Assume that adults have IQ scores that are normally distributed with a mean of and a standard deviation . Find the probability that a randomly selected adult has an IQ between 81 and 119 .

[tex]a=\dfrac{2}{3}\\r=\dfrac{1}{2}[/tex]

The sum exists if [tex]|r|<1[/tex]

[tex]\left|\dfrac{1}{2}\right|<1[/tex] therefore the sum exists

[tex]\displaystyle\\\sum_{k=0}^{\infty}ar^k=\dfrac{a}{1-r}[/tex]

[tex]\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{1}{6}+\ldots=\dfrac{\dfrac{2}{3}}{1-\dfrac{1}{2}}=\dfrac{\dfrac{2}{3}}{\dfrac{1}{2}}=\dfrac{2}{3}\cdot 2=\dfrac{4}{3}[/tex]

Answer:

x=3mod4

Means that when x is divided by 4 it gives an unknown integer and a remainder of 3.

x/4 = Z + 3/4

Z= (x-3)/4

Where Z is the integer

x=5 mod6

x/6 = Y + 5/6

Y = (x-5)/6

Where Y is the integer

Z-Y must be an integer on equal to zero

(x-3)/4 - (x-5)/6

3(x-3)/12 - 2(x-5)/12

(3x-9-2x+10)/12

(x+1)/12

If it is equal to 0

x=-1. But x should be positive

If it is equal to 1

x=11

Hence the smallest possible number is 11

Answer:

$198

Step-by-step explanation:

198x.07=13.86

198+13.86=211.86

Answer:

point A is the rays endpoint

Step-by-step explanation:

Answer:

The "endpoint" of a ray is the origin point of the ray, or the point at which the ray starts.

Step-by-step explanation:

A ray starts at a given point, the endpoint, and then goes in a certain direction forever ad infinitum.  The origin point of a ray is called "the endpoint".

Cheers.

Answer:

It is undefined.

Step-by-step explanation:

Let's take a look at the first equation- if we simplify and move the terms, it becomes sqrt of 6 = 0, which results in an undefined value of x. The second equation works with x=0 but not the first so the value of x is undefined.

Answer:

x = [tex]\frac{1}{2}[/tex]

Step-by-step explanation:

[tex]\frac{3x-2}{2}[/tex] = [tex]\frac{x-1}{2}[/tex]

Cross-multiply:

2(3x-2) = 2(x-1)

Simplify:

6x - 4 = 2x - 2

Subtract 2x from both sides:

4x - 4 = -2

Add 4 to both sides:

4x = 2

Divide both sides by 4:

x = [tex]\frac{1}{2}[/tex]

Answer:

3/10

Step-by-step explanation:

6+3+1=10

since there are 3 blue marbles, we put the 3 into the place of the numerator

and since there is 10 marbles in total it goes into the denominator

The probability that a randomly selected marble is not blue will be 0.70.

Its basic premise is that something will almost certainly happen. The percentage of favorable events to the total number of occurrences.

A bag contains 6 red marbles, 3 blue marbles and 1 green marble.

The total number of the event will be

Total event = 6 + 3 + 1

Total event = 10

Then the probability that a randomly selected marble is not blue will be

Favorable event = 7 {red, green}

Then the probability will be

P = 7 / 10

P = 0.70

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Answer:

the value of the 3 is 30

Step-by-step explanation:

the second digit to the left of a decimal is always tens column

Answer:

75

Step-by-step explanation:

4:3

4x25=100

3x25=75

Bold = changed words

1. We play tennis every Sunday.

2. I own two dogs and a cat. I love animals.

3. My suitcase weighs four kilos (kilograms).

4. When Mary came in, I talked to my mother on the phone. OR: I talked to Mother on the phone when Mary came in.

5. We passed the hotel two minutes ago. OR: We passed by the hotel two minutes ago.

Answer: 56.9 years to 63.1 years.

Step-by-step explanation:

Confidence interval for population mean (when population standard deviation is unknown):

[tex]\overline{x}\pm t_{\alpha/2}{\dfrac{s}{\sqrt{n}}}[/tex]

, where [tex]\overline{x}[/tex]= sample mean, n= sample size, s= sample standard deviation, [tex]t_{\alpha/2}[/tex]= Two tailed t-value for [tex]\alpha[/tex].

