Answer:
The answer is 9Step-by-step explanation:
14 + 5y
To solve the expression substitute the value of y that's - 1 into the expression
That's
14 + 5(-1)
= 14 - 5
= 9
Hope this helps you
Given: x - 5 > -2. Choose the solution set.
Answer: x>3
Step-by-step explanation:
x-5>2
x>+5-2
x>3
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] .
A recent study of the relationship between social activity and education for a sample of corporate executives showed the following results. Social Activity Education Above Average Average Below Average College 30 20 10 High School 20 40 90 Grade School 10 50 130 Using 0.05 as the significance level, what is the critical value for the test statistic
Answer:
9.488
Step-by-step explanation:
The critical value is found by first assessing which statistical test should be used.
We are interested in investigating relationship between social activity and education so chi-square test would be appropriate.
We have 3 rows and 3 columns. The degree of freedom for chi-square critical value is (r-1)(c-1)=(3-1)(3-1)=2*2=4
Chi-square critical value(0.05,4)= 9.488
the difference of 8 and 2, added to x"
Answer:
see below
Step-by-step explanation:
Difference is subtract
(8-2)
Then add this to x
(8-2) +x
6+x
The Freeman family is barbecuing veggie burgers, corn cobs, and mushroom caps in their local park. If 3 8 of the items barbecued are veggie burgers, and 1 3 of the items barbecued are corn cobs, what fraction of barbecued items are mushroom caps?
Answer:
The answer is below
Step-by-step explanation:
The Freeman family barbecued veggie burgers, corn cobs, and mushroom caps. 3/8 of the items barbecued are veggie burgers, and 1/3 of the items barbecued are corn cobs.
Let the total number of berbecued items be x. Therefore:
x = barbecued veggie burgers + barbecued corn cobs + barbecued mushroom caps
Barbecued veggie burgers = (3/8)x, barbecued corn cobs = (1/3)x, Let barbecued mushroom caps be y
Substituting:
x = (3/8)x + (1/3)x + y
Multiply through by 24
24x = 9x + 8x + 24y
24x = 17x + 24y
24y = 24x - 17x
24y = 7x
y = (7/24)x
barbecued mushroom caps = (7/24) of items
7/24 of the items barbecued are mushroom caps
Using fractions, it is found that the fraction of barbecued items that are mushroom caps is of [tex]\frac{7}{24}[/tex].
---------------------------
The total proportion of all products is 100% = 1.The fraction corresponding to veggie burgers is [tex]\frac{3}{8}[/tex].The fraction corresponding to corn cobs is [tex]\frac{1}{3}[/tex].The fraction corresponding to mushroom caps is x.---------------------------
Thus:
[tex]\frac{3}{8} + \frac{1}{3} + x = 1[/tex]
Solving for x, we find the fraction of mushroom caps.The least common multiple of 3 and 8 is 24.Then:
[tex]\frac{3\times3 + 8\times1 + 24x}{24} = 1[/tex]
[tex]\frac{17 + 24x}{24} = 1[/tex]
[tex]17 + 24x = 24[/tex]
[tex]24x = 7[/tex]
[tex]x = \frac{7}{24}[/tex]
The fraction of barbecued items that are mushroom caps is of [tex]\frac{7}{24}[/tex].
A similar problem is given at https://brainly.com/question/4231000
How to evaluate this help me out so lost?
Answer:
5443
Step-by-step explanation:
Order of Operations: BPEMDAS
Always left to right.
Step 1: Add 68 and 5042
68 + 5042 = 5110
Step 2: Add 5110 and 333
5110 + 333 = 5443
And we have our answer!
Complete the square: x2+7x+?=(x+?)2
Answer:
[tex] {x}^{2} + 7x + \frac{49}{4} = {(x + \frac{7}{2}) }^{2} [/tex]
Explanation:
[tex] {x}^{2} + 7x + a = {(x + b)}^{2} [/tex]
[tex] {x}^{2} + 7x + a = {x}^{2} + 2bx + {b}^{2} [/tex]
compare the x co-efficient
[tex] 7 = 2b[/tex]
[tex] b = \frac{7}{2} [/tex]
compare the constants
[tex]a = {b}^{2} [/tex]
[tex]a = {( \frac{7}{2}) }^{2} [/tex]
[tex]a = \frac{49}{4} [/tex]
HOPE IT HELPS....
