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3 years ago

Sn=7k=1Σ[1+ (k-1)(2)]

Mathematics

1 answer:

levacccp [35]3 years ago

5 0

Answer:

Step-by-step explanation:

I think I have read this right!

You let me know if you did not mean to write the following:

$\sum_{k=1}^{7}(1+(k-1)(2)$

Alright so the lower limit is 1 and the upper limit is 7.

All this means is we are going to use the expression 1+(k-1)(2) and evaluate it for each natural number between k=1 and k=7 and at both k=1 and k=7.

The sigma thing means we add those results.

So let's start.

Evaluating the expression at k=1: 1+(1-1)(2)=1+(0)(2)=1+0=1.

Evaluating the expression at k=2: 1+(2-1)(2)=1+(1)(2)=1+2=3.

Evaluating the expression at k=3: 1+(3-1)(2)=1+(2)(2)=1+4=5.

Evaluating the expression at k=4: 1+(4-1)(2)=1+(3)(2)=1+6=7.

Evaluating the expression at k=5: 1+(5-1)(2)=1+(4)(2)=1+8=9.

Evaluating the expression at k=6: 1+(6-1)(2)=1+(5)(2)=1+10=11.

Evaluating the expression at k=7: 1+(7-1)(2)=1+(6)(2)=1+12=13.

Now for the adding!

1+3+5+7+9+11+13

4+ 12+ 20+13

16+ 33

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Use the following image to assist in your answer. a = - 6 - 9 - 4

vfiekz [6]

Step-by-step explanation:

Pythagoras' theorem for the smallest one :

$c^2 = 4^2 + 6^2$

$c^2 = 16 + 36$

$c^{2}$ = 52

Pythagoras' theorem for the middle one :

$b^{2}$ = $6^{2}$ + $a^{2}$

Pythagoras' theorem for the biggest one :

$(4+a)^2 = c^2 + b^2$

$16 + 8a + a^2 = 52 + b^2$

Using the formula before (for $b^2$ ) it becomes :

$16 + 8a + a^2 = 52 + (6^2 + a^2)$

$16 + 8a + a^2 = 52 + 6^2 + a^2$

$16 + 8a = 52 + 6^2$

16 + 8a = 52 + 36

16+8a = 88

8a = 88-16

8a = 72

$a = \frac{72}{8}$

a = 9

Verifying :

$b^2 = 6^2 + a^2$

$b^2 = 36 + 81$

$b^2 = 117$

$b^{2}$ = 117

The biggest one :

$(4+a)^2 = c^2 + b^2$

$(4+9)^2 = 52 + 117$

$13^2 = 169$

True

8 0

3 years ago

Solve the 3x3 system shown below. Enter the values of x, y, and z.

slamgirl [31]

Answer:

x = 1

y = -1

z = 2

Step-by-step explanation:

You have the following system of equations:

$x+2y-z=-3\ \ \ \ \ (1)\\\\2x-y+z=5\ \ \ \ \ \ (2)\\\\x-y+z=4\ \ \ \ \ (3)$

First, you can subtract euqation (3) to equation (1):

x + 2y - z = -3

-x +y -z = - 4

0 3y -2z = -7 (4)

Next, you can multiply equation (3) by 2 and subtract it to equation (2):

2[ x -y + z = 4]

-2x +y -z = -5

0 -y + z= 3 (5)

You multiply equation (5) by 2 and sum (5) with (4):

2[ -y + z = 3]

3y -2z= -7

y + 0 = -1

Then y = -1

Next, you replace y=-1 in (5) to obtain z:

-(-1) + z = 3

z = 2

Finally, you can replace z and y in the equation (3) to obtain x:

x - (-1) + (2) = 4

x = 1

7 0

3 years ago

Most people Just most people in the nine states use feet in miles to measure distance in there every day lives why do you scient

babunello [35]

The third answer should be the right one

8 0

3 years ago

The director of admissions at Kinzua University in Nova Scotia estimated the distribution of student admissions for the fall sem

wariber [46]

Answer:

In order to calculate the expected value we can use the following formula:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And if we use the values obtained we got:

$E(X)=(1060*0.5) +(1400*0.1) +(1620*0.4)=1318$

Step-by-step explanation:

Let X the random variable that represent the number of admisions at the universit, and we have this probability distribution given:

X 1060 1400 1620

P(X) 0.5 0.1 0.4

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.

In order to calculate the expected value we can use the following formula:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And if we use the values obtained we got:

$E(X)=(1060*0.5) +(1400*0.1) +(1620*0.4)=1318$

3 0

3 years ago

Can someone. help me solve this pls .

olchik [2.2K]

Answer:

1.108

$729 = {n}^{6} \\ \sqrt[6]{729} = \sqrt[6]{ {n}^{6} } \\ n = 1.108$

7 0

3 years ago