Given: n= 24

degree of freedom = n- 1= 23

[tex]\overline{x}[/tex]= 60 years

s= 7.4 years

[tex]\alpha=0.05[/tex]

Two tailed t-critical value for significance level of [tex]\alpha=0.05[/tex] and degree of freedom 23:

[tex]t_{\alpha/2}=2.0687[/tex]

A 95% confidence interval on the true mean age:

[tex]60\pm (2.0686){\dfrac{7.4}{\sqrt{24}}}\\\\\approx60\pm3.1\\\\=(60-3.1,\ 60+3.1)\\\\=(56.9,\ 63.1)[/tex]

Hence, a 95% confidence interval on the true mean age. : 56.9 years to 63.1 years.

Answer:

248.26 cm²

Step-by-step explanation:

Surface area of the composite figure = (surface area of cuboid + surface area of hemisphere) - 2(base area of hemisphere)

Surface area of cuboid = [tex] 2(lw + lh + hw) [/tex]

Where,

l = 10 cm

w = 5 cm

h = 4 cm

Plug in the values into the formula:

[tex] SA = 2(10*5 + 10*4 + 4*5) [/tex]

[tex] SA = 2(50 + 40 + 20) [/tex]

[tex] SA = 2(110) = 220 cm^2 [/tex]

Surface area of hemisphere = 3πr²

Where,

π = 3.14

r = 3 cm

SA of hemisphere = 3*3.14*3² = 3*3.14*9 = 84.78 cm²

Base area of hemisphere = πr²

BA = 3.14*3² = 3.14*9 = 28.26 cm²

Surface area of the composite shape = (220 + 84.78) - 2(28.26)

= 304.78 - 56.52

SA = 248.26 cm²

Answer:

8 home games and 10 away games

Step-by-step explanation:

Total home goals

= 8+5+9+8

= 30

Number of home games

= 30/3.75

= 8

Total away game goals

= 7+8+4+5

= 24

Number of away games

= 24/2.4

= 10

Answer:

i think it is 8 home and 10 away matches

Step-by-step explanation:

the corresponding angles should be equal

so, [tex] 5x+15=90 \implies 5x=75\implies x=15^{\circ}[/tex]

Hey there! I'm happy to help!

We want to isolate y on one side of the equation to see what it equals. To do this, we use inverse operations to cancel out numbers on the y side and find the correct value.

1/3y+4=16

We subtract 4 from both sides, canceling out the +4 on the right but keeping the same y-value by doing the same to the other side.

1/3y=12

We divide both sides by 1/3 (which is multiplying both sides by 3) which will cancel out the 1/3 and tell us what y is equal to.

y=36

Now you know how to solve basic equations! Have a wonderful day! :D

Answer:

14 miles

Step-by-step explanation:

Since we know that the distance of the paths from Cedarburg to Allenville is 22 and 13/16 miles, and we know the distance from Cedarburg to Lakeside is 8 and 13/16 miles.

We know that the total distance is made up of the distance from C to L and L to A.

So 22 and 13/16 = 8 and 13/16 + L to A

We can subtract 22 and 13/16 by 8 and 13/16 to get 14 miles.

Hope this helps.

Answer:

Stratified Random sampling.

Step-by-step explanation:

As per the scenario, It is stratified random sampling as it divides students into strata which represent Sophomores, Juniors, and Seniors.

Simple random samples of the given sizes of the proportional to the size of the stratum which is to be taken from every stratum that is to be about 10 percent of students from every class that is selected here.

Hence, according to the given situation, the correct answer is a random stratified sampling.

Answer:

  C.  As x → –∞, y → +∞, and as x → +∞, y → +∞

Step-by-step explanation:

The leading coefficient of this even-degree function is positive, so y goes to +∞ when the magnitude of x gets large.

_____

When the function is even degree, its value for large magnitude x heads toward the infinity with the same sign as the leading coefficient.

When the function is odd degree, its value for large magnitude x will head toward the infinity with the sign that matches the product of the sign of x and the sign of the leading coefficient.