BRAINLIEST PLEASE ;-)The complete equation will be x^2+7x+49/4=(x+7/2)2
Given the quadratic function x^2 + 7x + ?
In order to complete the square using the completing the square method, we will add the square of the half of coefficient of x to both sides of the expression.
Coefficient of x = 7
Half of the coefficient = 7/2
Taking the square of the result = (7/2)² = 49/4
The constant that will complete the equation is 49/9. The equation becomes x^2 + 7x + (7/2)² = (x+7/2)²
Hence the complete equation will be x^2+7x+49/4=(x+7/2)2
Learn more here: https://brainly.com/question/13981588
Cancel the common factor of the numerator and the denominator and write specified expression
Step-by-step explanation:
Hello,
I hope you mean to cancel the common factor that exists in numerator and denominator,right.
so, Let's look for the common factor,
here, the expression is,
=4(x-2)/ (x+5)(x-2)
so, here we find the common factor is (x-2)
now, we have to cancel it. And after cancelling we get,
=4/(x+5)
Note:{ we cancel the common factor if the common factors are in multiply form.}
Hope it helps
Find the particular solution of the differential equation that satisfies the initial condition(s). (Remember to use absolute values where appropriate.) f ''(x) = 4 x2 , f '(1) = 2, f(1) = 5
Looks like either [tex]f''(x)=4x^2[/tex] or [tex]f''(x)=\frac4{x^2}[/tex]...
In the first case, integrate both sides twice to get
[tex]f''(x)=4x^2\implies f'(x)=\dfrac43x^3+C_1\implies f(x)=\dfrac13x^4+C_1x+C_2[/tex]
Then the initial conditions give
[tex]f'(1)=2\implies 2=\dfrac43\cdot1^3+C_1\implies C_1=\dfrac23[/tex]
[tex]f(1)=5\implies 5=\dfrac13\cdot1^4+C_1\cdot1+C_2\implies C_2=4[/tex]
so that the particular solution is
[tex]f(x)=\dfrac{x^4}3+\dfrac{2x}3+4[/tex]
If instead [tex]f''(x)=\frac4{x^2}[/tex], we have
[tex]f''(x)=\dfrac4{x^2}\implies f'(x)=-\dfrac4x+C_1\implies f(x)=-4\ln|x|+C_1x+C_2[/tex]
[tex]f'(1)=2\implies 2=-\dfrac41+C_1\implies C_1=6[/tex]
[tex]f(1)=5\implies 5=-4\ln|1|+C_1\cdot1+C_2\implies C_2=-1[/tex]
[tex]\implies f(x)=-4\ln|x|+6x-1[/tex]
Let f(x) = x - 1 and g(x) = x^2 - x. Find and simplify the expression. (f + g)(1) (f +g)(1) = ______
Answer:
The simplified answer of the given expression is 1.
Step-by-step explanation:
When you see (f + g)(x), then it means that you are going to add f(x) and g(x) together. So, we are going to add the terms together that are given in the problem. We are also given the value of x which is 1. So, we are going to combine this information together so we can simplify the expression.
(f + g)(1)
f(x) = x - 1
g(x) = x²
(f + g)(1) = (1 - 1) + (1²)
Simplify the terms in the parentheses.
(f + g)(1) = 0 + 1
Add 0 and 1.
(f + g)(1) = 1
So, (f + g)(1) will have a simplified answer of 1.
The chart shows a certain city's population by age. Assume that the selections are independent events. If 8 residents of this city are selected at random, find the probability that the first 2 are 65 or older, the next 3 are 25-44 years old, the next 2 are 24 or younger, and the last is 45-64 years old.
