees, r is the radius of the sector.

The central angle is 30 degrees and the radius is 6 inches.

Plugging these values into the formula:

Area of sector = (30/360) × π × 6²

= (1/12) × π × 36

= (1/12) × 3.14159 × 36

= 9.42478 square inches

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Answer:

tiStep-by-step explanaon:

Answer:

[tex]tan(\frac{11\pi}{6}) =-\frac{\sqrt{3} }{3}[/tex]

Step-by-step explanation:

Notice that [tex]\frac{11\pi}{6}[/tex] is an angle in the fourth quadrant (where the tangent is negative), and the angle is in fact equivalent to [tex]-\frac{\pi}{6}[/tex]. This is one of the special angles for which the sine and cosine functions, as well as the tangent function  have well know values:

Recall that the tangent is defined as

[tex]tan(\theta)=\frac{sin(\theta)}{cos(\theta)}[/tex]

and for this angle (  [tex]\frac{11\pi}{6}[/tex] ) the value of the sine and cosine functions are well known:

[tex]sin (\frac{11\pi}{6}) =-\frac{1}{2} \\cos( \frac{11\pi}{6}) =\frac{\sqrt{3} }{2}[/tex]

Then, the tangent would be:

[tex]tan(\theta)=\frac{sin(\theta)}{cos(\theta)}\\tan(\frac{11\pi}{6}) = \frac{-\frac{1}{2} }{\frac{\sqrt{3} }{2} } \\tan(\frac{11\pi}{6}) =-\frac{1}{\sqrt{3} } \\tan(\frac{11\pi}{6}) =-\frac{\sqrt{3} }{3}[/tex]

Answer:

3/4

Step-by-step explanation:

Answer:

3/4

Step-by-step

Since they are parallel they will have the same slope but not the same intercept

Answer:

  $6534.42

Step-by-step explanation:

Put the given values into the simple interest formula and solve for the remaining variable.

  A = P(1 +rt)

where P is the principal invested, r is the annual rate, and t is the number of years.

  $7000 = P(1 +0.095(9/12)) = 1.07125P

  $7000/1.07125 = P ≈ $6534.42

The value that must be invested is $6534.42.

Answer:

1)

A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552

C) P(X> 9)= 0.36317

D) IQR= 0.4422

2)

A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)

B) P( [tex]\frac{}{X}[/tex]<30)= 0.50

C) P₉₅= 32.60

D) P( [tex]\frac{}{X}[/tex]>36)= 0

E) Q₃: 31.0586

Step-by-step explanation:

Hello!

1)

The variable of interest is

X: pollutants found in waterways near a large city. (ppm)

This variable has a normal distribution:

X~N(μ;σ²)

μ= 8.5 ppm

σ= 1.4 ppm

A sample of 18 large cities were studied.

A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

The population mean is the same as the mean of the variable

μ= 8.5 ppm

The standard deviation is

σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108

So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B)

P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)

To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.

Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]

Then using the Z table you'll find the probability of

P(Z≤1.51)= 0.93448

Then

1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552

C)

In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:

P(X> 9)= 1 - P(X ≤ 9)

Z= (X-μ)/δ= (9-8.5)/1.44

Z= 0.347= 0.35

P(Z≤0.35)= 0.63683

Then

P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317

D)

The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:

Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25

Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:

P(Z≤z₁)= 0.25

Using the table you have to identify the value of Z that accumulates 0.25 of probability:

z₁= -0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)

z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)

[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5=  8.2789 ppm

The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

Using the table you have to identify the value of Z that accumulates 0.75 of probability:

z₃= 0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5=  8.7211 ppm

IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422

2)

A)

X ~ N(30,10)

For n=4

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

Population mean μ= 30

Population variance σ²/n= 10/4= 2.5

Population standard deviation σ/√n= √2.5= 1.58

[tex]\frac{}{X}[/tex] ~ N(30;2.5)

B)

P( [tex]\frac{}{X}[/tex]<30)

First you have to standardize the value and then look for the probability:

Z=  ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0

P(Z<0)= 0.50

Then

P( [tex]\frac{}{X}[/tex]<30)= 0.50

Which is no surprise since 30 y the value of the mean of the distribution.