Answer:

a). Function (4)

b). Function (2)

c). Function (3)

Step-by-step explanation:

Characteristics of the functions given,

Function (1),

Form the given graph,

Slope = [tex]\frac{\text{Rise}}{\text{Run}}[/tex]

          = [tex]-\frac{4}{1}[/tex]

          = -4

Y- intercept of the given function = 2

Function (2),

From he given table,

Slope = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]

         = [tex]\frac{5-3}{0+1}[/tex]

         = 2

y-intercept = 5 [Value of y for x = 0]

Function (3),

y = -x - 1

By comparing this equation with y = mx + b

Where 'm' = slope

and b = y-intercept

Slope = (-1)

y-intercept = (-1)

Function (4),

Slope = 5

y-intercept = (-4)

(a). Greatest slope of the function → Function (4)

(b). y-intercept greater than 3 → Function (2)

(c). Function with y-intercept closest to 0 → Function (3)

Answer:

(A) A 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval would increase.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval would decrease.

(D) A 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477]  .

Step-by-step explanation:

We are given that 19 students are randomly selected the sample mean was $390 and the standard deviation was $120.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, [tex]\bar X[/tex] = sample mean = $390

            s = sample standard deviation = $120

            n = sample of students = 19

            [tex]\mu[/tex] = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;

P(-2.101 < [tex]t_1_8[/tex] < 2.101) = 0.95  {As the critical value of t at 18 degrees of

                                               freedom are -2.101 & 2.101 with P = 2.5%}  

P(-2.101 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.101) = 0.95

P( [tex]-2.101 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.101 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

P( [tex]\bar X-2.101 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.101 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.101 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.101 \times {\frac{s}{\sqrt{n} } }[/tex] ]

                        = [ [tex]\$390-2.101 \times {\frac{\$120}{\sqrt{19} } }[/tex] , [tex]\$390+2.101 \times {\frac{\$120}{\sqrt{19} } }[/tex] ]

                        = [$332.16, $447.84]

(A)  Therefore, a 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval which is [tex]Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }[/tex] would increase because of an increase in the z value.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval which is [tex]Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }[/tex]  would decrease because as denominator increases; the whole fraction decreases.

(D) We are given that to estimate the proportion of students who purchase their textbooks used, 500 students were sampled. 210 of these students purchased used textbooks.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

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

where, [tex]\hat p[/tex] = sample proportion students who purchase their used textbooks = [tex]\frac{210}{500}[/tex] = 0.42    

            n = sample of students = 500

            p = population proportion

Here for constructing a 99% confidence interval we have used a One-sample z-test statistics for proportions

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99  {As the critical value of z at 0.5%

                                               level of significance are -2.58 & 2.58}  

P(-2.58 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 2.58) = 0.99

P( [tex]-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99

P( [tex]\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99

99% confidence interval for p = [ [tex]\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ]

= [ [tex]0.42 -2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }[/tex] , [tex]0.42 +2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }[/tex] ]

= [0.363, 0.477]

Therefore, a 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477]  .

Answer:

Yes, it is invertible

Step-by-step explanation:

We need to find in the matrix determinant is different from zero, since iif it is, that the matrix is invertible.

Let's use co-factor expansion to find the determinant of this 4x4 matrix, using the column that has more zeroes in it as the co-factor, so we reduce the number of determinant calculations for the obtained sub-matrices.We pick the first column for that since it has three zeros!

Then the determinant of this matrix becomes:

[tex]4\,*Det\left[\begin{array}{ccc}1&4&6\\0&3&8\\0&0&1\end{array}\right] +0+0+0[/tex]

And the determinant of these 3x3 matrix is very simple because most of the cross multiplications render zero:

[tex]Det\left[\begin{array}{ccc}1&4&6\\0&3&8\\0&0&1\end{array}\right] =1 \,(3\,*\,1-0)+4\,(0-0)+6\,(0-0)=3[/tex]

Therefore, the Det of the initial matrix is : 4 * 3 = 12

and then the matrix is invertible

Answer:

The  95% confidence interval is  [tex]0.503 < p < 0.535[/tex]

The  interpretation is that there is 95% confidence that the true population proportion lie within the confidence interval

Step-by-step explanation:

From the question we are told that

    The  sample size is n  =  1003

     The number that indicated television are a luxury is  k  =  521

Generally the sample mean is mathematically represented as

           [tex]\r p = \frac{k}{n}[/tex]

