Answer:
y = -1x + 2
Step-by-step explanation:
Complete the remainder
Answer:
-14 is the answer for the second term (?)
What type of model does the pattern show (linear or exponential)and explain please !!!!
Please and thanks
Explanation:
The first figure has 2 blocks.
The second figure has 4 blocks.
The third figure has 8 blocks
The pattern 2,4,8,... follows the rule "double the current value to get the next one". Because we have this going on, we have an exponential pattern here.
A linear pattern would be something like 2,4,6,8,10,... showing that we add on 2 each time, rather than multiply by 2 each time. So as you can guess, or already know, exponential patterns grow much quicker compared to linear ones.
please help me with this
Given:
d = 2
f = 4
To find:
Value of [tex]\frac{14(7)-d}{2f}[/tex]
Steps:
we need to substitute and then find the value,
[tex]= \frac{14(7)-2}{2(4)}\\ \\=\frac{98-2}{8} \\\\=\frac{96}{8}\\\\=12[/tex]
Therefore, the answer is option C) 12
Happy to help :)
If you need help, feel free to ask
A telescope contains both a parabolic mirror and a hyperbolic mirror. They share focus , which is 46feet above the vertex of the parabola. The hyperbola's second focus is 6 ft above the parabola's vertex. The vertex of the hyperbolic mirror is 3 ft below . Find the equation of the hyperbola if the center is at the origin of a coordinate system and the foci are on the y-axis. Complete the equation.
the center is at the origin of a coordinate system and the foci are on the y-axis, then the foci are symmetric about the origin.
The hyperbola focus F1 is 46 feet above the vertex of the parabola and the hyperbola focus F2 is 6 ft above the parabola's vertex. Then the distance F1F2 is 46-6=40 ft.
In terms of hyperbola, F1F2=2c, c=20.
The vertex of the hyperba is 2 ft below focus F1, then in terms of hyperbola c-a=2 and a=c-2=18 ft.
Use formula c^2=a^2+b^2c
2
=a
2
+b
2
to find b:
\begin{gathered} (20)^2=(18)^2+b^2,\\ b^2=400-324=76 \end{gathered}
(20)
2
=(18)
2
+b
2
,
b
2
=400−324=76
.
The branches of hyperbola go in y-direction, so the equation of hyperbola is
\dfrac{y^2}{b^2}- \dfrac{x^2}{a^2}=1
b
2
y
2
−
a
2
x
2
=1 .
Substitute a and b:
\dfrac{y^2}{76}- \dfrac{x^2}{324}=1
76
y
2
−
324
x
2
=1 .
The method of tree-ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution.
1,2851,1871,2221,1941,2681,3161,2751,3171,275
Required:
a. Use a calculator with mean and standard deviation keys to find the sample mean year x and sample standard deviations.
b. Find a 90% confidence interval for the mean of all tree-ring dates from this archaeological site. (Round your answers to the nearest whole number.)
Answer:
a) The sample mean is 1260 and the standard deviation is 48.
b) The 90% confidence interval for the mean of all tree-ring dates from this archaeological site is (1230, 1290).
Step-by-step explanation:
Question a:
Mean is the sum of all values divided by the number of values. So
[tex]\overline{x} = \frac{1285 + 1187 + 1222 + 1194 + 1268 + 1316 + 1275 + 1317 + 1275}{9} = 1260[/tex]
Standard deviation is the square root of the sum of the differences squared between each value and the mean, divided by the one less than the sample size. So
[tex]s = \sqrt{\frac{(1285-1260)^2 + (1187-1260)^2 + (1222-1260)^2 + (1194-1260)^2 + (1268-1260)^2 + ...}{8}} = 48[/tex]
The sample mean is 1260 and the standard deviation is 48.
