Answer:
D. 800,000
Step-by-step explanation:
It is D because you find the hundred thousand place which is the 8, the you go to the number next door which is 3, if the 3 is 5 or greater the 8 will become a 9 or if it is not then it will stay the same. And everything to the left stays the same, everything to the right turns into zeros.
The numbers of words defined on randomly selected pages from a dictionary are shown below. Find the mean, median, mode of the listed numbers. 30 31 64 59 57 33 54 77 56 41 What is the mean? Select the correct choice below and ,if necessary ,fill in the answer box within your choice.(around to one decimal place as needed)
Answer:
mean=502/10=50.2
median=(54+56)2=55
A standardized exam's scores are normally distributed. In a recent year, the mean test score was and the standard deviation was . The test scores of four students selected at random are , , , and . Find the z-scores that correspond to each value and determine whether any of the values are unusual. The z-score for is nothing. (Round to two decimal places as needed.) The z-score for is nothing. (Round to two decimal places as needed.) The z-score for is nothing. (Round to two decimal places as needed.) The z-score for is nothing. (Round to two decimal places as needed.) Which values, if any, are unusual? Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The unusual value(s) is/are nothing. (Use a comma to separate answers as needed.) B. None of the values are unusual.
Answer:
The z-score for 1880 is 1.34.
The z-score for 1190 is -0.88.
The z-score for 2130 is 2.15.
The z-score for 1350 is -0.37.
And the z-score of 2130 is considered to be unusual.
Step-by-step explanation:
The complete question is: A standardized exam's scores are normally distributed. In recent years, the mean test score was 1464 and the standard deviation was 310. The test scores of four students selected at random are 1880, 1190, 2130, and 1350. Find the z-scores that correspond to each value and determine whether any of the values are unusual. The z-score for 1880 is nothing. (Round to two decimal places as needed.) The z-score for 1190 is nothing. (Round to two decimal places as needed.) The z-score for 2130 is nothing. (Round to two decimal places as needed.) The z-score for 1350 is nothing. (Round to two decimal places as needed.) Which values, if any, are unusual? Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The unusual value(s) is/are nothing. (Use a comma to separate answers as needed.) B. None of the values are unusual.
We are given that the mean test score was 1464 and the standard deviation was 310.
Let X = standardized exam's scores
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean test score = 1464
[tex]\sigma[/tex] = standard deviation = 310
S, X ~ Normal([tex]\mu=1464, \sigma^{2} = 310^{2}[/tex])
Now, the test scores of four students selected at random are 1880, 1190, 2130, and 1350.
So, the z-score of 1880 = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{1880-1464}{310}[/tex] = 1.34
The z-score of 1190 = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{1190-1464}{310}[/tex] = -0.88
The z-score of 2130 = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{2130-1464}{310}[/tex] = 2.15
The z-score of 1350 = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{1350-1464}{310}[/tex] = -0.37
Now, the values whose z-score is less than -1.96 or higher than 1.96 are considered to be unusual.
According to our z-scores, only the z-score of 2130 is considered to be unusual as all other z-scores lie within the range of -1.96 and 1.96.
Given v(x) = g(x) (3/2*x^4 + 4x – 1), find v'(2).
Answer:
Step-by-step explanation:
Given that v(x) = g(x)×(3/2*x^4+4x-1)
Let's find V'(2)
V(x) is a product of two functions
● V'(x) = g'(x)×(3/2*x^4+4x-1)+ g(x) ×(3/2*x^4+4x-1)
We are interested in V'(2) so we will replace x by 2 in the expression above.
g'(2) can be deduced from the graph.
● g'(2) is equal to the slope of the tangent line in 2.
● let m be that slope .
● g'(2) = m =>g'(2) = rise /run
● g'(2) = 2/1 =2
We've run 1 square to the right and rised 2 squares up to reach g(2)
g(2) is -1 as shown in the graph.
■■■■■■■■■■■■■■■■■■■■■■■■■■
Let's derivate the second function.
Let h(x) be that function
● h(x) = 3/2*x^4 +4x-1
● h'(x) = 3/2*4*x^3 + 4
● h'(x) = 6x^3 +4
Let's calculate h'(2)
● h'(2) = 6 × 2^3 + 4
● h'(2) = 52
Let's calculate h(2)
●h(2) = 3/2*2^4 + 4×2 -1
●h(2)= 31
■■■■■■■■■■■■■■■■■■■■■■■■■■
Replace now everything with its value to find V'(2)
● V'(2) = g'(2)×h(2) + g(2)× h'(2)
● V'(2)= 2×31 + (-1)×52
●V'(2) = 61 -52
●V'(2)= 9
. En el triángulo ABC, la medida del ángulo exterior en el vértice B es el triple de la medida del ángulo C y la mediatriz de BC corta a AC en el punto F. sabiendo que FC=12. Calcular AB.
Responder:
| AB | = 12m
Explicación paso a paso:
Verifique el diagrama en el archivo adjunto.
En el diagrama, se puede ver que el lado FC es igual al lado FB de acuerdo con el triángulo isósceles FBC.
Además, el lado FB es igual a AB ya que son paralelos entre sí.
De la declaración anterior, | FC | = | FB | y | FB | = | AB |
Esto significa | FC | = | FB | = | AB |
Por lo tanto desde | FC | = 12 m, | AB | = 12 m ya que ambos lados son iguales.
De ahí el lado | AB | se mide 12m
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]
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,
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!
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:
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
2/3a - 1/6 =1/3 please help me
Answer:
[tex]a = \frac{3}{4}[/tex]
Step-by-step explanation:
Let's convert everything to sixths to make it easier to work with.
[tex]\frac{4}{6}a - \frac{1}{6} = \frac{2}{6}[/tex]
Add 1/6 to both sides:
[tex]\frac{4}{6}a = \frac{3}{6}[/tex].
Dividing both sides by 4/6:
[tex]a = \frac{3}{6} \div \frac{4}{6}\\\\a = \frac{3}{6} \cdot \frac{6}{4}\\\\a = \frac{18}{24}\\\\a = \frac{3}{4}[/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:
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
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
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°
A news article estimated that only 5% of those age 65 and older who prefer to watch the news, rather than to read or listen, watch the news online. This estimate was based on a survey of a large sample of adult Americans. Consider the population consisting of all adult Americans age 65 and older who prefer to watch the news, and suppose that for this population the actual proportion who prefer to watch online is 0.05. A random sample of n = 100 people will be selected from this population and p, the proportion of people who prefer to watch online, will be calculated.
(a) What are the mean and standard deviation of the sampling distribution of p? (Round your standard deviation to four decimal places.
(b) Is the sampling distribution of p approximately normal for random samples of size n 100? Explain.
i. The sampling distribution of p is approximately normal because np is less than 10.
ii. The sampling distribution of p is approximately normal because np is at least 10.
iii. The sampling distribution of p is not approximately normal because np is less than 10
iv. The sampling distribution of p is not approximately normal because np is at least 10
v. The sampling distribution of p is not approximately normal because n(1 - p) is less than 10.
(c) Suppose that the sample size is n = 400 rather than n = 100, what are the values for the mean and standard deviation when n=400?
Does the change in sample size affect the mean and standard deviation of the sampling distribution of p? If not, explain why not.
i. When the sample size increases, the mean increases.
ii. When the sample size increases, the mean decreases.
iii. When the sample size increases, the mean stays the same.
iv. The sampling distribution is always centered at the population mean, regardless of sample size.
v. When the sample size increases, the standard deviation increases.
vi. When the sample size increases, the standard deviation decreases.
Answer:
3.25
Step-by-step explanation:
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
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]
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.
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
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]
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
Grandma baked 969696 cookies and gave them to her grandchildren. One of the grandchildren, Cindy, received ccc fewer cookies than she would have received had all of the cookies been evenly divided among the 888 grandchildren.
Answer:
Answer:
96/8 - c
Step-by-step explanation:
96/8 = 12
The average amount of cookies each grandchild would get is 12. However, Cindy gets c less than the average amount. So, it would be 12 - c.
But, it's asking for the original expression. Therefore, the answer would be 96/8 - c.
Sarah has $30,000 in her bank account today. Her grand-father has opened this account for her 15 years ago when she was born. Calculate the money that was deposited in the account 15 years ago if money has earned 3.5% p.a. compounded monthly through all these years.
Answer:
Deposit value(P) = $17,760 (Approx)
Step-by-step explanation:
Given:
Future value (F) = $30,000
Number of Year (n) = 15 year = 15 × 12 = 180 month
rate of interest (r) = 3.5% = 0.035 / 12 = 0.0029167
Find:
Deposit value(P)
Computation:
[tex]A = P(1+r)^n\\\\ 30000 = P(1+0.0029167)^{180} \\\\ 30000 = P(1.68917) \\\\ P = 17760.2018[/tex]
Deposit value(P) = $17,760 (Approx)
Yo help me real quick?
Answer:
1,2 and 6
Step-by-step explanation:
pie symbol
2/3
0.333333....
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]
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, }
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!!!
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)
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
Please help. I’ll mark you as brainliest if correct.
Answer:
Infinite number of solutions.
Step-by-step explanation:
There are an infinite number of solutions. If you graph both lines, you find they are the same line. If you multiply the send equation by -4, you’ll end up with the first equation. I’m not sure what your teacher means by specifying their form.