          [tex]\r p = \frac{521}{1003}[/tex]

         [tex]\r p = 0.519[/tex]

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

       [tex]\alpha = 100 - 95[/tex]

       [tex]\alpha = 5\%[/tex]

       [tex]\alpha = 0.05[/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} } = 1.96[/tex]

The  margin of error is mathematically represented as

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

=>       [tex]E = 1.96 * \sqrt{ \frac{ 0.519 (1- 0.519 )}{1003} }[/tex]

=>       [tex]E = 0.016[/tex]

The  95%  confidence interval is mathematically represented as

       [tex]\r p -E < p < \r p +E[/tex]

=>   [tex]0.519 - 0.016 < p < 0.519 + 0.016[/tex]

=>    [tex]0.503 < p < 0.535[/tex]

Answer:

[tex]\huge\boxed{x+y=-2}[/tex]

Step-by-step explanation:

The standard form of an equation of a line:

[tex]Ax+By=C[/tex]

The point-slope form of an equation of a line:

[tex]y-y_1=m(x-x_1)[/tex]

where

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

We have two points (-2, 0) and (1, -3).

Substitute:

[tex]x_1=-2;\ y_1=0;\ x_2=1;\ y_2=-3[/tex]

[tex]m=\dfrac{-3-0}{1-(-2)}=\dfrac{-3}{1+2}=\dfrac{-3}{3}=-1\\\\y-0=-1(x-(-2))\\\\y=-(x+2)[/tex]

[tex]y=-x-2[/tex]         add x to both sides

[tex]x+y=-2[/tex]

Step-by-step explanation:

Hello, there!!!

Let ABC be a Right angled triangle,

where, AB = 3

BC= 10

and AC= x

now,

As the triangle is a Right angled triangle, taking angle C asrefrence angle. we get,

h= AC = x

p= AB = 3

b= BC= 10

now, by Pythagoras relation we get,

[tex]h = \sqrt{ {p}^{2} + {b}^{2} } [/tex]

[tex]or ,\: h = \sqrt{ {3}^{2} + {10}^{2} } [/tex]

by simplifying it we get,

h = 10.44030

Therefore, the answer is x= 10.

Hope it helps...

Answer:

Step-by-step explanation:

Given that:

[tex]\mathsf{f(x) = x^2 -5 } \\ \\ \mathsf{x_1 = 2}[/tex]

The derivative of the first function of (x) is:

[tex]\mathsf{f'(x) =2x }[/tex]

According to Newton's Raphson method for function formula:

[tex]{\mathrm{x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}}[/tex]

where;

[tex]\mathbf{x_1 =2}[/tex]

The first iteration is as follows:

[tex]\mathtt{f(x_1) = (2)^2 - 5} \\ \\ \mathbf{f(x_1) = -1}[/tex]

[tex]\mathtt{f'(x_1) = 2(2)} \\ \\ \mathbf{ = 4}[/tex]

[tex]\mathtt{\dfrac{f(x_1)}{f'(x_1)}} = \dfrac{-1}{4}}[/tex]

[tex]\mathbf{\dfrac{f(x_1)}{f'(x_1)} =-0.25}[/tex]

[tex]\mathtt{x_1 - \dfrac{f(x_1)}{f'(x_1)}} = \mathtt{2 - (-0.25)}}[/tex]

[tex]\mathbf{x_1 - \dfrac{f(x_1)}{f'(x_1)} = 2.25}[/tex]

Therefore;

[tex]\mathbf{x_2 = 2.25}[/tex]

For the second iteration;

[tex]\mathtt f(x_2) = (2.25)^2 -5}[/tex]

[tex]\mathtt f(x_2) = 5.0625-5}[/tex]

[tex]\mathbf{ f(x_2) =0.0625}[/tex]

[tex]\mathtt{f'(x_2)= 2(2.25)}[/tex]

[tex]\mathbf{f'(x_2)= 4.5}[/tex]

[tex]\mathtt{ \dfrac{f(x_2)}{f'(x_2)}} = \dfrac{0.0625}{4.5}}[/tex]

[tex]\mathbf{ \dfrac{f(x_2)}{f'(x_2)} = 0.01389}[/tex]

[tex]\mathtt{x_2 - \dfrac{f(x_2)}{f'(x_2)}} = \mathtt{2.25 -0.01389}}[/tex]

[tex]\mathbf{x_2 - \dfrac{f(x_2)}{f'(x_2)} = 2.2361}}[/tex]

Therefore, [tex]\mathbf{x_3 = 2.2361}[/tex]

Answer:

Step-by-step explanation:

a is expressed in terms of b as a = 144/(b - 4).