Question b:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom,which is the sample size subtracted by 1. So
df = 9 - 1 = 8
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 8 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.8595
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 1.8595\frac{48}{\sqrt{9}} = 30[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 1260 - 30 = 1230
The upper end of the interval is the sample mean added to M. So it is 1260 + 30 = 1290
The 90% confidence interval for the mean of all tree-ring dates from this archaeological site is (1230, 1290).
please give me correct answer
Answer:
36 = 17+19 ---> They are twin primes and their sum is 3684 = 41+43 ---> They are twin Primes and sum is 84120 = 59+61 ---> They also are twin primes and their sum is 120144 = 71+73 ---> They are also twin primes and the sum is 144A particular variety of watermelon weighs on average 22.4 pounds with a standard deviation of 1.36 pounds. Consider the sample mean weight of 64 watermelons of this variety. Assume the individual watermelon weights are independent.
Required:
a. What is the expected value of the sample mean weight?
b. What is the standard deviation of the sample mean weight?
c. What is the approximate probability the sample mean weight will be less than 22.02?
d. What is the value c such that the approximate probability the sample mean will be less than c is 0.9?
Answer:
a) 22.4 pounds.
b) 0.17 pounds.
c) 0.0127 = 1.27% approximate probability the sample mean weight will be less than 22.02.
d) c = 22.62
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Average 22.4 pounds with a standard deviation of 1.36 pounds.
This means that [tex]\mu = 22.4, \sigma = 1.36[/tex]
Consider the sample mean weight of 64 watermelons of this variety.
This means that [tex]n = 64, s = \frac{1.36}{\sqrt{64}} = 0.17[/tex]
a. What is the expected value of the sample mean weight?
By the Central Limit Theorem, 22.4 pounds.
b. What is the standard deviation of the sample mean weight?
By the Central Limit Theorem, 0.17 pounds.
c. What is the approximate probability the sample mean weight will be less than 22.02?
This is the p-value of Z when X = 22.02. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{22.02 - 22.4}{0.17}[/tex]
[tex]Z = -2.235[/tex]
[tex]Z = -2.235[/tex] has a p-value of 0.0127.
0.0127 = 1.27% approximate probability the sample mean weight will be less than 22.02.
d. What is the value c such that the approximate probability the sample mean will be less than c is 0.9?
This is the 90th percentile, that is, [tex]X = c[/tex] when z has a p-value of 0.9, so X when Z = 1.28.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]1.28 = \frac{c - 22.4}{0.17}[/tex]
[tex]c - 22.4 = 1.28*0.17[/tex]
[tex]c = 22.62[/tex]
The domain for all variables in the expressions below is the set of real numbers. Determine whether each statement is true or false.(i)∀x ∃y(x+y≥0)
The domain of a set is the possible input values the set can take.
It is true that the domain of ∀x ∃y(x+y≥0) is the set of real numbers
Given that: ∀x ∃y(x+y≥0)
Considering x+y ≥ 0, it means that the values of x + y are at least 0.
Make y the subject in x+y ≥ 0
So, we have:
[tex]\mathbf{y \le -x}[/tex]
There is no restriction as to the possible values of x.
This means that x can take any real number.
Hence, it is true that the domain of ∀x ∃y(x+y≥0) is the set of real numbers.
Read more about domain at:
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Which of the following behaviors would best describe someone who is listening and paying attention? a) Leaning toward the speaker O b) Interrupting the speaker to share their opinion c) Avoiding eye contact d) Asking questions to make sure they understand what's being said
Answer:
D
Step-by-step explanation:
I explained why 5 minutes ago on a different question
Hey community I thank you guys fir your help
Answer:
A, B, and E.
Step-by-step explanation:
A. 5^x * 5^x
= 5^x+x
=5^(2)(x)
=25^x
B. 5^2x
=5^(2)(x)
=25^x
C. 5*5^2x
=5^1+2x
D. 5*5^x
=5^1+x
E. (5*5)^x
=5^x*5^x
=5^(2)(x)
=25^x
F. 5^2*5^x
=5^2+x
What are 3 ratios that are equivalent to 8 :5
Answer:
Step-by-step explanation:
8/5 = 16/10 = 24/15
8:5 = 16:10 = 24:15
What is the product?
Answer:
Step-by-step explanation:
[tex]\begin{bmatrix}3 & 6 & 1\\ 2 & 4& 0\\ 0 & 6 & 2\end{bmatrix}\times\begin{bmatrix}2\\ 0\\ 1\end{bmatrix}[/tex]
Multiply the terms of the rows of the first matrix with the terms given in the column of the second matrix.
