Which of the following rules represent the function graphted
F(x)=1/2x
F(x)=x
F(X) = x + 2

The first term, a1, in the geometric series is -3.

What is Geometric Series?

A geometric series is a series for which the ratio of two consecutive terms is a constant function of the summation index. The more general case of a ratio and a rational sum-index function produces a series called a hypergeometric series. For the simplest case of a ratio equal to a constant, the terms have the form

To find the first term, a1, in a geometric series given the sum, Sn = 93, the common ratio, r = 2, and the number of terms, n = 5, we can use the formula for the sum of a geometric series:

Sn = a1 * (1 - r^n) / (1 - r)

Plugging in the given values, we have:

93 = a1 * (1 - 2^5) / (1 - 2)

Simplifying the expression:

93 = a1 * (1 - 32) / (-1)

93 = a1 * (-31)

Now we can solve for a1 by dividing both sides of the equation by -31:

a1 = 93 / -31

a1 = -3

Therefore, the first term, a1, in the geometric series is -3.

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(a) The function g(x) = -8\(x^2\) + 5x - 2 doesn't have two x-intercepts because ac > 0.

(b) The function h(x) = (-1/3)\(x^2\) + 3x doesn't have two x-intercepts because ac = 0.

(c) The function k(x) = 8\(x^2\) - 5x - 7 have two x-intercepts because ac < 0.

In the theorem, a, b, and c are real numbers.

Theorem - If f is a quadratic function of the form f(x) = a\(x^2\) + bx + c and ac < 0, then the function f has two x-intercepts.

(a) The function is g(x) = -8\(x^2\) + 5x - 2

On comparing the equation by f(x) = a\(x^2\) + bx + c, we get a = -8, b = 5 and c = -2.

To check the two x-intercepts, we put values in ac and compare with ac<0

ac = (-8)(-2)

ac = 16 > 0

As the value of ac is greater than 0. So we can say that it doesn't have two x-intercepts.

(b) The function is h(x) = (-1/3)\(x^2\) + 3x

On comparing the equation by f(x) = a\(x^2\) + bx + c, we get a = -1/3, b = 3 and c = 0.

To check the two x-intercepts, we put values in ac and compare with ac<0

ac = (-1/3)(0)

ac = 0 = 0

As the value of ac is equal to 0. So we can say that it doesn't have two x-intercepts.

(c) The function is k(x) = 8\(x^2\) - 5x - 7

On comparing the equation by f(x) = a\(x^2\) + bx + c, we get a = 8, b = -5 and c = -7.

To check the two x-intercepts, we put values in ac and compare with ac<0

ac = 8(-7)

ac = -56

ac < 0

As the value of ac is less than 0. So we can say that it have two x-intercepts.

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Answer:

y=2x-7

Step-by-step explanation:

2 would equal M because m is your slope.

-7 would be your y intercept

following y=mx+b

The required equation of the line is given as y = 2x -7.

Given that,
The slope of 2 and y-intercept of −7

The slope of the line is a tangent angle made by line with horizontal. i.e. m =tanx where x in degrees.

Here,
The equation of the line is given as.
y = mx + c
Substitute the value in the equation,
y = 2x - 7

Thus, the required equation of the line is given as y = 2x -7.

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Answer:

4

Step-by-step explanation:

t = d/r

t = 280/70

t = 4 hours

Answer:

37.7 inches

Step-by-step explanation:

u have the formula so with r being radius just multiply by 2 then multiply that answer by pie and round to the nearest tenth.

In this simulation experiment, the population distribution represents the lifetimes of a certain type of electronic control and the density curve in the figure represents the population distribution.

The distribution is skewed and follows a lognormal distribution with an expected value of ln(X) equal to 3 and a variance of ln(X) equal to 0.16. The experiment consisted of 500 replications and examined different sample sizes: n = 5, n = 10, n = 20, and n = 30.

The density curve in the figure represents the population distribution, with an expected value of E(A) equal to 21.7584 and a variance of V(X) equal to 82.1449. The histograms shown illustrate the distribution of sample means for the different sample sizes. Additionally, a normal probability plot from MINITAB is included for the x-values based on a sample size of n = 30, allowing for an assessment of the normality assumption for the sample means.

These visual representations provide valuable insights into the characteristics of the simulated experiment's population distribution, as well as the behavior of the sample means for different sample sizes.

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The complete question is:

Consider a simulation experiment in which the population distribution is quite skewed. The figure below shows the density curve for lifetimes of_a certain type of electronic control [this is actually a lognormal distribution with E(ln(X)) = 3 and V(ln(X)) = 0.16]. The statistic of interest is the sample mean X. The experiment utilized 500 replications and considered these sample sizes: n = 5, n = 10, n = 20, and n = 30. The resulting histograms along with a normal probability plot from MINITAB for the x values based on n = 30 are shown in the figures below. Density curve for the simulation experiment (E(A) = 21.7584, V(X) = 82.1449].

It is not a field because it has zero divisors. Hence, congruence class ring Z5[x]/(x3-x2+2x+1) is not a field.

No, the congruence class ring Z5[x]/(x3-x2+2x+1) is not a field.

In a ring, if there is a nonzero divisor, it will not be a field.

For this to be a field, the congruence class ring Z5[x]/(x3-x2+2x+1) has to have no zero divisors.

However, in Z5[x]/(x3-x2+2x+1), x3-x2+2x+1 is irreducible over Z5[x], which means that (x3-x2+2x+1) is a prime ideal in Z5[x].

Hence, Z5[x]/(x3-x2+2x+1) is a domain since (x3-x2+2x+1) is a prime ideal.

However, it is not a field because it has zero divisors.

Hence, congruence class ring Z5[x]/(x3-x2+2x+1) is not a field.