C)

P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95

P( Z≤ z₀)= 0.95

z₀= 1.645

Now you have to reverse the standardization:

z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)

z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)

[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60

P₉₅= 32.60

D)

P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0

E)

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

z₃= 0.67

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586

Q₃: 31.0586

Answer:

a)[tex]A(t)=2000e^{0.085t}[/tex]

b)[tex]A'(t)=170e^{0.085t}[/tex]

c)$3059.1808

d)t=4.77 years

e) The balance growing is $254.99/year

Step-by-step explanation:

We are given that Two thousand dollars is deposited into a savings account at 8.5​% interest compounded continuously.

Principal = $2000

Rate of interest = 8.5%

a) What is the formula for​ A(t), the balance after t​ years? ​

Formula [tex]A(t)=Pe^{rt}[/tex]

So,[tex]A(t)=2000e^{0.085t}[/tex]

B)What differential equation is satisfied by​ A(t), the balance after t​ years?

So, [tex]A'(t)=2000 \times 0.085 e^{0.085t}[/tex]

[tex]A'(t)=170e^{0.085t}[/tex]

c)How much money will be in the account after 5 ​years? ​

Substitute t = 5 in the formula "

[tex]A(t)=2000e^{0.085t}\\A(5)=2000e^{0.085(5)}\\A(5)=3059.1808[/tex]

d)When will the balance reach ​$3000​?

Substitute A(t)=3000

So, [tex]3000=2000e^{0.085t}[/tex]

t=4.77

The balance reach $3000 in 4.77 years

e)How fast is the balance growing when it reaches ​$3000​?

Substitute the value of t = 4.77 in derivative formula :

[tex]A'(t)=170e^{0.085t}\\A'(t)=170e^{0.085 \times 4.77}\\A'(t)=254.99[/tex]

Hence the balance growing is $254.99/year

Answer:

Second option is the correct choice. See the explanation below.

Step-by-step explanation:

[tex]A=\frac{bh}{2}\\\\\mathrm{Switch\:sides}:\\\\\frac{bh}{2}=A\\\\\mathrm{Multiply\:both\:sides\:by\:}2\\\\\frac{2bh}{2}=2A\\\\hb=2A\\\\\mathrm{Divide\:both\:sides\:by\:}b;\\\\\frac{hb}{b}=\frac{2A}{b}\\\\h=\frac{2A}{b}[/tex]

Best Regards!

Answer:

[tex]h = \frac{2A}{b} [/tex]

Option B is the right option.

Solution,

[tex]a = \frac{bh}{2} \\ 2a = bh(cross \: multiplication) \\ 2a = b \times h \\ h = \frac{2A}{b} [/tex]

hope this helps...

Good luck on your assignment..

Answer:

50,063,860 different samples can be chosen

Step-by-step explanation:

The order in which the microchips are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

How many different samples can be chosen

We choose 6 microchips from a set of 60. So

[tex]C_{60,6} = \frac{60!}{6!(60-6)!} = 50063860[/tex]

50,063,860 different samples can be chosen

Answer:

The numbers of doors that will have no blemishes will be about 6065 doors

Step-by-step explanation:

Let the number of counts by the  worker of each blemishes on the door be (X)

The distribution of blemishes followed the Poisson distribution with parameter  [tex]\lambda =0.5[/tex] / door

The probability mass function on of a poisson distribution Is:

[tex]P(X=x) = \dfrac{e^{- \lambda } \lambda ^x}{x!}[/tex]

[tex]P(X=x) = \dfrac{e^{- \ 0.5 }( 0.5)^ x}{x!}[/tex]

The probability that no blemishes occur is :

[tex]P(X=0) = \dfrac{e^{- \ 0.5 }( 0.5)^ 0}{0!}[/tex]

[tex]P(X=0) = 0.60653[/tex]

P(X=0) = 0.6065

Assume the number of paints on the door by q = 10000

Hence; the number of doors that have no blemishes  is = qp

=10,000(0.6065)

= 6065

Answer:

A' - (8,2)

B' - (5,1)

C' - (4,2)

D' - (4,5)

Step-by-step explanation:

See attachment for the missing figure.

We can see that the vertices of the polygon ABCD have coordinates A(4,6), B(5,3), C(4,2) and D(1,2)

Polygon ABCD is rotated 90° clockwise about point C to form polygon A′B′C′D′ (see attached diagram), then

A'(8,2);

B'(5,1);

C' is the same as C, thus, C'(4,2);

D'(4,5).