[tex]=\begin{bmatrix}(3\times 2+6\times 0+1\times 1)\\ (2\times 2+4\times 0+0\times1)\\ (0\times 2+6\times 0+2\times 1)\end{bmatrix}[/tex]
[tex]=\begin{bmatrix}7\\ 4\\ 2\end{bmatrix}[/tex]
Can someone help me? I figured out part B however, I am struggling with part A and I would be so happy if any of you helped me. Thank you for your help.
Note: you may need to delete the comma
============================================================
Explanation:
The info "5 months" is never used to compute the mean. We could easily replace it with "6 months" or "7 months" or any stretch of time, and still get the same answer. So we'll ignore this value.
What we'll do is add up the 21 items given to us, and then divide by 21.
Because there are so many values, and it's easy to get lost, I'm going to add up across the rows
Row One: 256,229+253,657+218,747+246,163+235,626+288,694 = 1,499,116 Row Two: 316,265+196,721+319,620+285,077+215,152+253,291 = 1,586,126 Row Three: 315,011+199,901+265,443+291,806+303,556+215,359 = 159,1076 Row Four: 258,554+293,658+289,935 = 842,147Those subtotals then add up to this
1,499,116+1,586,126+1,591,076+842,147 = 5,518,465
This is the same as adding up all 21 values.
Finally, we divide that sum over 21 as there are 21 values in this list
(5,518,465)/21 = 262,784.047619048
That value then rounds up to 262,785
If your teacher just wanted things to the nearest whole number (without rounding up), then the answer would be 262,784
Side note: using a spreadsheet program would probably be the most efficient/fastest method for this type of problem.
What is the smallest number that becomes 600 when rounded to the nearest hundred?
A. 545
B. 550
C. 555
D. 590
Answer:
B. 550
Step-by-step explanation:
550 is the smallest number that becomes 600 when rounded to the nearest hundred
[tex]\lim_{x\to \ 0} \frac{\sqrt{cos2x}-\sqrt[3]{cos3x} }{sinx^{2} }[/tex]
Answer:
[tex]\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}[/tex]
General Formulas and Concepts:
Calculus
Limits
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
L'Hopital's Rule
Differentiation
DerivativesDerivative NotationBasic Power Rule:
f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Step-by-step explanation:
We are given the limit:
[tex]\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}[/tex]
When we directly plug in x = 0, we see that we would have an indeterminate form:
[tex]\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}[/tex]
This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:
[tex]\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}[/tex]
Plugging in x = 0 again, we would get:
[tex]\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}[/tex]
Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:
[tex]\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}[/tex]
Substitute in x = 0 once more:
[tex]\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}[/tex]
And we have our final answer.
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Limits
HELP PLEASE! I tried everything from adding to dividing, subtracting, multiplying but still no correct answer. Can someone help me out here please?
Answer:
46%
Step-by-step explanation:
Divde the smaller # by the bigger # to get the precentage
An average San Francisco customer uses what percent of electricity used by an average Houston customer?
In other words, San Francisco is what part of Houston?
---Just like, 7 is what part of 49? These are the same questions and would be solved in the same way
San Francisco / Houston
6753 / 14542
0.4644 = 46.44%
ANSWER: 46%
Hope this helps!
how do i establish this identity?
RHS
[tex]\\ \sf\longmapsto \frac{2 \tan( \theta) }{ \sin(2 \theta) } \\ \\ \sf\longmapsto \frac{ \frac{2 \sin( \theta) }{2 \cos( \theta) } }{ \sin(2 \theta) } \\ \\ \sf\longmapsto \frac{1}{ \cos {}^{2} ( \theta) } \\ \\ \sf\longmapsto {sec}^{2} \theta[/tex]
Your help is very much appreciated I will mark brainliest:)
Answer:
B. Yes. By SSS~
Step-by-step explanation:
From the diagram given, we have the corresponding sides of both triangles as follows:
RQ/KL = 24/20 = 6/5
QP/LM = 18/15 = 6/5
RP/KM = 12/10 = 6/5
From the above, we can see that the ratio of the corresponding side lengths of both triangles are equal. This means that all three sides of one triangle are proportional to all corresponding sides of the other triangle.
