illustrate the distributive property to solve 144/8

Answer:

hello your question is incomplete below is the complete question

Find the minimum sample size n needed to estimate μ For the given values of​ c, σ​, and E. c=0.98​, σ=6.5​, and E=22 Assume that a preliminary sample has at least 30 members.

Answer : 48

Step-by-step explanation:

Given data:

E = 2.2,

std ( σ ) = 6.5

c ( level of confidence ) = 0.98

To find the minimum sample size

we have to first obtain the value of  [tex]Z_{a/2}[/tex]  

note : a can be found using this relation :

( 1 - a ) = 0.98 ----- equation 1

a = 1 - 0.98 = 0.02

hence:  a/2 = 0.01

This means that P( Z ≤ z ) = 0.99  the value of z can be found using the table of standard normal distribution. from the table the value of z = 2.33

P( Z ≤ 2.33 ) = 0.99

To obtain the sample size n

[tex]n = (\frac{std*z}{E} )^{2}[/tex]

n = [tex](\frac{6.5*2.33}{2.2} )^2[/tex] =  (6.88409)^2

Therefore n ≈ 48

Answer:

Nine thousandths = 0.009

Step-by-step explanation:

thousandths =  1/1000 = 0.001

nine thousandths = 9/1000 = 0.009

Answer:

.009

Step-by-step explanation:

9 thousandths as a decimal is 9/1000.  Which is the same 0.009

Answer:

38.911≤p≤41.089

Step-by-step explanation:

The formula for calculating confidence interval for a population mean us as shown below;

CI = xbar ± Z×S/√N where;

xbar is the sample mean = 40

Z is the z score at 95% confidence interval = 1.96

S is the standard deviation = 5

N is the sample size = 81

Substituting this parameters in the formula we have;

CI = 40±1.96×5/√81

CI = 40±(1.96×5/9)

CI = 40±(1.96×0.556)

CI = 40±1.089

CI = (40-1.089, 40+1.089)

CI = (38.911, 41.089)

The 95% confidence interval for the population mean is 38.911≤p≤41.089

Answer:

38.9 ≤ U ≤ 41.1

Step-by-step explanation:

Mean, m = 40; standard deviation, α = 5; Confidence limit, U = 95% or 0.95

N = 81

The standard error, α(m) = α/√(N) = 5/√81 =5/9

Using table: 0.95 = 0.0379

Z(0.95) = 2 - 0.0379 = 1.9621 or 1.96

Hence, confidence interval = { m - 1.96(α/√N) ≤ U ≤ m +1.96(α/√N)}

But, 1.96(α/√N) = 1.96 X 5/9 = 1.96 X 0.56 = 1.1

(40 - 1.1 ≤ U ≤ 40 + 1.1)

∴ the confidence interval = 38.9 ≤ U ≤ 41.1

Answer:

Below

Step-by-step explanation:

The two given expressions are:

● √(2p-7) = 3

● 7√(3q-1) = 2

We are told to evaluate p+q^2

To do that let's find the values of p and q^2

■■■■■■■■■■■■■■■■■■■■■■■■■■

Let's start with p.

● √(2p-7) = 3

Square both sides

● (2p-7) = 3^2

● 2p-7 = 9

Add 7 to both sides

● 2p-7+7 = 9+7

● 2p = 16

Divide both sides by 2

● 2p/2 = 16/2

● p = 8

So the value of p is 8

■■■■■■■■■■■■■■■■■■■■■■■■■■

Let's find the value of q^2

● 7√(3q-1) = 2

Square both sides

● 7^2 × (3q-1) = 2^2

● 49 × (3q-1) = 4

● 49 × 3q - 49 × 1 = 4

● 147q - 49 = 4

Add 49 to both sides

● 147q -49 +49 = 4+49

● 147q = 53

Divide both sides by 147

● 147q/147 = 53/147

● q = 53/ 147

Square both sides

● q^2 = 53^2 / 147^2

● q^2 = 2809/21609

■■■■■■■■■■■■■■■■■■■■■■■■■

● p+q^2 = 8 +(2809/21609)

● p+q^2 = (2809 + 8×21609)/21609

● p+q^2 = 175681 / 21609

● p + q^2 = 8.129

Round it to the nearest unit

● p+ q^2 = 8

Answer:

50hours

Step-by-step explanation:

Given that there are 400 pages in Sheila's favorite book.