Answer:
0.000014
Step-by-step explanation:
The chart is not provided so i will use an example chart to explain the answer. Here is a sample chart:
City X's Population by Age
0-24 years old 33%
25-44 years old 22%
45-64 years old 21%
65 or older 24%
In order to find probability of independent events we find the probability of each event occurring separately and then multiply the calculated probabilities together in the following way:
P(A and B) = P(A) * P(B)
probability that the first 2 are 65 or older
Let A be the event that the first 2 are 65 or older
The probability of 65 or older 24% i.e. 0.24
So the probability that first 2 are 65 or older is:
0.24(select resident 1) * 0.24(select resident 2)
P(A) = 0.24 * 0.24
= 0.0576
P(A) = 0.0576
probability that the next 3 are 25-44 years old
Let B be the event that the next 3 are 25-44 years old
25-44 years old 22% i.e. 0.22
So the probability that the next 3 are 25-44 years old is:
0.22 * 0.22* 0.22
P(B) = 0.22 * 0.22 * 0.22
= 0.010648
P(B) = 0.010648
probability that next 2 are 24 or younger
Let C be the event that the next 2 are 24 or younger
0-24 years old 33% i.e. 0.33
So the probability that the next 2 are 24 or younger is:
0.33 * 0.33
P(C) = 0.33 * 0.33
= 0.1089
P(C) = 0.1089
probability that last is 45-64 years old
Let D be the event that last is 45-64 years old
45-64 years old 21% i.e. 0.21
So the probability that last is 45-64 years old is:
0.21
P(D) = 0.21
So probability of these independent events is computed as:
P(A and B and C and D) = P(A) * P(B) * P(C) * P(C)
= 0.0576 * 0.010648 * 0.1089 * 0.21
= 0.000014
g A slot machine has three slots; each will show a cherry, a lemon, a star, or a bar when spun. The player wins if all three slots show the same three items. a. How many simple events are in the sample space
Answer:
64
Step-by-step explanation:
Let us consider E_abc to be the event that a, b and c appear on the first, second and third slot of the spin machine.
Now, we are told that each slot has 4 possibilities which are a cherry, a lemon, a star, or a bar when spun.
Thus, from mn rule in probability, the total number of simple events in the sample space is = 4³ = 64
Consider F and C below.
F(x, y) = x2 i + y2 j
C is the arc of the parabola y = 2x2 from (−1, 2) to (2, 8)
(a) Find a function f such that F = ∇f. f(x, y) =
(b) Use part (a) to evaluate C ∇f · dr along the given curve C.
(a)
[tex]\dfrac{\partial f}{\partial x}=x^2\implies f(x,y)=\dfrac{x^3}3+g(y)[/tex]
[tex]\dfrac{\partial f}{\partial y}=\dfrac{\mathrm dg}{\mathrm dy}=y^2\implies g(y)=\dfrac{y^3}3+C[/tex]
[tex]\implies f(x,y)=\dfrac{x^3+y^3}3+C[/tex]
(b)
[tex]\displaystyle\int_C\nabla f\cdot\mathrm d\mathbf r=f(2,8)-f(-1,2)=\boxed{171}[/tex]
If you use a 5/8 inch drill bit instead of a 3/16 that the project called for ,your hole will be too . by inches
Combine like terms. What is a simpler form of each expression? 4c-4d+8c-3d
Answer:
12c-7d
Step-by-step explanation:
[tex]4c-4d+8c-3d=0\\4c+8c=3d+4d\\12c=7d\\12c-7d[/tex]
===============================================
Explanation:
The terms 4c and 8c are one pair of like terms that combine to 4c+8c = 12c. We add 4 and 8 to get 12, then tack a 'c' at the end
The other pair of like terms are -4d and -3d. They combine to -7d for similar reasoning.
12c and -7d are not like terms, so we can't combine them and we stop here.
-----------
One way to think of combining like terms is consider simplifying 2c+3c. You could say that 2c represents having 2 cups while 3c is having 3 cups. Writing 2c+3c means we start with 2 cups and add on 3 more getting a total of 2+3 = 5 cups. Symbolically we would then write 5c. Therefore 2c+3c = 5c.