The SSS similarity theorem states that if all sides of one triangle are proportional to all corresponding sides of another, then both triangles are similar to each other.
Therefore, ∆KLM ~ ∆RQP by SSS similarity.
I need to solve for x and z if you could explain as well. Thank you
Answer:
x = 6
z = 60
Step-by-step explanation:
Solve for x
(6x + 84) = 120
- 84 -84
6x = 36
6x/6 = 36/6
x = 6
Then solve for z
120 + z = 180
-120 -120
z = 60
Does the function ƒ(x) = (1∕2) + 25 represent exponential growth, decay, or neither?
A) Exponential growth
B) Impossible to determine with the information given.
C) Neither
D) Exponential decay
Answer:
A) Exponential growth
Step-by-step explanation:
A value meal package at Ron's Subs consists of a drink, a sandwich, and a bag of chips. There are 66 types of drinks to choose from, 33 types of sandwiches, and 44 types of chips. How many different value meal packages are possible
36 different value meal packages are possible
Step-by-step explanation:
To answer this question, multiply all given numbers together.
4*3*3
12*3
36
HELP PLEASE!!!!
The median age for a first marriage in the United States for women was 25.9 in 2009 and 26.1 in 2010. Use an exponential model to predict the median age for women in 2019, where x is the number of years since 2009.
A) 23.9
B) 28.3
C) 28.0
D) 24.0
Answer:
29.1 Is the answer for 2009
Step-by-step explanation:
A population is equally divided into three class of drivers. The number of accidents per individual driver is Poisson for all drivers. For a driver of Class I, the expected number of accidents is uniformly distributed over [0.2, 1.0]. For a driver of Class II, the expected number of accidents is uniformly distributed over [0.4, 2.0]. For a driver of Class III, the expected number of accidents is uniformly distributed over [0.6, 3.0]. For driver randomly selected from this population, determine the probability of zero accidents.
Answer:
Following are the solution to the given points:
Step-by-step explanation:
As a result, Poisson for each driver seems to be the number of accidents.
Let X be the random vector indicating accident frequency.
Let, [tex]\lambda=[/tex]Expected accident frequency
[tex]P(X=0) = e^{-\lambda}[/tex]
For class 1:
[tex]P(X=0) = \frac{1}{(1-0.2)} \int_{0.2}^{1} e^{-\lambda} d\lambda \\\\P(X=0) = \frac{1}{0.8} \times [-e^{-1}-(-e^{-0.2})] = 0.56356[/tex]
For class 2:
[tex]P(X=0) = \frac{1}{(2-0.4)} \int_{0.4}^{2} e^{-\lambda} d\lambda\\\\P(X=0) = \frac{1}{1.6} \times [-e^{-2}-(-e^{-0.4})] = 0.33437[/tex]
For class 3:
[tex]P(X=0) = \frac{1}{(3-0.6)} \int_{0.6}^{3} e^{-\lambda} d\lambda\\\\P(X=0) = \frac{1}{2.4} \times [-e^{-3}-(-e^{-0.6})] = 0.20793[/tex]
The population is equally divided into three classes of drivers.
Hence, the Probability
[tex]\to P(X=0) = \frac{1}{3} \times 0.56356+\frac{1}{3} \times 0.33437+\frac{1}{3} \times 0.20793=0.36862[/tex]
If A is the center of the circle, then which statement explains how segment GH is related to segment FH? Circle A with inscribed triangle EFG; point D is on segment EF, point H is on segment GF, segments DA and HA are congruent, and angles EDA and GHA are right angles.
Answer:
I can say for sure that the answer is not segment GH ≅ segment FH because the tangents that create the segment FG share a common endpoint. I believe the answer is segment GH ≅ segment FH because arc EF ≅ arc GF.
Step-by-step explanation:
Again, I'm not sure about the correct answer but I know for sure it isn't segment GH ≅ segment FH because the tangents that create the segment FG share a common endpoint.
