consider the distribution of monthly social security (OASDI) payments. Assume a normal distribution with a standard deviation of $116. if one-fourth of payments are above $1214,87 what is the mean monthly payment?

Answer:

(assuming all members of the gang get equal percentages)

For a gang of 8: 12.5% (100/8)

If 2 are killed: 16.67% (100/6), so percentage increases by roughly 4.2% (16.67-12.5)

Percent that belongs to 3: 12.5%*3=37.5%

Percent that belongs to 3 if 2 are killed: 16.67%*3=50.0%

Each robber's share still increases by 4.2% (50%-37.5%=12.5%, 12.5%/3=4.2%)

If you would like additional help with math or another subject, check out growthinyouth.org!

Step-by-step explanation:

Answer:

Step-by-step explanation:

(3,-2)

plug in 3 for x and -2 for y

-2< -1/2x3+2

-2<-1.5+2

-2<0.5

Answer:

Step-by-step explanation:

(4x³ - 5x² + 3x ) - 4(5x² - 7x³ - 8x)

Remove the brackets and simplify.

We have

4x³ - 5x² + 3x - 20x² + 28x³ + 32x

Group like terms and simplify

That's

4x³ + 28x³ - 5x² - 20x² + 3x + 32x

We have the final answer as

- 3 - ( 4x + 3y - 2z ) - 4 + 2( 5x - 4y + 6z)

Remove the brackets and simplify

That's

- 3 - 4x - 3y + 2z - 4 + 10x - 8y + 12z

Group like terms and simplify

- 4x + 10x - 3y - 8y + 2z + 12z - 3 - 4

We have the final answer as

Hope this helps you

Answer:

Problem 1: We conclude that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement.

Problem 2: We conclude that the volume of Google stock has changed.

Step-by-step explanation:

Problem 1:

We are given that in a recent survey conducted by Pew Research, it was found that 156 of 295 adult Americans without a high school diploma were worried about having enough saved for retirement.

Let p = proportion of adult Americans without a high school diploma who are worried about having enough saved for retirement

So, Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 50%    {means that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement}

Alternate Hypothesis, [tex]H_A[/tex] : p > 50%     {means that a majority of adult Americans without a high school diploma are worried about having enough saved for retirement}

This is a right-tailed test.

The test statistics that would be used here is One-sample z-test for proportions;

                       T.S.  =  [tex]\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of adult Americans who were worried about having enough saved for retirement = [tex]\frac{156}{295}[/tex] = 0.53

           n = sample of adult Americans = 295

So, the test statistics =  [tex]\frac{0.53-0.50}{\sqrt{\frac{0.50(1-0.50)}{295} } }[/tex]

                                    =  1.03

The value of z-test statistics is 1.03.

Also, the P-value of the test statistics is given by;

              P-value = P(Z > 1.03) = 1 - P(Z [tex]\leq[/tex] 1.03)

                           = 1 - 0.8485 = 0.1515

Now, at a 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is less than the critical value of z as 1.03 < 1.645, so we insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement.

Problem 2:

We are given that a random sample of 35 trading days in 2014 resulted in a  sample mean of 3.28 million shares with a standard deviation of 1.68 million shares.

Let [tex]\mu[/tex] = mean daily volume in Google stock

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 5.44 million shares    {means that the volume of Google stock has not changed}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] [tex]\neq[/tex] 5.44 million shares     {means that the volume of Google stock has changed}

This is a two-tailed test.

The test statistics that would be used here is One-sample t-test statistics because we don't know about the population standard deviation;

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

where, [tex]\bar X[/tex] = sample mean volume in Google stock = 3.28 million shares

            s = sample standard deviation = 1.68 million shares

           n = sample of trading days = 35

So, the test statistics =  [tex]\frac{3.28-5.44}{\frac{1.68}{\sqrt{35} } }[/tex]  ~ [tex]t_3_4[/tex]

                                    =  -7.606

The value of t-test statistics is -7.606.

Also, the P-value of the test statistics is given by;

              P-value = P([tex]t_3_4[/tex] < -7.606) = Less than 0.05%

Now, at a 0.05 level of significance, the t table gives a critical value of -2.032 and 2.032 at 34 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the volume of Google stock has changed.

Answer:

[tex]\r p = 0.11[/tex]

[tex]\r q = 0.89[/tex]

n = 2000

[tex]E = \pm 5[/tex]

p - population proportion

[tex]\alpha = 5[/tex]%

Step-by-step explanation:

From the question we are told that  

   The  sample size is  [tex]n = 2000[/tex]

   The  proportion of the population for chocolate pie is  [tex]p_c = 0.11[/tex]

   The  margin of error is  [tex]E = \pm 5[/tex]%  

Now in the question we are asked to provide the meaning of  

       [tex]\r p , \r q , n , E , and\ p[/tex]

Now  [tex]\r p[/tex] is the sample proportion of the population that those  chocolate​ pie as  favorite pie of   which is given from the question as 0.11

Now

         [tex]\r q[/tex] is the sample  proportion of the population that choose other pies apart from chocolate pie as their favorite and it is evaluated as

      [tex]\r q = 1 - \r p[/tex]

      [tex]\r q = 1 - 0.11[/tex]

       [tex]\r q = 0.89[/tex]

       n is the sample size which is given as  n =  2000

       E is the margin of  error which given as  [tex]E = \pm 5[/tex]%

       p is the population proportion

Given that the confidence level is  95 %  then the level of significance is  mathematically evaluated as

        [tex]\alpha =(100 - 95)[/tex]%

        [tex]\alpha =[/tex]5%

Answer:

Total amount to be paid back = $29971

Step-by-step explanation:

Formula used to calculate the final amount of the loan to be paid,

Total amount to be paid = [tex]P(1+\frac{r}{n})^{nt}[/tex]

Where P = Principal amount of loan taken

r = rate of interest

n = Number of compounding in a year

t = duration of investment

By substituting the values in the formula,

Total amount to be paid after loan maturity = [tex]18000(1+\frac{0.04}{1})^{13\times 1}[/tex]

= [tex]18000(1.04)^{13}[/tex]

= 18000(1.66507)

= $29971.32

≈ $29971

Total amount to be paid after loan maturity will be $29971.

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

Work Shown:

x = speed of train A

y = speed of train B

"train A travels 4/5 the speed of train B" (I'm assuming "train one" is supposed to read "train B"). So this means x = (4/5)y

distance = rate*time

d = x*7

d = (4/5)y*7 = (28/5)y represents the distance train A travels

d = y*7 = 7y represents the distance train B travels

summing those distances will give us 693

(28/5)y + 7y = 693

5*(  (28/5)y + 7y ) = 5*693

28y + 35y = 3465

63y = 3465

y = 3465/63

y = 55

Train B's speed is 55 mph

4/5 of that is (4/5)y = (4/5)*55 = 4*11 = 44 mph

Train A's speed is 44 mph

To make it easier, you can convert 7/20 to a decimal, and as a decimal it is 0.35. 0.42*0.35=0.147, so .42*7/20=0.147.

Answer:

0.147

Step-by-step explanation:

0.42 * 7/20

Well, 0.42 = 42/100. Now we have both numbers as fractions.

We can simplify 42/100 to 21/50

Therefore we have 7/20 * 21/50

To multiply fractions multiply the numerators together and multiply the denominators together.

This gives: 147 / 1000

Which is equal to 0.147

Therefore 0.42 * 7/20 = 0.147

Answer:

[tex]\huge\boxed{z=3}[/tex]

Step-by-step explanation:

If two polygons are similar, then corresponding sides are in proportion.

The corresponding sides:

4 → x

y → 15

3 → w

2 → 6

z → 9

therefore:

[tex]\dfrac{z}{9}=\dfrac{2}{6}[/tex]         cross multiply

[tex](z)(6)=(9)(2)[/tex]

[tex]6z=18[/tex]         divide both sides by 6

[tex]z=3[/tex]

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

When the input is 4, the output of f(x) = x + 21 is f(4).

Substitute x = 4 to f(x):

f(4) = 4 + 21 = 25

Answer:

25

Step-by-step explanation:

We can find the output by plugging in 4 as x into the function:

f(x) = x + 21

f(4) = 4 + 21

f(4) = 25

you can use the value of sin\

sin(theta) = 12/16

Now solve for theta, and do inverse sine

so theta would be 48.59037789 , or around 50 degrees

Answer:

Step-by-step explanation:

Since, it is a right triangle.