Find the maximum and minimum values of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤100. The maximum value of f(x,y) is:
First find the critical points of f :
[tex]f(x,y)=2x^2+3y^2-4x-5=2(x-1)^2+3y^2-7[/tex]
[tex]\dfrac{\partial f}{\partial x}=2(x-1)=0\implies x=1[/tex]
[tex]\dfrac{\partial f}{\partial y}=6y=0\implies y=0[/tex]
so the point (1, 0) is the only critical point, at which we have
[tex]f(1,0)=-7[/tex]
Next check for critical points along the boundary, which can be found by converting to polar coordinates:
[tex]f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t[/tex]
Find the critical points of g :
[tex]\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0[/tex]
[tex]\implies\sin t=0\text{ OR }1+5\cos t=0[/tex]
[tex]\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi[/tex]
where n is any integer. We get 4 critical points in the interval [0, 2π) at
[tex]t=0\implies f(10,0)=155[/tex]
[tex]t=\cos^{-1}\left(-\dfrac15\right)\implies f(-2,4\sqrt6)=299[/tex]
[tex]t=\pi\implies f(-10,0)=235[/tex]
[tex]t=2\pi-\cos^{-1}\left(-\dfrac15\right)\implies f(-2,-4\sqrt6)=299[/tex]
So f has a minimum of -7 and a maximum of 299.
An analyst takes a random sample of 25 firms in the telecommunications industry and constructs a confidence interval for the mean return for the prior year. Holding all else constant, if he increased the sample size to 30 firms, how are the standard error of the mean and the width of the confidence interval affected
Answer:
The standard error decreases and the width of the confidence interval also decreases.
Step-by-step explanation:
The standard error of a distribution (E) is given as:
[tex]E=z_{\frac{\alpha}{2} }*\frac{\sigma}{\sqrt{n} }[/tex]
Where n is the sample size, [tex]z_{\frac{\alpha}{2} }[/tex] is the z score of he confidence and [tex]\sigma[/tex] is the standard deviation.
The sample size is inversely proportional to the standard error. If the sample size is increased and everything else is constant, the standard would decrease since they are inversely proportional to each other. The confidence interval = μ ± E = (μ - E, μ + E). μ is the mean
The width of the confidence interval = μ + E - (μ - E) = μ + E - μ + E = 2E
The width of the confidence interval is 2E, therefore as the sample size increase, the margin of error decrease and since the width of the confidence interval is directly proportional to the margin of error, the width of the confidence interval also decreases.
Megan has 12 pounds of cheesecake. On Monday, she and her friends eat 4 pounds. On Tuesday, she and her friends eat another 3 pounds. On Wednesday, her friend Mark gives her some more cheesecake so that she has 3 times as much as she had at the end of Tuesday. On Thursday, some of her cheesecake goes bad, so she has the amount that she had at the end of Wednesday, but divided by 5. On Friday, she gives 3 pounds to her dog. On Saturday, her mom gives her one more pound. On Sunday, how many pounds of cheesecake does Megan have left?
Answer:
Step-by-step explanation:
First we start with 12 pounds
On Monday, she and her friends eat 4 pounds. So we have 8 now.
On Tuesday, she and her friends eat another 3 pounds. So we gave 5 now.
On Wednesday, her friend Mark gives her some more cheesecake so that she has 3 times as much as she had at the end of Tuesday. 5 * 3 = 15
On Thursday, some of her cheesecake goes bad, so she has the amount that she had at the end of Wednesday, but divided by 5. She had 15 at the end of Wednesday. 15/5 = 3.
On Friday, she gives 3 pounds to her dog. 5 - 3 = 2.
On Saturday, her mom gives her one more pound. 2 + 1 = 3.
On Sunday, she finally has 3 pounds.
Answer:
nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
Step-by-step explanation:
At a sale, dresses were sold for $39 each. This price was 65% of a dress's original price. How much did a dress originally cost?
Answer:
Hey there!
We can write the equation:
0.65x=39
x=60
The dress originally sold for 60 dollars.
Hope this helps :)
what should be added to 66.778 get 78.2
Answer:
11.422
Step-by-step explanation:
[tex]78.2 - 66.778 \\ = 11.422[/tex]
Select the function that represents a parabola with zeros at x = –2 and x = 4, and y-intercept (0,–16). A ƒ(x) = x2 + 2x – 8 B ƒ(x) = 2x2 + 4x – 16 C ƒ(x) = x2 – 2x – 8 D ƒ(x) = 2x2 – 4x – 16
Answer:
D. [tex]f(x) = 2\cdot x^{2}-4\cdot x -16[/tex]
Step-by-step explanation:
Any parabola is modelled by a second-order polynomial, whose standard form is:
[tex]y = a\cdot x^{2}+b\cdot x + c[/tex]
Where:
[tex]x[/tex] - Independent variable, dimensionless.
[tex]y[/tex] - Dependent variable, dimensionless.