The segment GH and the segment FH are equal to each other because the line AH is coming from the center of the circle and is bisecting the line GF.
What is a circumscribed circle?
The circumcenter of a triangle can be constructed by drawing any two of the three perpendicular bisectors. For three non-collinear points, these two lines cannot be parallel, and the circumcenter is the point where they cross.
Any point on the bisector is equidistant from the two points that it bisects, from which it follows that this point, on both bisectors, is equidistant from all three triangle vertices
Hence the segment GH and the segment FH are equal to each other because the line AH is coming from the center of the circle and is bisecting the line GF.
To know more about a Circumscibed circle follow
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Which answer is it I’m confused ... ???
Answer:
the answer is D
Step-by-step explanation:
v=πr²h
divide both side by πh
r²=v/πh
square both sides
r=√v/πh
plz help I will give Brianly
A pair of linear equations is shown below: y = −x + 1 y = 2x + 4 Which of the following statements best explains the steps to solve the pair of equations graphically? (4 points) Select one: a. On a graph, plot the line y = −x + 1, which has y-intercept = −1 and slope = 1, and y = 2x + 4, which has y-intercept = 2 and slope = 4, and write the coordinates of the point of intersection of the two lines as the solution. b. On a graph, plot the line y = −x + 1, which has y-intercept = 1 and slope = 1, and y = 2x + 4, which has y-intercept = 1 and slope = 4, and write the coordinates of the point of intersection of the two lines as the solution. c. On a graph, plot the line y = −x + 1, which has y-intercept = 1 and slope = −1, and y = 2x + 4, which has y-intercept = −2 and slope = 2, and write the coordinates of the point of intersection of the two lines as the solution. d. On a graph, plot the line y = −x + 1, which has y-intercept = 1 and slope = −1, and y = 2x + 4, which has y-intercept = 4 and slope = 2, and write the coordinates of the point of intersection of the two lines as the solution.
Answer:
the answer is 89
Step-by-step explanation:
this is a hard one to salve but basically if you know and lern hout to do it it is not that hard
Can anyone help with this math equation please?
X,and z are midpoints.find the length of each segment
Answers:
MZ = 10ZO = 10MO = 20XZ = 9YZ = 7===========================================
Explanation:
Side MO is twice as long as the midsegment XY. Note how XY and MO are parallel.
This makes
MO = 2*XY = 2*10 = 20
Side MO breaks into two equal halves MZ and ZO
Each of MZ and ZO are 20/2 = 10 units long.
Put another way: XY, MZ and ZO are all the same length (all 10 units long).
---------------
The diagram shows that segment NO is 18 units long, which cuts in half to 18/2 = 9. This is the length of NY, YO and XZ
Also, MN = 14 which cuts in half to 7. This means MX, XN and YZ are all 7 units each.
For a 13-person team, how does the actual weekly labor cost compare to the targeted labor cost?
The actual labor cost is $600 over the targeted labor cost.
Given that,
Work done by each person per week = 40 hours
Required labor hours per week = 600 hours
No. of workers in the team = 13
To find,
Actual weekly labor cost = ?
Procedure:
Actual weekly labor cost = No. of workers * no. of hours performed by them
[tex]= 13 * 40[/tex]
[tex]= 520 hours[/tex]
Given that,
[tex]Regular rate = $ 15.00 per hour[/tex]
[tex]Overtime rate = $ 22.50 per hour[/tex]
Thus,
Actual labor cost = (regular hours worked * regular rate) + (overtime * overtime rate)
[tex]= (520 * 15) + ([600 - 520] * 22.50)[/tex]
[tex]= $ 7,800 + $ 1,800[/tex]
[tex]= $ 9600[/tex]
Targeted Labor cost = $ 9,000 per week
Thus, option C i.e. the actual labor cost is $ 600 over the targeted labor cost.
Learn more about 'Labor Cost' here:
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the tangent of theta is 1, the terminal side of theta lies in the 3rd quadrant. what is a possible value for theta? give your answer in radians or degrees
Answer:
5π/4 radians or 225°
Step-by-step explanation: