Answer:
The circumference is 50.24 in. and the area is 200.96 in².
Step-by-step explanation:
The circumference formula is C = 2πr where C = Circumference, π = pi and r = radius. We know that r = 8 and π = 3.14 and that we're solving for C, so we can substitute those values into the equation to get C = 2 * 3.14 * 8 = 50.24 in.
The area formula is A = πr² where A = Area, π = pi and r = radius. Again, we're solving for A and we know that r = 8 and π = 3.14 so A = 3.14 * 8² = 3.14 * 64 = 200.96 in².
Answer:
The circumference is 50.24 in. and the area is 200.96 in².
Step-by-step explanation:
MARK SNOG AS BRAINLIEST
When x€Q, what is the solution of 3x-2/2=x-1/2 ?
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]
A rectangular parcel of land has an area of 6,000 ft2. A diagonal between opposite corners is measured to be 10 ft longer than one side of the parcel. What are the dimensions of the land, correct to the nearest foot? ft (smaller value) by ft (larger value)
Answer:
50ft by 120ft
Step-by-step explanation:
Area of a rectangle = L × W
6000ft² = L × W
L = 6000/W
When a diagonal line divides a rectangle into 2 right angled triangles, the diagonal line = Hypotenuse of either of the triangle and it is the longest side.
The formula for a right angle triangle =
a² + b² = c²( c = hypotenuse)
We are told in the question that:
A diagonal between opposite corners is measured to be 10 ft longer than one side of the parcel
Let us assume the side that the hypotenuse is longer than = Width
Hence, the Diagonal = (W + 10)²
Therefore
L² + W² = (W + 10)²
Since L = 6000/W
W² + (6000/W)² = (W + 10)²
W² + (6000/W)² = (W + 10) (W + 10)
W² + (6000/W)² = W² + 10W + 10W + 100
W² + (6000/W)² = W² + 20W + 100
W² - W² + (6000/W)² = 20W+ 100
6000²/W² = 20W + 100
6000² = W²( 20W + 100)
6000² = 20W³ + 100W²
20W³ + 100W² - 6000² = 0
20W³ + 100W² - 36000000 = 0
20(W³ + 5W² - 1800000) = 0
Factorising the quadratic equation,
20(W − 120)(W² + 125W + 15000) = 0
W - 120 = 0
W = 120
Therefore,
W(Width) = 120feet
Since the Width = 120 feet
We can find the length
6000ft² = L × W
L = 6000/W
L = 6000/120
L = 50 feet
The dimensions of the land, correct to the nearest foot is 50ft by 120ft
The amount of money spent on textbooks per year for students is approximately normal.
A. To estimate the population mean, 19 students are randomly selected the sample mean was $390 and the standard deviation was $120. Find a 95% confidence for the population meam.
B. If the confidence level in part a changed from 95% 1 to 1999%, would the margin of error for the confidence interval:
1. decrease.
2. stay the same.
3. increase not.
C. If the sample size in part a changed from 19% 10 to 22, would the margin of errot for the confidence interval:
1. decrease.
2. stay the same.
3. increase
D. To estimate the proportion of students who purchase their textbookslused, 500 students were sampled. 210 of these students purchased used textbooks. Find a 99% confidence interval for the proportion of students who purchase used text books.
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] .
Let A = {June, Janet, Jill, Justin, Jeffrey, Jelly}, B = {Janet, Jelly, Justin}, and C = {Irina, Irena, Arena, Arina, Jelly}. Find the given set. A ∪ C a. {June, Janet, Jill, Justin, Jeffrey, Jelly, Irina, Irena, Arena, Arina} b. {June, Justin, Irina, Irena, Arena, Arina, Jelly} c. {June, Janet, Jill, Justin, June, Jelly} {Jelly} d. ∅
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, }
The blood platelet counts of a group of women have a bell-shaped distribution with a mean of 256.3 and a standard deviation of 66.8. (All units are 1000 cells/μL.) Using the empirical rule, find each approximate percentage below. a. What is the approximate percentage of women with platelet counts within 3 standard deviations of the mean, or between 55.9 and 456.7? b. What is the approximate percentage of women with platelet counts between 122.7 and 389.9?
Answer:
a) In the interval ( 55,9 ; 456,7 ) we will find 99,7 % of all values
b) In the interval ( 122,7 ; 389,9 ) we find 95,4 % of all values
Step-by-step explanation:
For a Normal distribution N (μ ; σ ) the Empirical rule establishes that the intervals:
( μ ± σ ) contains 68,3 % of all values
( μ ± 2σ ) contains 95,4 % of all values
( μ ± 3σ ) contains 99,7 % of all values
If N ( 256,3 ; 66,8 )
σ = 66,8 ⇒ 3*σ = 3 * 66,8 = 200,4
Then: 256,3 - 200,4 = 55,9
And 256,3 + 200,4 = 456,7
a) In the interval ( 55,9 ; 456,7 ) we will find 99,7 % of all values
b) 2*σ = 2 * 66,8 = 133,6
Then 256,3 - 133,6 = 122,7
And 256,3 + 133,6 = 389,90
Then in the interval ( 122,7 ; 389,9 ) we find 95,4 % of all values
Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer. [Start 4 By 4 Matrix 1st Row 1st Column 4 2nd Column 5 3rd Column 7 4st Column 5 2nd Row 1st Column 0 2nd Column 1 3rd Column 4 4st Column 6 3rd Row 1st Column 0 2nd Column 0 3rd Column 3 4st Column 8 4st Row 1st Column 0 2nd Column 0 3rd Column 0 4st Column 1 EndMatrix ]
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
An investigator claims, with 95 percent confidence, that the interval between 10 and 16 miles includes the mean commute distance for all California commuters. To have 95 percent confidence signifies that
Answer:
Hello the options to your question is missing below are the options
A) if sample means were obtained for a long series of samples, approximately 95 percent of all sample means would be between 10 and 16 miles
B.the unknown population mean is definitely between 10 and 16 miles
C.if these intervals were constructed for a long series of samples, approximately 95 percent would include the unknown mean commute distance for all Californians
D.the unknown population mean is between 10 and 16 miles with probability .95
Answer : if these intervals were constructed for a long series of samples, approximately 95 percent would include the unknown mean commute distance for all Californians ( c )
Step-by-step explanation:
95% confidence
interval = 10 to 16 miles
To have 95% confidence signifies that if these intervals were constructed for a long series of samples, approximately 95 percent would include the unknown mean commute distance for all Californians
confidence interval covers a range of samples/values in the interval and the higher the % of the confidence interval the more precise the interval is,
A box contains 40 identical discs which are either red or white if probably picking a red disc is 1/4. Calculate the number of;
1. White disc.
2. red disc that should be added such that the probability of picking a red disc will be 1/4
Solve for y: 1/3y+4=16
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
Question 15
FLAG QUESTION
You bought 9.5 pounds of chicken for $12.73. Find the cost of one pound of chicken.
Answer:
One pound of chicken is $1.34
Step-by-step explanation:
So, if 9.5 pounds of chicken is $12.73, all you have to do it divide 12.73 and 9.5. That way, you can see how much each pound is separately! Hopefully this helps!
Answer:
[tex]\huge\boxed{1\ pound\ of\ chicken = \$1.34}[/tex]
Step-by-step explanation:
9.5 pounds of chicken = $12.73
Dividing both sides by 9.5
1 pound of chicken = $12.73 / 9.5
1 pound of chicken = $1.34
Please answer this correctly without making mistakes I need to finish this today as soon as possible
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.
Show the distributive property can be used to evaluate 7x8 4/5
Answer:
308/5
Step-by-step explanation:
It can be Written as
= 7 ×(8 + 4/5 )
= 7 × 8 + 7 × 4/5
= 56 + 28/5
= 280+28
________
5
= 308/5
which of the following are possible values of r?
