find the 5th term in the sequence an=n÷n+1

You said    - 1/3 - 3/5 x  =  1/2

Multiply each side by 3 :

- 1 - 9/5 x  =  3/2

Multiply each side by 5 :

- 5 - 9x  =  15/2

Multiply each side by 2 :

- 10 - 18x = 15

Add 10 to each side :

- 18x  =  25

Divide each side by -18 :

x = - 25/18

or  x = - 1 and 7/18 (same thing)

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:

7

Step-by-step explanation:

10+4 = 14

14/2  = 7

x = 7

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:

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:

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:

Daniel sold 54 and Jack sold 42

Step-by-step explanation:

D = number of tickets that Daniel sold

J = number of tickets that Jack sold

D + J = 96

D = 12+ J

Substitute the second equation into the first equation

12 + J + J = 96

Combine like terms

12 + 2J = 96

Subtract 12 from each side

2J = 84

Divide by 2

J = 42

D = J+12

D = 54

Daniel sold 54 and Jack sold 42

Answer:

Jack sold 42 & Daniel sold 54.

Step-by-step explanation:

96 - 12 = 84

84 / 2 = 42

Jack sold 42.

42 + 12 = 54

Daniel sold 54.

42 + 54 = 96

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:

you could stand at 5.0 ft and still be completely in the shadow of the tree

Step-by-step explanation:

From the diagram attached below;

We consider;

[tex]\overline {BC}[/tex] to be the height of the tree and [tex]\overline {DE}[/tex] to be the height of how tall you could be and still be completely in the shadow of the tree.

∠D = ∠B = 90°

Also;

ΔEAD = ΔBAC   (similar triangles)

Therefore, their sides will also be proportional

i.e

[tex]\dfrac{\overline {DE}}{ \overline {BC}}= \dfrac{\overline{AD}}{ \overline{AC}}[/tex]

[tex]\dfrac{x}{ 45}= \dfrac{225-220}{225}[/tex]

[tex]\dfrac{x}{ 45}= \dfrac{25}{225}[/tex]

By cross multiply

225x = 45 × 25

[tex]x = \dfrac{45 \times 25}{225}[/tex]

[tex]x = \dfrac{1125}{225}[/tex]

x = 5.0 ft

Therefore, you could stand at 5.0 ft and still be completely in the shadow of the tree

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:

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:

[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]

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:

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:

[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:

√5

Step-by-step explanation:

[tex](2 ,-5) = (x_1,y_1)\\(3,-7)=(x_2,y_2)\\\\d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ \\d = \sqrt{(3-2)^2 +(-7-(-5))^2}\\ \\d = \sqrt{(1)^2+(-7+5)^2}\\ \\d = \sqrt{(1)^2 + (-2)^2}\\ \\d = \sqrt{1 +4}\\ \\d = \sqrt{5}[/tex]

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:

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:

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:

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:

  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:

nd is c

Step-by-step explanation:

Answer:

mean=87

median=87

Step-by-step explanation:

mean=sum of test score/number of subject

mean=79+91+93+85+86+88/6

mean=522/6

mean=87

Literal meaning of median is medium.

To find the number which lies in the medium, we must rearrange the number in ascending.

79, 91, 93, 85, 86, 88

79, 85, 86, 88, 91, 93

86+88/2=87

Hope this helps ;) ❤❤❤

Let me know if there is an error in my answer.

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:

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:

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: When the addition of two or more equations leads to the elimination of one of the variables.

Step-by-step explanation:

When we have a system of equations, the addition method seems to be useful only when adding the equations will lead to the elimination of one of the variables:

An example of this can be, for the variables x and y:

3*x + x*y - 2*y = 3

x^2 + x*y - 2y = 42

now we can "add" (actually subtract) the equations and get (eq2 minus eq1)

(x^2 + x*y - 2y) - (3*x + x*y - 2*y ) = 42 - 3

x^2 - 3*x = 39

x^2 - 3*x - 39 = 0

And now we can solve it for x, and then find the value of y.

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:

11/(x + 1) thus d: is the answer

Step-by-step explanation:

Simplify the following:

(11 (x + 8))/((x + 1) (x + 8))

(11 (x + 8))/((x + 1) (x + 8)) = (x + 8)/(x + 8)×11/(x + 1) = 11/(x + 1):

Answer: 11/(x + 1)