Answer:

The degrees of freedom are given by:

[tex] df = n_A +n_B -2 = 18 +15-2= 31[/tex]

And the 90% confidence interval for this case is:

[tex] -1.90 \leq \mu_A -\mu_B \leq 0.13[/tex]

And for this case since the confidence interval contains the value 0 we can conclude that:

A. We are 90% confident that, on average, there is no difference in operating hours between toothbrushes from Company A compared to those from Company B.

Step-by-step explanation:

We know the following info given:

[tex] \bar X_A= 119.7[/tex] sample mean for A

[tex] s_A = 1.74[/tex] sample deviation for A

[tex] n_A = 18[/tex] sample size from A

[tex] \bar X_B= 120.6[/tex] sample mean for B

[tex] s_B = 1.72[/tex] sample deviation for B

[tex] n_B = 15[/tex] sample size from B

The degrees of freedom are given by:

[tex] df = n_A +n_B -2 = 18 +15-2= 31[/tex]

And the 90% confidence interval for this case is:

[tex] -1.90 \leq \mu_A -\mu_B \leq 0.13[/tex]

And for this case since the confidence interval contains the value 0 we can conclude that:

A. We are 90% confident that, on average, there is no difference in operating hours between toothbrushes from Company A compared to those from Company B.

Answer:

The 95% confidence interval is 2.5 < u <3.1.

Step-by-step explanation:

The provided sample mean is X = 2.8 and the sample standard deviation is s = 1.2, and the sample size is n = 64.

1. Null and Alternative Hypotheses:

The following null and alternative hypotheses need to be tested:

H0  u = 3

Ha: u < 3

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

2. Rejection Region Based

on the information provided, the significance level is alpha = 0.05, and the critical value for a left-tailed test is t c = -1.669.

The rejection region for this left-tailed test is R = t : t < -1.669

3. Test Statistics

The t-statistic is computed as follows:

t = (X - uo)/[s/n^(1/2)] =

replacing

t = (2.8 - 3)/ [1.2/64 ^(1/2)]

t =-1.333  

4. Decision about the null hypothesis

Since it is observed that t = -1.333 > t c = -1.669, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.0936, and since p= 0.0936 => 0.05, it is concluded that the null hypothesis is not rejected.

5. Conclusion It is concluded that the null hypothesis H0 is not rejected. Therefore, there is not enough evidence to claim that the population mean u is less than 3, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval is 2.5 < u <3.1.

Answer:

A. [tex](3b+8)^2[/tex]

Step-by-step explanation:

[tex]9b^2+48b +64\\=(3b)^2 + 2\times 3b\times 8 +(8)^2\\=(3b+8)^2[/tex]

Answer:

91 2/3 pi cubic units

Step-by-step explanation:

The formula for the volume of cone is [tex]\dfrac{1}{3}\pi r^2 h[/tex]. Plugging in the given numbers, you get:

[tex]\dfrac{1}{3}\cdot \pi \cdot 5^2 \cdot 11= 91 \ 2/3 \pi[/tex]

Hope this helps!

Answer:

[tex]Volume=\frac{1}{3} \,275\,\pi[/tex] cubic units

Notice that this answer doesn't agree with any of the first three in the list provided via the screenshot

Step-by-step explanation:

Recall the formula for the volume of a cone:

[tex]Volume=\frac{1}{3} Base\,*\,Height[/tex]

In this case the Height is 11 units, and they also give us the radius of the circular base (5 units) from which we can find the circle's base area:

[tex]Area_{circle} = \pi\,R^2\\Area_{circle}=\pi\,(5)^2\\Area_{circle}=25 \pi[/tex]

therefore the total volume becomes:

[tex]Volume=\frac{1}{3} Base\,*\,Height\\Volume=\frac{1}{3} 25\,\pi\,*\,11\\\\Volume=\frac{1}{3} \,275\,\pi[/tex]

*see attachment showing the 2 boxes

Answer:

3 ft²

Step-by-step explanation:

==>Given:

Box J with the following dimensions:

L = 4.5ft

W = 3ft

H = 2ft

Box F:

L = 3ft

W = 3ft

H = 3ft

==>Required:

Difference between the surface area of box J and box F

==>Solution:

Surface area = 2(WL + HL + HW)

=>S.A of box J = 2(3*4.5 + 2*4.5 + 2*3)

= 2(13.5 + 9 + 6)

= 2(28.5)

S.A of box J = 57 ft²

=>S.A of box F = 2(3*3 + 3*3 + 3*3)

= 2(9 + 9 + 9)

= 2(27)

S.A of box F = 54 ft²

Difference between box J and box F = 57 - 54 = 3 ft²

The mean and the standard deviation of the sampling distribution of the sample means are 36 and 1.94 respectively.

Hence option B is correct.

Given Data

Population Mean, μ = 36.0

Population Standard Deviation, σ = 8

Sample size n = 17

Mean standard deviation, μ = 36.0

[tex]\sigma_\bar x[/tex] = [tex]\sigma/\sqrt{n}[/tex]

= [tex]8/\sqrt{17}[/tex]

= 1.94

The mean and the standard σ/√n of the sampling distribution of the sample means are 36 and 1.94 respectively.

Therefore correct option is B .

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Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as [tex]e=\frac{c}{a}[/tex]

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

[tex]\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)[/tex]

b) Eccentricity =5

[tex]5=\frac{c}{a} \:c=5a[/tex]

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

[tex]If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0[/tex]

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

[tex]a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)[/tex]

[tex]Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1[/tex]

Answer:

41 word/min

Step-by-step explanation:

Before noon Ali works:

She types:

After lunch she works:

She types:

Total Ali works= 4+4= 8 hours= 480 min

Total Ali types= 11520+8160= 19680 words

Average typing rate= 19680 words/480 min= 41 word/min

Answer:

[tex]\frac{1}{6}[/tex]

Step-by-step explanation:

Answer:

Step-by-step explanation:

[tex]\frac{5}{6}-\frac{2}{3}=\frac{5}{6}-\frac{2*2}{3*2}\\\\=\frac{5}{6}-\frac{4}{6}\\\\=\frac{5-4}{6}\\\\=\frac{1}{6}[/tex]

Answer:

Step-by-step explanation:

Given that,

[tex]n_1=9,x=520,s^2_x=38\\\\n_2=8,y=540,s^2_y=20[/tex]

a) Under null hypothesis H₀ : there is no difference between the variability of night shift and day shift workers

i.e [tex]H_0:\sigma^2_x=\sigma^2_y=\sigma^2[/tex]

Alternative hypothesis [tex]H_1:\sigma_x^2>\sigma_y^2[/tex]

Level of significance = 5% = 0.05

b) The test statistic

[tex]F=\frac{S^2_x}{S_y^2} \sim F(n_1-1,n_2-1)\\\\=\frac{38}{20}\\\\=1.9[/tex]

Table value of [tex]F_{0.05}(n_1-1,n_2-1)[/tex]

[tex]=F_{0.05}(9-1,8-1))\\\\=F_{0.05}(8,7)\\\\=3.726[/tex]

[tex]\therefore F_{calculated}=1.9<Tab F_{0.05}(8,7)=3.726[/tex]

[tex]H_0[/tex] is accepted at 5% level of significance

Therefore ,there is no difference between the variability of night shift and day shift worker

c) The P-value is 0.206356

The result is not significant at P < 0.05

d) At 95% confidence interval

[tex]\frac{\frac{S^2_x}{S^2_y} }{F_{t-\alpha/2(8,7)} } <\frac{\sigma^2_x}{\sigma^2_y}\frac{\frac{S^2_x}{S^2_y} }{F_{\alpha/2(8,7)} } \\\\\Rightarrow\frac{1.9}{F_{t-(0.05/2)}} <\frac{\sigma^2_x}{\sigma^2_y} <\frac{1.9}{F_{(0.05/2)}} \\\\\Rightarrow\frac{1.9}{F_{0.975}} <\frac{\sigma^2_x}{\sigma^2_y} <\frac{1.9}{F_{(0.025)}} \\\\\Rightarrow\frac{1.9}{F_{3.726}} <\frac{\sigma^2_x}{\sigma^2_y} <\frac{1.9}{F_{(0.286)}}[/tex]

Variance -ratio lies betwee (0.51,6.643)

Conclusion: There is not sufficient evidence to support the claim  that the night shift worker show more variability in their output levels, than the day workers at α =0.05

Answer:

Step-by-step explanation:

Hello!

The claim is that the variability of the output levels of the night-shift is greater than the variability in the output levels of the day workers.

Be

X₁: output level of night shift workers.

n₁= 9

X[bar]₁= 520

S₁= 38

X₂: output level of day shift workers.

n₂= 8

X[bar]₂= 540

S₂= 20

Considering both variables have a normal distribution, the parameters of interest are the population variances.

a)

H₀: σ₁² ≤ σ₂²

H₁: σ₁² > σ₂²

b)

To compare both variances you have to conduct a variance ratio test with statistic:

[tex]F= (\frac{S^2_1}{S^2_2} )*(\frac{Sigma^2_1}{Sigma^2_2} )~~F_{n_1-1;n_2-1}[/tex]

[tex]F_{H_0}= (\frac{1444}{400} )*1=3.61[/tex]

c)

The test is one tailed  to the right, the p-value will have the same direction, i.e. it will be in the right tail of the distribution. The F distribution has degrees of freedom:

n₁ - 1= 9 - 1= 8

n₂ - 1= 8 - 1= 7

P(F₈,₇ ≥ 3.61) = 1 - P(F₈,₇ < 3.61) = 1 - 0.9461= 0.0539

The p-value of this test is 0.0539

d)

The CI for the variance ratio is:

[tex][\frac{S^2_1/S_2^2}{F_{n_1-1;n_2-1;1-\alpha /2}}; \frac{S^2_1/S_2^2}{F_{n_1-1;n_2-1;\alpha /2}}][/tex]

[tex]F_{n_1-1;n_2-1;1-\alpha /2}= F_{8;7;0.975}= 4.90[/tex]

[tex]F_{n_1-1;n_2-1;\alpha /2}= F_{8;7;0.025}= 0.22[/tex]

[tex][\frac{1444/400}{4.90}}; \frac{1444/400}{0.22}}][/tex]

[0.736; 16.409]

Using the level of significance complementary to the confidence level of the interval, you can compare it to the p-value calculated in item c.

p-value: 0.0539

α: 0.05

The p-value is less than the significance level, the decision is to reject the null hypothesis. Using a 5% significance level you can conclude that the variance in the output levels of the night shift workers is greater than the variance in the output levels of the day shift workers.

Answer:

2.86=28 tenths+6 hundredths.

Step-by-step explanation:

2.86=2 ones+8 tenths+6 hundredths.

2.86=28 tenths+6 hundredths.

Answer:

see

Step-by-step explanation:

ones .  tenths  hundredths

2      .     8          6

8 tenths

If we are not using the ones place

we have 28 tenths

Answer:

Option C is correct.

The sampling distribution with sample size n=100 will have less variability.

Step-by-step explanation:

Complete Question

Consider random samples selected from the population of all female college soccer players in the United States. Assume the mean height of female college soccer players in the United States is 66 inches and the standard deviation is 3.5 inches. Which do you expect to have less variability (spread): a sampling distribution with sample size n = 100 or a sample size of n = 20.

A. Both sampling distributions will have the same variability.

B.The sampling distribution with sample size n=20 will have less variability

C. The sampling distribution with sample size n =100 will have less variability

Solution

The central limit theorem allows us to say that as long as

- the sample is randomly selected from the population distribution with each variable independent of each other and with the sample having an adequate enough sample size.

- the random sample is normal or almost normal which is guaranteed if the population distribution that the random sample was extracted from is normal or approximately normal,

1) The mean of sampling distribution (μₓ) is approximately equal to the population mean (μ)

μₓ = μ = 66 inches

2) The standard deviation of the sampling distribution or the standard error of the sample mean is related to the population standard deviation through

σₓ = (σ/√N)

where σ = population standard deviation = 3.5 inches

N = Sample size

And the measure of variability for a sampling distribution is the magnitude of the standard deviation of the sampling distribution.

For sampling distribution with sample size n = 100

σₓ = (3.5/√100) = 0.35 inch

For sampling distribution with sample size n = 20

σₓ = (3.5/√20) = 0.7826 inch

The standard deviation of the sampling distribution with sample size n = 20 is more than double that of the sampling distribution with sample size n = 100, hence, it is evident that the bigger the sample size, the lesser the standard deviation of the sampling distribution and the lesser the variability that the sampling distribution shows.

Hope this Helps!!!

Answer:

[tex]A.\ y = x^2 - 6x + 13[/tex] is the correct answer.

Step-by-step explanation:

We know that vertex equation of a parabola is given as:

[tex]y = a(x-h)^2+k[/tex]

where [tex](h,k)[/tex] is the vertex of the parabola and

[tex](x,y)[/tex] are the coordinate of points on parabola.

As per the question statement:

The parabola opens upwards that means coefficient of [tex]x^{2}[/tex] is positive.

Let [tex]a = +1[/tex]

Minimum of parabola is at x = 3.

The vertex is at the minimum point of a parabola that opens upwards.

[tex]\therefore[/tex] [tex]h = 3[/tex]

Putting value of a and h in the equation:

[tex]y = 1(x-3)^2+k\\\Rightarrow y = (x-3)^2+k\\\Rightarrow y = x^2-6x+9+k[/tex]

Formula used: [tex](a-b)^2=a^{2} +b^{2} -2\times a \times b[/tex]

Comparing the equation formulated above with the options given we can observe that the equation formulated above is most similar to option A.

Comparing [tex]y = x^2 - 6x + 13[/tex] and [tex]y = x^2-6x+9+k[/tex]

13 = 9+k

k = 4

Please refer to the graph attached.

Hence, correct option is [tex]A.\ y = x^2 - 6x + 13[/tex]

Answer:

A. y = x^2 -6x + 13

Step-by-step explanation:

Answer:

slope = -4/5

Step-by-step explanation:

A line passes two points (x1, y1) and (x2, y2).

The slope of this line can be calculate by the formula:

s = (y2 - y1)/(x2 - x1)

=>The line that passes A(-4, 8) and B(-9, 12) has the slope:

s = (12 - 8)/(-9 - -4) = 4/(-5) = -4/5

Hope this helps!

Answer:

The probability that demand during lead-time will exceed 20 bails is 0.2033.

Step-by-step explanation:

We are given that it has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails.

Let X = demand during the lead-time

So, X ~ Normal([tex]\mu=15, \sigma^{2} = 6^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

                               Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu=[/tex] population mean demand = 15 bails

           [tex]\sigma[/tex] = standard deviation = 6 bails

Now, the probability that demand during lead-time will exceed 20 bails is given by = P(X > 20 bails)

       P(X > 20 bails) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{20-15}{6}[/tex] ) = P(Z > 0.83) = 1 - P(Z [tex]\leq[/tex] 0.83)

                                                             = 1 - 0.7967 = 0.2033

Answer:

210

Complete question found at brainly(ID): 18678557 is stated below.

There is more than one integer, greater than 1, which leaves a remainder of 1 when divided by each of the four smallest primes. What is the difference between the two smallest such integers?

Step-by-step explanation:

Prime numbers are numbers that can only be divided by itself and 1

The smallest of the prime numbers we have = 2, 3, 5, 7

Since the integers greater than 1 are said to be divided by the four smallest prime numbers, we would assume the number of integers are 4 in total.

Let the integers be T

From the question:

Integer/(prime number) = quotient + (remainder/prime number)

Integer/(prime number) = Q + R/P

Let the different quotients derived from all 4 prime number = w, x, y, z

For prime 2:

T/2 = w + 1/2

T/2 - 1/2 = w

(T-1)/2 = w

T = 2w + 1

T-1 = 2w

Following the above solution

For prime 3:

T = 3x + 1

T-1 = 3x

For prime 5:

T = 5y + 1

T-1 = 5y

For prime 7:

T = 7z + 1

T-1 = 7z

T-1= T-1 = T-1 = T-1

2w = 3x = 5y = 7z

T-1 = LCM of all the prime numbers

T- 1 = 2×3×5×7

T-1 = 210

T = 210+1 = 211

T = 211

The smallest of the integer greater than 1 that leaves a remainder of 1 = 1(T-1) + 1 = 211

The next after the smallest number: 2(T-1) +1= 2(210) + 1 = 421

The two smallest number = 1(T-1) + 1 and 2(T-1) +1 respectively

The difference between the two smallest such integers = 421-211 = 210