If A is inversely proportional to (b - 4), we can express this relationship using the formula:

A = k/(b - 4),

where k is the constant of proportionality.

To determine the value of k, we can use the given information when a = 8 and b = 22:

8 = k/(22 - 4).

Simplifying the equation:

8 = k/18.

To isolate k, we multiply both sides of the equation by 18:

8 * 18 = k.

k = 144.

Now that we know the value of k, we can rewrite the equation in terms of b:

A = 144/(b - 4).

Therefore, a is expressed in terms of b as a = 144/(b - 4).

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Answer:

Please see the Step-by-step explanation for the answers

Step-by-step explanation:

1)

∑[tex]\left \ {{5} \atop {j=1}} \right.[/tex] 2j

The sum of series from j=1 to j=5 is:

∑ = 2(1) + 2(2) + 2(3) + 2(4) + 2(5)

  =  2 + 4 + 6 + 8 + 10

∑ = 30

2)

This question is not given clearly so i assume the following series that will give you an idea how to solve this:

∑[tex]\left \ {{4} \atop {k=1}} \right.[/tex] 2k²

The sum of series from k=1 to j=4 is:

∑ = 2(1)² + 2(2)² + 2(3)² + 2(4)²

  = 2(1) + 2(4) + 2(9) + 2(16)

  =  2 + 8 + 18 + 32

∑ = 60

∑[tex]\left \ {{4} \atop {k=1}} \right.[/tex] (2k)²

∑ = (2*1)² + (2*2)² + (2*3)² + (2*4)²

  = (2)² + (4)² + (6)² + (8)²

  = 4 + 16 + 36 + 64

∑ = 120

∑[tex]\left \ {{4} \atop {k=1}} \right.[/tex] (2k)²- 4

∑ = (2*1)²-4 + (2*2)²-4 + (2*3)²-4 + (2*4)²-4

  = (2)²-4 + (4)²-4 + (6)²-4 + (8)²-4

  = (4-4) + (16-4) + (36-4) + (64-4)

  = 0 + 12 + 32 + 60

∑ = 104

∑[tex]\left \ {{4} \atop {k=1}} \right.[/tex] 2k²- 4

∑ = 2(1)²-4 + 2(2)²-4 + 2(3)²-4 + 2(4)²-4

  = 2(1)-4 + 2(4)-4 + 2(9)-4 + 2(16)-4

  = (2-4) + (8-4) + (18-4) + (32-4)

  = -2 + 4 + 14 + 28

∑ = 44

3)

∑[tex]\left \ {{6} \atop {k=3}} \right.[/tex] (2k-10)

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)  

  = (6-10) + (8-10) + (10-10) + (12-10)

  = -4 + -2 + 0 + 2  

∑ = -4

4)

1+1/2+1/4+1/8+1/16+1/32+1/64

This is a geometric sequence where first term is 1 and the common ratio is 1/2 So

a = 1

This can be derived as

1/2/1 = 1/2 * 1 = 1/2

1/4/1/2 = 1/4 * 2/1 = 1/2

1/8/1/4 = 1/8 * 4/1  = 1/2

1/16/1/8 = 1/16 * 8/1  = 1/2

1/32/1/16 = 1/32 * 16/1  = 1/2

1/64/1/32 = 1/64 * 32/1  = 1/2

Hence the common ratio is r = 1/2

So n-th term is:

[tex]ar^{n-1}[/tex] = [tex]1(\frac{1}{2})^{n-1}[/tex]

So the answer that represents the series in sigma notation is:

∑[tex]\left \ {{7} \atop {j=1}} \right.[/tex] [tex](\frac{1}{2})^{j-1}[/tex]

5)