[tex]a[/tex], [tex]b[/tex], [tex]c[/tex] - Coefficients, dimensionless.
In addition, a system of three linear equations is constructed by using all known inputs:
(-2, 0)
[tex]4\cdot a -2\cdot b + c = 0[/tex] (Eq. 1)
(4, 0)
[tex]16\cdot a + 4\cdot b +c = 0[/tex] (Eq. 2)
(0,-16)
[tex]c = -16[/tex] (Eq. 3)
Then,
[tex]4\cdot a - 2\cdot b = 16[/tex] (Eq. 4)
[tex]16\cdot a + 4\cdot b = 16[/tex] (Eq. 5)
(Eq. 3 in Eqs. 1 - 2)
[tex]a - 0.5\cdot b = 4[/tex] By Eq. 4 (Eq. 4b)
[tex]a = 4 + 0.5\cdot b[/tex]
Then,
[tex]16\cdot (4+0.5\cdot b) + 4\cdot b = 16[/tex] (Eq. 4b in Eq. 5)
[tex]64 + 12\cdot b = 16[/tex]
[tex]12\cdot b = -48[/tex]
[tex]b = -4[/tex]
The remaining coeffcient is:
[tex]a = 4 + 0.5\cdot b[/tex]
[tex]a = 4 + 0.5\cdot (-4)[/tex]
[tex]a = 2[/tex]
The function that represents a parabola with zeroes at x = -2 and x = 4 and y-intercept (0,16) is [tex]f(x) = 2\cdot x^{2}-4\cdot x -16[/tex]. Thus, the right answer is D.
Answer:
D ƒ(x) = 2x2 – 4x – 16
Step-by-step explanation:
Can someone do this assuming that it is infinite and as well as assuming it's not infinite? Thanks!
Answer:
see below
Step-by-step explanation:
4,7,12,19
We are adding 3,5,7,9..... each time
The sequence is not arithmetic because we are not adding a constant. It is not geometric since we are not multiplying by a constant term each time
There is no common difference or common ratio.
The explicit formula is
an =n^2 +3
The recursive formula is
(n+1)^2 +3 - (n^2 +3)
n^2 +2n+1+3 - ( n^2+3)
2n+1
a sub(n+1) = a sub( n) + 2n+1
The 10th term
an = n^2 +3
Let n=10
an = 10^2+3
= 100+3
= 103
summation
see image
since the numbers are increasing and greater than 1 the sum does not exist
What is the answer, what are the steps to solve this, and what do the parts of the equation represent?
Step-by-step explanation:
Just sub 4 into where n is
5/3 x 6/7 real quick plz
Answer:
10/7 or 1 3/7. I hope this helps,
Step-by-step explanation:
Plz answer last question and im lost!
Answer:
[tex]\pi[/tex] radian
Step-by-step explanation:
We know that angle for a full circle is 2[tex]\pi[/tex]
In the given figure shape is semicircle
hence,
angle for semicircle will be half of angle of full circle
thus, angle for given figure = half of angle for a full circle = 1/2 * 2[tex]\pi[/tex] = [tex]\pi[/tex]
Thus, answer is [tex]\pi[/tex] radian
alternatively, we also know that angle for a straight line is 180 degrees
and 180 degrees is same as [tex]\pi[/tex] radian.
The expression −50x+100 represents the balance, in dollars, of a bank account after x months. What is the rate of change, in dollars per month, of the bank account balance?
Answer:
-50
Step-by-step explanation:
Basically get two slopes -50(1)+100 will get you 1,50 (1 is x and 50 is y since its the answer)
-50(0)+100 (0,100) Y₂-Y₁/X₂-X₁ 50-100/1-0
Rate of change per month = -$50
A local mattress manufacturer wants to know if its manufacturing process is in or out of control and has hired you, a statistics expert in the field, to analyze its process. Specifically, the business has run 20 random samples of size 5 over the past month and has determined the mean of each sample.
a. Determine the estimate of the mean when the process is in control.
b. Assuming the process standard deviation is .50 and the mean of the process is the estimate calculated in part a, determine the Upper Control Limit (UCL) and the Lower Control Limit (LCL) for the manufacturing process.
c. Explain the results to the vice-president of the mattress manufacturer focusing on whether, based on the results, the process is in or out of control.