Perpendicular (p) = 12

hypotenuse ( h) = 16

now,

[tex]sin \: (x \: degree) = \frac{perpendicular}{hypotenuse} [/tex]

[tex]sin \: x \: = \frac{12}{16} [/tex]

[tex] x = {sin}^{ - 1} ( \frac{12}{16} )[/tex]

[tex]x = 48.59 \: degree[/tex]

hope this helps...

good luck on your assignment..

Answer:

B: The mean will increase

Step-by-step explanation: The outlier is 46, which is way below all the other numbers, which is the definition of an outlier. If we remove a really low number from the set, then the mean(average) will increase.

Answer:

The confidence interval is  [tex]25.16 < \mu < 26.85[/tex]

Step-by-step explanation:

From  the question we are given a data set

     25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

The mean of the this sample data is  

       [tex]\= x = \frac{\sum x_i}{n}[/tex]

where is the sample size with values  n =  10

         [tex]\= x = \frac{25.5+ 26.1+ 26.8+23.2+ 24.2+ 28.4+ 25.0+ 27.8+ 27.3+ 25.7}{10}[/tex]

          [tex]\= x = 26[/tex]

The standard deviation is evaluated as

             [tex]\sigma = \sqrt{\frac{\sum (x-\= x)}{n} }[/tex]

substituting values

     [tex]= \sqrt{\frac{ ( 25.5-26)^2, (26.1-26)^2, (26.8-26)^2, (23.2-26)^2}{10} }[/tex]

                  [tex]\cdot \ \cdot \ \cdot \sqrt{\frac{ ( 24.2-26)^2, (28.4-26)^2+( 25.0-26)^2+ (27.8-26)^2+( 27.3-26)^2+( 25.7-26)^2}{10} }[/tex]

     [tex]\sigma = 1.625[/tex]

The  confidence level is given as 90% hence the level of significance is calculated as

    [tex]\alpha = 100 -90[/tex]

    [tex]\alpha =10[/tex]%

   [tex]\alpha = 0.10[/tex]

Now the critical values of [tex]\frac{\alpha }{2}[/tex] is obtained from the normal distribution table as

      [tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]

The reason  we are obtaining  the critical values of [tex]\frac{\alpha }{2}[/tex] instead of  [tex]\alpha[/tex] is because  we are considering two tails of the area under the normal curve

  The margin of error is evaluated as

            [tex]MOE = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]

substituting values

           [tex]MOE = 1.645 * \frac{1.625 }{\sqrt{10} }[/tex]

           [tex]MOE = 0.845[/tex]

The  90%, two sided confidence interval is mathematically evaluated as

           [tex]\= x - MOE < \mu < \= x + MOE[/tex]

           [tex]26 - 0.845 < \mu < 26 + 0.845[/tex]

           [tex]25.16 < \mu < 26.85[/tex]

Given that the lower and the upper limit is greater than  25 then we can assure the manufactures  that the battery life exceeds 25 hours

Answer:

The probability that the cycle time exceeds 65 minutes if it is known that the cycle time exceeds 55 minutes, should be 1 / 3.

Step-by-step explanation:

It is known that the cycle times for a truck hauling concrete is uniformly distributed over a time interval of ( 50, 70 ). If c = cycle time, according to the question the probability that the cycle exceeds 65 minutes, respectively exceed 55 minutes should be the following - ' [tex]Probability( c > 65 | c > 55 )[/tex]. '

_____

[tex]f( c ) = \left \{ {{1 / 20,} \atop {0}} \right. \\50< c<70 - ( elsewhere )[/tex]

We know that the formula for Probability( A | B ) is P( A ∩ B ) / P( B ),

[tex]P( c > 65 | c > 55 ) =[/tex] [tex]P( c > 55[/tex]  ∩ [tex]c > 65 )[/tex] /  [tex]P( c > 55 )[/tex],

And now we come to the formula [tex]P( a < c < b )[/tex] = [tex]\int\limits^{70}_{65} {f(x)} \, dc[/tex]. Substitute known values to derive two solutions, forming a fraction that represents the probability we desire.

[tex]P( 65<c<70) = \int\limits^{70}_{65} {f(y)} \, dy\\ = \int\limits^{70}_{65} {(1/20)} \, dy\\ \\= 0.25[/tex]

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

[tex]P( 55<c<70) = \int\limits^{70}_{55} {f(y)} \, dy\\ = \int\limits^{70}_{65} {(1/20)} \, dy\\ \\= 0.75[/tex]

Take 0.25 over 0.75, 0.25 / 0.75, simplified to the fraction 1 / 3, which is our solution.