The average number of words per page in the book is 300

She types an average rate of 40words per minute.

So to type 400pages of the book

Total number of words in the pages = 400×300 = 120000 words

Typing rate : 40words ------- 1minute

120000 words ----------- x minutes

Hence we have 40 × X mins = 120000 × 1min

Make X the subject

40X = 120000minutes

X = 120000/40

X = 3000minutes

Since 60minutes = 1hour

3000minutes = 3000minutes/60

= 50hours

Hence it took her 50hours to type 400pages

Solution:

The total number of words in the book is 400 x 300. Sheila types at a rate of 40 words per minute, or 40 x 60 words per hour. The number of hours it takes her is equal to the number of words divided by her rate of typing, or 400x300/40x60 = 50 hours.

Answer:

7

Step-by-step explanation:

10+4 = 14

14/2  = 7

x = 7

Answer:

x = -5/2

Step-by-step explanation:

1+2x=6x+11

Subtract 2x from each side

1+2x-2x=6x-2x+11

1 = 4x+11

Subtract 11 from each side

1-11 = 4x

-10 =4x

Divide by 4

-10/4 = 4x/4

-5/2 =x

Answer:

[tex]\boxed{x=-\frac{5}{2}}\\[/tex]

Step-by-step explanation:

To begin, get the variable on one side of the equation - preferably the left for standard solution notation (for this equation, it is easier to place it on the right side to avoid negative values). Do this by subtracting 2x from both sides of the equation. Then, subtract 11. Finally, divide by 4 and get the answer in terms of x.

1 + 2x = 6x + 11

1 = 4x + 11

-10 = 4x

[tex]\boxed{x=-\frac{5}{2}}[/tex]

Answer:

The answer is option D

Step-by-step explanation:

Volume of a sphere is given by

[tex]V = \frac{4}{3} \pi {r}^{3} [/tex]

where r is the radius

From the question to calculate the radius we use the formula

radius = diameter / 2

diameter =70mm

radius = 70/2 = 35 mm

So the volume of the sphere is

[tex]V = \frac{4}{3} \pi \times {35}^{3} [/tex]

[tex]V = \frac{171500\pi}{3} [/tex]

We have the final answer as

Hope this helps you

Answer:

5/7

Step-by-step explanation:

There are a couple ways to solve this.  One would be by finding the least common denominator for each one with 2/3, subtracting, and seeing what is left over.  Another way is converting to decimals.

2/3=0.666666

————————-

7/8=0.875

8/9=0.88888

4/5=0.8

5/7=0.7143

They are all greater than 2/3 (0.6666666), but 5/7 is the closest, so would have the least waste.

Answer:

(A) The shape of the distribution of the sample mean is bell-shaped.

(B) The standard error of the distribution of the sample mean is 1.1.

(C) The proportion of the samples that have a mean useful life of more than 36 hours is 0.1814.

(D) The proportion of the sample that has a mean useful life greater than 34.5 hours is 0.6736.

(E) The proportion of the sample that has a mean useful life between 34.5 and 36.0 hours is 0.4922.

Step-by-step explanation:

We are given that Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 35.0 hours and a standard deviation of 5.5 hours.

As a part of its quality assurance program, Power +, Inc. tests samples of 25 batteries.

Let [tex]\bar X[/tex] = sample mean life of these batteries

(A) The shape of the distribution of the sample mean will be bell-shaped because the sample mean also follows the normal distribution as it is taken from the population data only.