−3+(−1)+1+3+5

This is an arithmetic sequence where the first term is -3 and the common difference is 2. So  

a = 1

This can be derived as

-1 - (-3) = -1 + 3 = 2

1 - (-1) = 1 + 1 = 2

3 - 1 = 2

5 - 3 = 2

Hence the common difference d = 2

The nth term is:

a + (n - 1) d

= -3 + (n−1)2

= -3 + 2(n−1)

= -3 + 2n - 2

= 2n - 5

So the answer that represents the series in sigma notation is:

∑[tex]\left \ {{5} \atop {j=1}} \right.[/tex] (2j−5)

For (1) the sum is 30, for (2) the sum is 90, for (3) the sum is -4, for(4) the sigma notation is  [tex]\rm \sum j = 1(\frac{1}{2})^{j-1}\\[/tex]  where j = 1 to j = 7, and for (5) the sigma notation is  [tex]\rm\sum j = (2j-5)[/tex]  where j = 1 to j = 5.

We have different series in the question.

It is required to find the sum of all series.

In mathematics, a series can be defined as a group of data that followed certain rules of arithmetic.

1) We have:

[tex]\rm \sum j=2j[/tex]   where j = 1 to j = 5

After expanding the series, we get:

= 2(1)+2(2)+2(3)+2(4)+2(5)

=2(1+2+3+4+5)

= 2(15)

=30

2) We have:

[tex]\rm \sum k=(2k^2-4)[/tex]  where k = 1 to k = 4

After expanding the series, we get:

[tex]\rm = (2(1)^2-4)+(2(2)^2-4)+(2(3)^2-4)+(2(4)^2-4)+(2(5)^2-4)\\[/tex]

[tex]\rm = 2[1^2+2^2+3^2+4^2+5^2]-4\times5\\\\\rm=2[55]-20\\\\\rm = 90[/tex]

3) We have:

[tex]\rm \sum k= (2k-10)[/tex]  where k = 3 to k = 6

After expanding the series, we get:

[tex]= (2(3)-10)+(2(4)-10)+(2(5)-10)+(2(6)-10)\\\\=2[3+4+5+6] - 10\times4\\\\=2[18] - 40\\\\= -4[/tex]

4) The series given below:

[tex]1, \frac{1}{2} ,\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64}[/tex]

It is a geometric progression:

[tex]\rm n^t^h[/tex] for the geometric progression is given by:

[tex]\rm a_n = ar^{n-1}[/tex]

[tex]\rm a_n = 1(\frac{1}{2})^{n-1}\\\\\rm a_n = (\frac{1}{2})^{n-1}\\[/tex]

In sigma notation we can write:

[tex]\rm \sum j = 1(\frac{1}{2})^{j-1}\\[/tex]  where j = 1 to j = 7

5) The given series:

−3+(−1)+1+3+5, it is arithmetic series.

[tex]\rm n^t^h[/tex] for the arithmetic progression is given by:

[tex]\rm a_n = a+(n-1)d[/tex]

[tex]\rm a_n = -3+(n-1)(2)\\\\\rm a_n = 2n-5[/tex]

In sigma notation we can write:

[tex]\rm\sum j = (2j-5)[/tex]  where j = 1 to j = 5

Thus, for (1) the sum is 30, for (2) the sum is 90, for (3) the sum is -4, for(4) the sigma notation is  [tex]\rm \sum j = 1(\frac{1}{2})^{j-1}\\[/tex]  where j = 1 to j = 7, and for (5) the sigma notation is  [tex]\rm\sum j = (2j-5)[/tex]  where j = 1 to j = 5.

Learn more about the series here:

https://brainly.com/question/10813422

Answer:

{June, Janet, Jill, Justin, Jeffrey, Jelly,Irina, Irena, Arena, Arina, }

Step-by-step explanation:

A ∪ C

This means union so we join the sets together

A = {June, Janet, Jill, Justin, Jeffrey, Jelly} + C = {Irina, Irena, Arena, Arina, Jelly}

A U C =  {June, Janet, Jill, Justin, Jeffrey, Jelly,Irina, Irena, Arena, Arina, Jelly}

We get rid of repeats

A U C =  {June, Janet, Jill, Justin, Jeffrey, Jelly,Irina, Irena, Arena, Arina, }