_____

Probability: 1 / 3

Answer:

When the price of DVD player increases, the demand will decrease.

Step-by-step explanation:

The price and demand function have an inverse relationship. When the price of good increases its demand will decrease. This is law of demand. If the price of goods increase to a certain level its demand will start decreasing. The people will start buying lesser good and will consume less. The propensity to consume will decline and less economic goods will be consumed. When the price of DVD player is increased then the demand will decline for DVD players and people will drop the decision to buy the DVD player.

Answer:

term to term rule is +2

The first term is -8

80th term is 150

Step-by-step explanation:

70 to 72 is adding 2

72 to 74 is adding 2

term to term rule is +2

The equation for a sequence like this is

xn = a + d(n−1)

where n is the term number  a is the first term and d is the term to term rule

Using the 40th term is 70

70 = a + 2( 40-1)

70 = a + 2( 39)

70 = a + 78

Subtract 78 from each side

-8 = a

The first term is -8

80th term

xn = a + d(n−1)

x80 = -8 +2 (80-1)

     = -8+2(79)

    = -8+158

   = 150

Answer:

true

Step-by-step explanation:

Answer:

B. 168 students

Step-by-step explanation:

Given that there are a total of 600 students.

28% of the students pack their lunch.

To find:

Total number of students who pack their lunch = ?

Solution:

Percentage of a given number is calculated using the following method.

[tex]y\%[/tex] of a number [tex]x[/tex] is given by:

[tex]x \times \dfrac{y}{100}[/tex]

i.e. multiply the number by percentage to be found and divide by 100.

So, we have to find 28% of 600 here, to find the answer to the question.

[tex]\therefore[/tex] Number of students who pack their lunch is given as: (Multiply the given number 600 with 28 and divide by 100.)

[tex]600 \times \dfrac{28}{100}\\\Rightarrow 6 \times 28\\\Rightarrow \bold{168}[/tex]

So, the correct answer is:

B. 168

Answer:

Step-by-step explanation:

Perimeter of a rectangle = 2l + 2w

where

l is the length

w is the width

Perimeter = 128 feet

The width is 24 feet less than the length is written as

w = l - 24

128 = 2l + 2( l - 24)

128 = 2l + 2l - 48

Group like terms

4l = 176

Divide both sides by 4

l = 44

The length is 44 feet

Substitute l = 44 into w = l - 24

w = 44 - 24

w = 20

The width is 20 feet

Hope this helps you

Answer:

[tex]\boxed{\sf Option \ E}[/tex]

Step-by-step explanation:

(1) 3.25 kg

(2) 1 kg = 1000 g

(3) 1000 mg = 1g

= (4) 3.25 mg

The first problem occurred in column (4) where we actually had to multiply it by 1000 but we divided it by 1000 and so our conversion was incorrect.

The correct form of it is:

1) 3.25 kg

2) 1 kg = 1000 g ; 3.25 kg = 3250 g

3) 1 g = 1000 mg ; 3250 * 1000 = 3,250,000 mg

4) Answer = 3,250,000 mg

Answer:

[tex]\boxed{\sf Option \ E}[/tex]

Step-by-step explanation:

1 kg = 1000 g

1000 mg = 1g

3.25 kilograms is to be multiplied by 1000 to get the value in grams. Then the result should be multiplied again by 1000 to get the value in milligrams.

3.25 kg ⇒ 3250 g ⇒ 3250000 mg

The first problem occurs in column (4). The multiplication and division was performed incorrectly.

Answer:

The estimate is  [tex]P__{hat}} \pm E = 0.37 \pm 0.0348[/tex]

Step-by-step explanation:

From the question we are told that  

    The sample size is  n =  522

    The sample proportion of students  would like to talk about school is  [tex]\r p__{hat}} = 0.37[/tex]

  Given that the confidence level is  90 % then the level of significance can be mathematically evaluated as

                  [tex]\alpha = 100 - 90[/tex]

                  [tex]\alpha = 10\%[/tex]

                  [tex]\alpha = 0.10[/tex]

Next we obtain the critical value of  [tex]\frac{\alpha }{2}[/tex] from the normal distribution table, the value is  

               [tex]Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } = 1.645[/tex]

Generally the margin of error can be mathematically represented as

               [tex]E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }[/tex]

=>            [tex]E = 1.645 * \sqrt{\frac{0.37 (1- 0.37 )}{522 } }[/tex]

=>             [tex]E = 0.0348[/tex]

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is  

                       [tex]P__{hat}} \pm E[/tex]

substituting values

                     [tex]0.37 \pm 0.0348[/tex]

Answer: [tex]\dfrac{60}{143}[/tex]

Step-by-step explanation:

Given,  A class has five boys and nine girls.