(B) The standard error of the distribution of the sample mean is given by;

            Standard error =  [tex]\frac{\sigma}{\sqrt{n} }[/tex]

Here, [tex]\sigma[/tex] = standard deviation = 5.5 hours

         n = sample of batteries = 25

So, the standard error =  [tex]\frac{5.5}{\sqrt{25} }[/tex]  = 1.1.

(C) The z-score probability distribution for the sample mean is given by;

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

where, [tex]\mu[/tex] = population mean life of battery = 35.0 hours

            [tex]\sigma[/tex] = standard deviation = 5.5 hours

            n = sample of batteries = 25

Now, the proportion of the samples that will have a mean useful life of more than 36 hours is given by = P([tex]\bar X[/tex] > 36 hours)

       P([tex]\bar X[/tex] > 36 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{36-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z > 0.91) = 1 - P(Z [tex]\leq[/tex] 0.91)

                                                               = 1 - 0.8186 = 0.1814

(D) The proportion of the samples that will have a mean useful life of more than 34.5 hours is given by = P([tex]\bar X[/tex] > 34.5 hours)

       P([tex]\bar X[/tex] > 34.5 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{34.5-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z > -0.45) = P(Z [tex]\leq[/tex] 0.45)

                                                                    = 0.6736

(E) The proportion of the samples that will have a mean useful life between 34.5 and 36.0 hours is given by = P(34.5 hrs < [tex]\bar X[/tex] > 36 hrs)

     P(34.5 hrs < [tex]\bar X[/tex] < 36 hrs) = P([tex]\bar X[/tex] < 36 hrs) - P([tex]\bar X[/tex] [tex]\leq[/tex] 34.5 hrs)

     P([tex]\bar X[/tex] < 36 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{36-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z < 0.91) = 0.8186

     P([tex]\bar X[/tex] [tex]\leq[/tex] 34.5 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{34.5-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.45) = 1 - P(Z [tex]\leq[/tex] 0.45)

                                                                    = 1 - 0.6736 = 0.3264                              

Therefore, P(34.5 hrs < [tex]\bar X[/tex] < 36 hrs) = 0.8186 - 0.3264 = 0.4922.

Answer:

[tex]\huge\boxed{4 ( 20 + 21)}[/tex]

Step-by-step explanation:

4(20) + 84

Resolve Parenthesis

80 + 84

Taking 4 common as both are the multiples of 4

4 ( 20 + 21)

Answer:

Student t-distribution.

Step-by-step explanation:

In this scenario, a test is being conducted to test the difference between two population "means" using data that are gathered from a matched pairs experiment. If the paired differences are normal, then the distribution used for testing is the student t-distribution.

In Statistics and probability, a student t-distribution can be defined as the probability distribution which can be used to estimate population parameters when the population variance is not known (unknown) and the sample population is relatively small. The student t-distribution is a statistical distribution which was published in 1908 by William Sealy Gosset.

A student t-distribution has a similar curve with the normal distribution curve, except that it is fatter and a little bit shorter.

Answer:

a) 40 dollars

b) 480 dollars

Step-by-step explanation:

Given the average property tax on a three bedroom home in a certain community modelled by the equation T(x) =20x²+40x+600, the rate at which the property tax is increasing with respect to time in 2008 can be derived by solving for the function T'(x) at x=0

T'(x) = 2(20)x¹ + 40x° + 0

T'(x) = 40x+40

At x = 0,

T'(0) = 40(0)+40

T'(0) = 40

Hence the property tax was increasing at a rate of 40dollars with respect to the initial year (2008).

b) There are 4 years between 2008 and 2012. To know how much that the tax change between the years 2008 and 2012, we will find T(4) - T(0)

Given T(x) =20x²+40x+600

T(4) =20(4)²+40(4)+600

T(4) = 320+160+600

T(4) = 1080 dollars

Also T(0) =20(0)²+40(0)+600

T(0) = 0+0+600

T(0)= 600 dollars

T(4) - T(0) = 1080 - 600

T(4) - T(0) = 480 dollars

Hence, the tax has changed by $480 between 2008 and 2012

Answer:

The number of ways is  [tex]\left n} \atop {}} \right. P_r = 336[/tex]

Step-by-step explanation:

From the question we are told that

   The number of tickets are   [tex]n = 8[/tex]

    The number of finalist are [tex]r =3[/tex]

Generally the number of way by which this winners can be drawn and arrange in the order of   [tex]1^{st} , \ 2nd , \ 3rd[/tex]    is mathematically represented as

             [tex]\left n} \atop {}} \right. P_r = \frac{n\ !}{(n-r) !}[/tex]

substituting values

             [tex]\left n} \atop {}} \right. P_r = \frac{ 8!}{(8-3) !}[/tex]

           [tex]\left n} \atop {}} \right. P_r = \frac{ 8* 7*6*5*4*3*2*1}{ 5*4*3*2*1}[/tex]

           [tex]\left n} \atop {}} \right. P_r = 336[/tex]

Answer:

It sold 14 cans boxes of food and 12 cans of food.

Step-by-step explanation:

The factor for the food cans depend upon every seven food boxes .So, the same no. of sets of food cans will be sold.

Let the no. of sets of food boxes be x.

According to the question,

6x+7x=26

13x=26

x=26/13

x=2

No. of food cans =6x=6×2=12 cans

No. of food boxes=7x=7×2=14 boxes

Please mark brainliest ,if it is truly the best ! Thank you!

Answer:

D. the coefficient of determination is .81.

Step-by-step explanation:

SST = SSE + SSR

where

SST is the summation of square total

SSE is the summation of squared error estimate = 608

SSR is the summation of square of residual = 2593

with these in mind we put the values into the formula

= 2592 + 608

=3200

Coefficient of determination = SSR/SST

= 2592/3200

= 0.81

Therefore option D is the correct answer to the question.

Answer:

[tex]\large \boxed{\mathrm{34.2}}[/tex]

Step-by-step explanation:

[tex]\sf B= arcsin (\frac{b \times sin(A)}{a} )[/tex]

[tex]\sf B= arcsin (\frac{7 \times sin(40\°)}{8} )[/tex]

[tex]\sf B = 0.59733 \ rad = 34.225\°[/tex]

Answer:

A)0.126775 B)0.000004325376 C) 0.07548

Step-by-step explanation:

Given the following :

A.) a. n = 15, p = 0.4, find P(4 successes)

a = number of trials p=probability of success

P(4 successes) = P(x = 4)

USING:

nCx * p^x * (1-p)^(n-x)

15C4 * 0.4^4 * (1-0.4)^(15-4)

1365 * 0.0256 * 0.00362797056

= 0.126775

B)

b. n = 12, p = 0.2, find P(2 failures),

P(2 failures) = P(12 - 2) = p(10 success)

USING:

nCx * p^x * (1-p)^(n-x)

12C10 * 0.2^10 * (1-0.2)^(12-10)

66 * 0.0000001024 * 0.64

= 0.000004325376

C) n = 20, p = 0.05, find P(at least 3 successes)

P(X≥ 3) = p(3) + p(4) + p(5) +.... p(20)

To avoid complicated calculations, we can use the online binomial probability distribution calculator :

P(X≥ 3) = 0.07548

Answer:

The correct option is  A

Step-by-step explanation:

From the question we are told that

    The  population is  [tex]\mu = 6.6[/tex]

     The level of significance is [tex]\alpha = 5\% = 0.05[/tex]

      The sample data is  9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds

The Null hypothesis is [tex]H_o : \mu = 6.6[/tex]

 The Alternative hypothesis is  [tex]H_a : \mu > 6.6[/tex]

The critical value of the level of significance obtained from the normal distribution table is

                       [tex]Z_{\alpha } = Z_{0.05 } = 1.645[/tex]