Sample no. Mean of Sample
1 95.72
2 95.44
3 95.40
4 95.50
5 95.56
6 95.72
7 95.60
8 95.24
9 95.46
10 95.44
11 95.80
12 95.20
13 94.82
14 95.78
15 95.18
16 95.32
17 95.08
18 95.22
19 95.04
20 95.
Answer:
Answer to question a = 95.4
Answer to question b = UCL = 96.07
LCL = 94.73
Answer to question c = Process is still in control
Step-by-step explanation:
a. The computation of estimate mean is as shown below:-
= 95.4
b. The computation of Upper Control Limit (UCL) and the Lower Control Limit (LCL) for the manufacturing process is shown below:-
= 95.4 + 0.67082
= 96.07
= 95.4 - 0.67082
= 94.73
c. The explanation is shown below:-
From the above calculation we can see that the sample lies between LCL AND UCL that is (94.73 ,96.07) ,
The Process is still in control
Sherina wrote and solved the equation. x minus 56 = 230. x minus 56 minus 56 = 230 minus 56. x = 174. What was Sherina’s error?
Answer:
subtracting 56 instead of adding (or adding wrong)
Step-by-step explanation:
She wrote ...
x - 56 = 230
x - 56 - 56 = 230 -56 . . . . correct application of the addition property*
x = 230 -56 . . . . . . . . . . . . incorrect simplification
Correctly done, the third line would be ...
x -112 = 174
This would have made Sherina realize that the error was in subtracting 56 instead of adding it. The correct solution would be ...
x - 56 + 56 = 230 + 56 . . . using the addition property of equality
x = 286 . . . . . . . . . . . . . . . . correct simplification on both sides
__
There were two errors:
1) incorrect strategy --- subtracting 56 instead of adding
2) incorrect simplification --- simplifying -56 -56 to zero instead of -112
We don't know whether you want to count the error in thinking as the first error, or the error in execution where the mechanics of addition were incorrectly done.
_____
* The addition property of equality requires the same number be added to both sides of the equation. Sherina did that correctly. However, the number chosen to be added was the opposite of the number that would usefully work toward a solution.
Answer:
D: Sherina should have added 56 to both sides of the equation.
Step-by-step explanation:
I got a 100% on my test.
I hope this helps.
Last question of the day!!
Answer:
Correct options are 2, 5 and 7.
Step-by-step explanation:
Consider the given vertices of triangle are A(-3,-3), B(-3,2) and C(1,2).
Distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using distance formula, we get
[tex]AB=\sqrt{(-3-(-3))^2+(2-(-3))^2}[/tex]
[tex]AB=\sqrt{(0)^2+(5)^2}[/tex]
[tex]AB=\sqrt{25}[/tex]
[tex]AB=5[/tex]
Similarly,
[tex]BC=\sqrt{(1-(-3))^2+(2-2)^2}=4[/tex]
[tex]AC=\sqrt{(1-(-3))^2+(2-(-3))^2}=\sqrt{16+25}=\sqrt{41}[/tex]
From the above calculation it is clear that AC>AB and AC>BC.
According to Pythagoras theorem, in a right angle triangle, the square of largest side is equal to the sum of squares of two small sides.
[tex]hypotenuse^2=base^2+perpendicular^2[/tex]
[tex]AC^2=(\sqrt{41})^2=41[/tex]
[tex]AB^2+BC^2=(5)^2+4^2=24+16=41=AC^2[/tex]
So, given triangle is a right angle triangle and AC is its hypotenuse.
Therefore, the correct options are 2, 5 and 7.
Assume that thermometer readings are normally distributed with a mean of 0C and a standard deviation of 1.00C. A thermometer is randomly selected and tested. For the case below, draw a sketch, and find the probability of the reading. (The given values are in Celsius degrees.) Between and
Answer: 0.0546 and 0.9829
Step-by-step explanation:
solution:
= P( 1.50< Z <2.25 )
= P(Z <2.25 ) - P(Z <1.50 )
Using z table,
= 0.9878-0.9332
=0.0546
b.
= P( -2.12< Z <3.73 )
= P(Z <3.73) - P(Z <-2.12 )
Using z table,
= 0.9999-0.0170
=0.9829