[tex] {r}^{2 } = \frac{3}{16} [/tex]
Answer:
[tex]r=\frac{\sqrt{3} }{4}[/tex] and [tex]r=-\frac{\sqrt{3} }{4}[/tex]
Step-by-step explanation:
when you solve for r in the given equation, you need to apply the square root property, which gives positive and negative answers (both should therefore be considered):
[tex]r^2=\frac{3}{16} \\r=+/-\sqrt{\frac{3}{16}} \\r=+/-\frac{\sqrt{3} }{4}[/tex]
then you need to include these two possible solutions:
[tex]r=\frac{\sqrt{3} }{4}[/tex] and [tex]r=-\frac{\sqrt{3} }{4}[/tex]
Benjamin decides to treat himself to breakfast at his favorite restaurant. He orders chocolate milk that costs $3.25. Then, he wants to buy as many pancakes as he can, but he wants his bill to be at most $30 before tax. The restaurant only sells pancakes in stacks of 44 pancakes for $5.50 . Let S represent the number of stacks of pancakes that Benjamin buys. 1) Which inequality describes this scenario?2) What is the largest number of pancakes that Benjamin can afford?
Answer:
3.25 + 5.50S ≤ 30
Step-by-step explanation:
Given:
Chocolate milk cost $3.25.
Maximum bill that Benjamin wants = $30
Cost of a stack pancake(44) = $5.50
Let number of stacks of pancakes bought = S
Benjamin will spend all the money available on 1 chocolate milk and S number of stacks of pancakes.
Cost of 1 pancake = $5.50
Cost of S number of stacks of pancakes = S*5.50
=5.50S
Total money spent =$3.25+5.50S
The total money spent should either be lesser than or equal to $30
The inequality is
3.25 + 5.50S ≤ 30
Largest number of pancakes Benjamin can afford
3.25 + 5.50S ≤ 30
5.50S ≤ 30-3.25
5.50S ≤ 26.75
Divide both sides by 5.50
S ≤ 4.86
1 stack=44 pancakes
4.86 stacks= 4.86 *44
=213.84 pancakes
Answer:
3.25+5.50S≤30
and 16 pancakes
Step-by-step explanation:
What is the difference between a line graph and a scatter plot?
Step-by-step explanation:
scatter plot s are similar to line graphs in that they start with mapping quantitive data points. The difference is that with a scatter plot, the decision is made the the individual points should not be connected directly together with a line but, instead express a trend
Find X so that m is parallel to n. Identify the postulate or theorem you used. Please help with these 3 problems, I don’t understand it at all
the corresponding angles should be equal
so, [tex] 5x+15=90 \implies 5x=75\implies x=15^{\circ}[/tex]
x
Find the value
of x. Show
3
10
your work.
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...
The ratio of the number of Anne's pencils to the number of jason's pencils is 4:3 Anne has 100 pencils how many pencils does jason have
Answer:
75
Step-by-step explanation:
4:3
4x25=100
3x25=75
Calculate two iterations of Newton's Method for the function using the given initial guess. (Round your answers to four decimal places.) f(x) = x2 − 5, x1 = 2n xn f(xn) f '(xn) f(xn)/f '(xn) xn − f(xn)/f '(xn)1 2
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]
The diagonals of a rhombus bisect each other of measures 8cm and 6cm .Find its perimeter. please help !!!!!!!!!!!!!!!!!!!!!!!!
Answer:
20 cm
Step-by-step explanation:
20 cm
8/2 = 4
6/2 = 3
3 and 4 are the sides of the triangle (four triangles in rhombus)
a²+b²=c²
4³+3²=c²
c = 5
5 x 4 = 20
Hope this helped
Answer:
perimeter = 20 cm
Step-by-step explanation:
consider breaking the rhombus into four equal parts.
and that gives you a triangle.
(refer to image attached for more clarification)
let a = 3, b = 4
to get the side c, use Pythagorean theorem = c² = a² + b²
c = sqrt (3² + 4²)
side c = 5
therefore,
perimeter = 4 x sides (c)
perimeter = 4 x 5
perimeter = 20 cm
graph 3x-y-2=0 using the x- and y-intercepts
Step-by-step explanation:
I used an app called DESMOS It Is usually super helpful!!!
what is the end point of a ray
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.
The energy E (in ergs) released by an earthquake is approximated by log E= 11.8 + 1.5M. Where M is the magnitude of the earthquake. What is the energy released by the 1906 San Francisco quake, which measured 8.3 on the Richter scale? This energy, it is estimated, would be sufficient to provide the entire world's food requirements for a day. Answer in ergs.