Generally the sample mean is mathematically evaluated as

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

substituting values

      [tex]\=x = \frac{9.0 + 7.3 + 6.0+ 8.8+ 6.8+ 8.4+6.6 }{7}[/tex]

      [tex]\=x = 7.5571[/tex]

The standard deviation is mathematically evaluated as

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

substituting values

          [tex]\sigma = \sqrt{\frac{ [ 9.0-7.5571]^2 + [7.3 -7.5571]^2 + [6.0-7.5571]^2 + [8.8- 7.5571]^2 + [6.8- 7.5571]^2 + [8.4 - 7.5571]^2+ [6.6- 7.5571]^2 }{7} }[/tex][tex]\sigma = 1.1774[/tex]

Generally the test statistic is mathematically evaluated as

            [tex]t = \frac{\= x - \mu } { \frac{\sigma }{\sqrt{n} } }[/tex]

substituting values

           [tex]t = \frac{7.5571 - 6.6 } { \frac{1.1774 }{\sqrt{7} } }[/tex]

            [tex]t = 1.4274[/tex]

Looking at the value of  t and  [tex]Z_{\alpha }[/tex]   we see that [tex]t < Z_{\alpha }[/tex] hence we fail to reject the null hypothesis

  What this implies is that there is no sufficient evidence to state that the sample data show as significant increase in the average birth rate

The conclusion is that the mean is  [tex]\mu = 6.6 \ lb[/tex]

Answer:

40

Step-by-step explanation:

In the question only lies the answer:

"and a total of forty students plan to participate in cross country."

Answer:

2

Step-by-step explanation:

2

Answer:

  3x +4 = -2x -1

Step-by-step explanation:

The line that goes up to the right has a y-intercept of +4. This is where it crosses the y-axis. It's slope (rise/run) is 3/1 = 3, so its equation in slope-intercept form is ...

  y = mx +b . . . . where m is the slope, b is the y-intercept

  y = 3x +4

The other line has a negative slope and a y-intercept of -1. The slope of that line is rise/run = -2/1 = -2, so its equation is ...

  y = -2x -1

__

The solution point will have the x-coordinate that is the solution of the equation ...

  y = y

  3x +4 = -2x -1 . . . . . . substituting the above expressions for y.

Answer:

  linear with a common first difference of 2

Step-by-step explanation:

On the face of it, you can reject answers that ascribe a common ratio to a linear or quadratic function. (A common ratio is characteristic of an exponential function.)

You can also reject the answer that ascribes a common first difference to a quadratic function. (A quadratic function has a common second difference.)

After you reject the nonsense answers, there is only one remaining choice. It is also the correct one:

  linear with a common first difference of 2

_____

The ratio of change in y to change in x is ...

  (0 -(-2))/(-2 -(-3)) = 2

  (4 -0)/(0 -(-2)) = 2

  (12 -4)/(4 -0) = 2

That is, y increases by 2 when x increases by 1. The common first difference is 2.

Answer:

  see the attachment

Step-by-step explanation:

When you have a number of function evaluations to do, it is convenient to let a graphing calculator or spreadsheet do them. That avoids the tedium and the mistakes in arithmetic.

Here's your completed table.

Answer:

x = 6

Step-by-step explanation:

In the third line of the solution on right side of the equal sign, middle term should be 8x instead of 4x.

The final value of x will be 6.

[tex] PQ^2 + QO^2 = PO^2 \\

x^2 + 8^2 = (4+x)^2 \\

x^2 + 64 = 16 + 8x + x^2 \\

64 = 16 + 8x \\

64 - 16 = 8x \\

48 = 8x \\

6 = x\\[/tex]

Answer:

Step-by-step explanation:

Lets, turn this into words and use order of operations, First, we look for multiplication and division.

the sum of one fourth of 5 times of 8 and 10 gets you 1/4(5*8) + 10 = 20

what is the number if 4 is subtracted from the sum

20 - 4 = 16

Answer:

Linear

Step-by-step explanation:

Given

Height of a tree grows by 2.5 feet

Required

Determine the type of relationship

Take for instance, the height of the tree at year 1 is x

At year 2, it will be x + 2 * 1

At year 3, it will be x + 2 * 2

At year 4, it will be x + 2 * 3

Following same pattern

At year n, it will be x + 2 *(n - 1)

Hence, growth rate = x + 2(n -1)

From the list of given options, the correct answer is Linear because the derived formula above is an example of a linear equation

Answer:

The probability  is  [tex]P(x < 13) = 0.8732[/tex]

Step-by-step explanation:

From the question we are told that

    The  probability of success is    p = 0.70

     The  sample size is  [tex]n = 15[/tex]

Generally the distribution of U.S. households have vcrs follow a binomial distribution given that there are only two outcome (household having vcrs or household not having vcrs )

The probability of failure is mathematically evaluated as

       [tex]q = 1- p[/tex]

substituting values

      [tex]q = 1- 0.70[/tex]

      [tex]q = 0.30[/tex]

The probability that fewer than 13 have vcrs is mathematically represented as

          [tex]P(x < 13) = 1- [P(13) + P(14) + P(15)][/tex]

=>     [tex]P(x < 13) = 1-[( \left 15 } \atop {}} \right. C_{13} *p^{13}* q^{15-13})+ (\left 15 } \atop {}} \right. C_{14} *p^{14}* q^{15-14}) +( \left 15 } \atop {}} \right. C_{15} *p^{15}* q^{15-15}) ][/tex]

 Here  [tex]\left 15 } \atop {}} \right. C_{13}[/tex] means  15 combination 13 and the value is  105 (obtained from calculator)

 Here  [tex]\left 15 } \atop {}} \right. C_{14}[/tex] means  15 combination 14 and the value is  15 (obtained from calculator)

 Here  [tex]\left 15 } \atop {}} \right. C_{15}[/tex] means  15 combination 15 and the value is  1 (obtained from calculator)

So

 [tex]P(x < 13) = 1-[(105 *p^{13}* q^{2})+ (15 *p^{14}* q^{1}) +(1*p^{15}* q^{0}) ][/tex]

substituting values      

 [tex]P(x < 13) = 1-[(105 *(0.70)^{13}* (0.30)^{2})+ (15 *(0.70)^{14}* (0.30)^{1}) +(1*(0.70)^{15}* (0.30)^{0}) ][/tex]

 [tex]P(x < 13) = 0.8732[/tex]

Answer:

7/6

Step-by-step explanation:

The LCD of these three fractions is 6; the denominators 3, 3 and 6 divide evenly into 6.

Therefore we have:

4/6 + 2/6 + 1/6 = 7/6

Answer:

[tex]f(-2)=33\\f(-1)=12\\f(0)=1\\f(1)=0\\f(2)=9[/tex]

Step-by-step explanation:

[tex]f(x)=5x^{2} -6x+1\\f(-2)=5(-2)^{2} -6(-2)+1\\f(-2)=5(4)+12+1\\f(-2)=20+13\\f(-2)=33[/tex]

[tex]f(x)=5x^{2}-6x+1\\f(-1)=5(-1)^{2} -6(-1)+1\\f(-1)=5(1)+6+1\\f(-1)=5+7\\f(-1)=12[/tex]

[tex]f(x)=5x^{2}-6x+1\\f(0)=5(0)^{2}-6(0)+1\\f(0)=5(0)-0+1\\f(0)=0+1\\f(0)=1[/tex]

[tex]f(x)=5x^{2}-6x+1\\f(1)=5(1)^{2}-6(1)+1\\f(1)=5(1)-6+1\\f(1)=5-5\\f(1)=0[/tex]

[tex]f(x)=5x^{2}-6x+1\\f(2)=5(2)^{2}-6(2)+1\\f(2)=5(4)-12+1\\f(2)=20-11\\f(2)=9[/tex]