Total students = 5+9=14

Number of ways to choose 6 students out of 14= [tex]^{14}C_6[/tex]  [Using combinations]

Number of ways to choose 4 girls out of 6 (4 girls + 2 boys = 6 ) = [tex]^{9}C_4\times\ ^{5}C_2[/tex]

If the teacher randomly picks six students, then the probability that he will pick exactly four girls:-

[tex]\dfrac{^{9}C_4\times \ ^{5}C_2}{^{14}C_6}[/tex]

[tex]=\dfrac{\dfrac{9!}{4!5!}\times\dfrac{5!}{2!3!}}{\dfrac{14!}{6!8!}}\\\\=\dfrac{1260}{3003}\\\\=\dfrac{60}{143}[/tex]

hence, the required probability = [tex]\dfrac{60}{143}[/tex] .

Answer:

Step by step explanation

Total length of pole = 21.3 m

Length of pole inside the ground = 0.2 m

Let length of pole outside the ground be X,

So, according to the Question,

[tex]x + 0.2 = 21.3[/tex]

Move constant to R.H.S and change its sign

[tex]x = 21.3 - 0.2[/tex]

Calculate the difference

[tex]x = 21.1 \: m[/tex]

Hope this helps...

Good luck on your assignment...

Answer:  B) 10 three-point questions and 14 five-point questions

Step-by-step explanation:

x represents three-point questions

y represents five-point questions

3x + 5y = 100  →  1(3x + 5y = 100)  =  3x + 5y = 100

 x  +  y  = 24   → -3(x  +  y  = 24)   = -3x  -3y  = -72

                                                                  2y = 28

                                                                    y = 14  (five-point questions)

x +  y = 24

x + 14 = 24

     x = 10  (three-point questions)

Answer: Smokers pay less attention to driving while trying to light a cigarette.

Step-by-step explanation:

( A ) reasons for, why the given statement was misleading.

Following are the reasons for accidents

(i) Drunk Driving.

(ii) Driving during the night.

(iii) Smoking.

(iv) Unsafe lane changes.

(v) Driving the wrong way.

(vi) Bad weather e.g fog.

From the list above, it is shown that, smoking is also one of the major cause of car accidents on the roads.

( B.) Smokers tends to pay less attention to driving why trying to light a cigarette.

Answer:

366 Minutes

Step-by-step explanation:

Answer:

$0.26

Step-by-step explanation:

To find the unit price, divide the cost by the amount you have.

$2.86/11 = $0.26

The unit price is $0.26.

Answer:

a)P [ z > 1,38 ] = 0,08379

b) P [ 1,233 < z < 2,43 ]  = 0,1012

c)  P [ z > -2,43 ]  = 0,99245

Step-by-step explanation:

a) P [ z > 1,38 ] = 1 -  P [ z < 1,38 ]

From z-table  P [ z < 1,38 ] = 0,91621

P [ z > 1,38 ] = 1 - 0,91621

P [ z > 1,38 ] = 0,08379

b)  P [ 1,233 - 2,43 ]  must be  P [ 1,233 < z < 2,43 ]

P [ 1,233 < z < 2,43 ]  = P [ z < 2,43 ] - P [ z > 1,233 ]

P [ z < 2,43 ]  = 0,99245

P [ z > 1,233 ] = 0,89125    ( approximated value  without interpolation)

Then

P [ 1,233 < z < 2,43 ]  = 0,99245 - 0,89125

P [ 1,233 < z < 2,43 ]  = 0,1012

c) P [ z > -2,43 ]

Fom z-table

P [ z > -2,43 ] = 1 - P [ z < -2,43 ]

P [ z > -2,43 ] = 1 - 0,00755

P [ z > -2,43 ]  = 0,99245