Answer:
[tex]\large \boxed{3.4 \times 10^{10}\text{ ergs }}[/tex]
Step-by-step explanation:
[tex]\begin{array}{rcl}\log E & = & 11.8 + 1.5M\\& = & 11.8 + 1.5 \times 8.3\\& = & 11.8 + 12.45\\& = & 24.25\\E & = & e^{24.25}\\& = & \mathbf{3.4 \times 10^{10}} \textbf{ ergs}\\\end{array}\\\text{ The energy released was $\large \boxed{\mathbf{3.4 \times 10^{10}}\textbf{ ergs }}$}[/tex]
Please help me understand this question!
Answer:
C
Step-by-step explanation:
The first sentence basically sets up the equation which is given, so we can read it for knowledge but it is not crucial to solve the problem.
We start here:
we are given: $120 - 0.2($120)
= 120 - (0.2)(120) (factoring out 120)
= 120 (1 - 0.2)
= 120 (0.8)
= 0.8 (120) (answer c)
Evaluate 2/3 + 1/3 + 1/6 + … THIS IS CONTINUOUS. It is NOT as simple as 2/3 + 1/3 + 1/6.
[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]
Put the following equation of a line into slope-intercept form, simplifying all
fractions.
3x + 3y = -9
Answer:
[tex]y = -x - 3[/tex]
Step-by-step explanation:
We are trying to get the equation [tex]3x + 3y = -9[/tex] into the form [tex]y = mx+b[/tex], aka slope-intercept form.
To do this we are trying to isolate y.
[tex]3x + 3y = -9[/tex]
Subtract 3x from both sides:
[tex]3y = -9 - 3x[/tex]
Rearrange the terms:
[tex]3y = -3x - 9[/tex]
Divide both sides by 3:
[tex]y = -x - 3[/tex]
Hope this helped!
How do you write 5.44 in words?
Answer:
five and forty-four hundredths
Step-by-step explanation:
Answer:
five point four four
Step-by-step explanation:
Help with number 50 please. Thanks.
Answer:
[tex] d = 7 + 3\sqrt{3} [/tex] and
[tex] d = 7 - 3\sqrt{3} [/tex]
Step-by-step explanation:
To solve the equation, [tex] d^2 - 14d - 22 = 0 [/tex], using the quadratic formula,
Recall: quadratic formula = [tex] \frac{-b ± \sqrt{b^2 - 4ac}}{2a} [/tex]
Where,
a = 1
b = -14
c = 22
Plug in your values into the formula and solve:
[tex] \frac{-(-14) ± \sqrt{(-14)^2 - 4(1)(22)}}{2(1)} [/tex]
[tex] \frac{14 ± \sqrt{196 - 88}}{2} [/tex]
[tex] \frac{14 ± \sqrt{108}}{2} [/tex]
[tex] d = \frac{14 + \sqrt{108}}{2} [/tex]
[tex] d = \frac{14 + 6\sqrt{3}}{2} [/tex]
[tex] d = (\frac{2(7 + 3\sqrt{3})}{2} [/tex]
[tex] d = 7 + 3\sqrt{3} [/tex]
And
[tex] d = \frac{14 - \sqrt{108}}{2} [/tex]
[tex] d = \frac{14 - 6\sqrt{3}}{2} [/tex]
[tex] d = (\frac{2(7 - 3\sqrt{3})}{2} [/tex]
[tex] d = 7 - 3\sqrt{3} [/tex]
Need a little help thanks :D
Answer:
71°
Step-by-step explanation:
Consider triangle BDH. x is the external angle that is remote to internal angles B and D, so is equal to their sum:
x° = 41° +30°
x° = 71°
1-What is the sum of the series? ∑j=152j Enter your answer in the box.
2-What is the sum of the series? ∑k=14(2k2−4) Enter your answer in the box.
3-What is the sum of the series? ∑k=36(2k−10)
4-Which answer represents the series in sigma notation? 1+12+14+18+116+132+164 ∑j=1712(j+1) ∑j=172j−1 ∑j=1712j+1 ∑j=17(12)j−1
5-Which answer represents the series in sigma notation? −3+(−1)+1+3+5 ∑j=155j−1 ∑j=15(3j−6) ∑j=15(2j−5) ∑j=15−3(13)j−1
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.
What is a 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