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Using eIiminatiοn, NataIie is 13 years οId and Frankie is 31 years οId.

An equatiοn is a mathematicaI statement that shοws that twο expressiοns are equaI. It cοnsists οf twο sides separated by an equaIs sign (=). The expressiοns οn bοth sides οf the equatiοn can cοntain numbers, variabIes, and mathematicaI οperatiοns such as additiοn, subtractiοn, muItipIicatiοn, and divisiοn.

Let's use variabIes tο represent NataIie's and Frankie's ages.

Let's use the variabIe "N" tο represent NataIie's age and "F" tο represent Frankie's age.

We can write twο equatiοns based οn the infοrmatiοn given in the prοbIem:

Equatiοn 1: N + F = 44 (the sum οf their ages is 44)

Equatiοn 2: 3N + F = 70 (Frankie's age added tο 3 times NataIie's age equaIs 70)

We can use the eIiminatiοn methοd tο sοIve the system οf equatiοns.

First, we'II eIiminate "F" by muItipIying Equatiοn 1 by -1 and adding it tο Equatiοn 2:

-1(N + F = 44) => -N - F = -44

3N + F = 70

2N = 26

N = 13

Nοw we can substitute N = 13 intο either equatiοn tο find Frankie's age:

N + F = 44

13 + F = 44

F = 31

Therefοre, NataIie is 13 years οId and Frankie is 31 years οId.

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The answers to the questions involving trigonometry are: 90, BC/AB ÷ BC/AB = 1, g = 6.5, <I = 62 degrees, h= 13.8, 12.0, x = 6.8, x = 66.4, 160.6, The pole = 6.7

Trigonometric ratios are special measurements of a right triangle, defined as the ratios of the sides of a right-angled triangle.  There are three common trigonometric ratios: sine, cosine, and tangent

For page 3 question 3,

a) <A + <B = 90 since <C = right angle

b) SinA = BC/AB and CosB = BC/AB

The ratio of the two angles BC/AB ÷ BC/AB = 1

I notice that the ratio of sinA and cosB gives 1

b) The ratio of CosA and SinB will give

BC/AB ÷ BC/AB

= BC/AB * AB/BC = 1

For page 5 number 3

Tan28 = g/i

g/12.2 = tan28

cross multiplying to have

g = 12.2*tan28

g = 12.2 * 0.5317

g = 6.5

b) the angle I is given as 90-28 degrees

<I = 62 degrees

To find the side h we use the Pythagoras theorem

h² = (12.2)² + (6.5)²

h² = 148.84 +42.25

h²= 191.09

h=√191.09

h= 13.8

For page 6

1) Sin42 = x/18

x=18*sin42

x = 18*0.6691

x = 12.0

2) cos28 = 6/x

xcos28 = 6

x = 6/cos28

x [= 6/0.8829

x = 6.8

3)  Tan63 = x/34

x = 34*tan63

x= 34*1.9526

x = 66.4

4) Sin50 123/x

xsin50 = 123

x = 123/sin50

x = 123/0.7660

x =160.6

5)  Sin57 = P/8

Pole = 8sin57

the pole = 8*0.8387

The pole = 6.7

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Answer:

360

Step-by-step explanation:

The random process Y(t) is not wide-sense stationary (WSS) because the phase term, ϕ, is uniformly distributed between 0 and π/4. In a WSS process, the statistical properties, such as mean and autocorrelation, should be independent of time.

To determine if the random process Y(t) is wide-sense stationary (WSS), we need to examine its statistical properties. A WSS process has two main characteristics: time-invariance and finite second-order moments.

Let's analyze the given process: Y(t) = A sin(wet + ϕ), where A is the amplitude, ω is the angular frequency, et is the time, and ϕ is uniformly distributed between 0 and π/4.

1. Time-Invariance: A WSS process should exhibit statistical properties that are independent of time. In this case, the phase term ϕ is uniformly distributed between 0 and π/4. As time progresses, the phase term ϕ changes randomly, leading to time-dependent variations in the process Y(t). Therefore, the process is not time-invariant and does not satisfy the first condition for WSS.

2. Finite Second-Order Moments: A WSS process should have finite mean and autocorrelation functions. Let's examine the mean and autocorrelation of Y(t):

Mean: E[Y(t)] = E[A sin(wet + ϕ)] = A E[sin(wet + ϕ)]

Since ϕ is uniformly distributed between 0 and π/4, its expected value is E[ϕ] = (0 + π/4) / 2 = π/8.

E[Y(t)] = A E[sin(wet + ϕ)] = A E[sin(wet + π/8)]

The expected value of sin(wet + π/8) is not zero, and it varies with time. Therefore, the mean of Y(t) is time-dependent, violating the WSS condition.

Autocorrelation: R_Y(t1, t2) = E[Y(t1)Y(t2)] = E[A sin(wet1 + ϕ)A sin(wet2 + ϕ)]

Expanding this expression and taking expectations, we have:

R_Y(t1, t2) = A^2 E[sin(wet1 + ϕ)sin(wet2 + ϕ)]

The product of two sine terms can be expanded using trigonometric identities. The resulting expression will involve cosines and sines of the sum and difference of the angles. Since ϕ is uniformly distributed, these trigonometric terms will also vary with time, making the autocorrelation function time-dependent.

Hence, we can conclude that the random process Y(t) is not wide-sense stationary (WSS) due to the time-dependent phase term ϕ, which violates the time-invariance property required for WSS processes.

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The different quotient for f(x)=8x³+1 is 8(-x³+(h+x)³)/h.

it is given that, g(x)=8x³+1

The difference quotient is given by f(x+h)-f(x)/h

to find f(x+h), put x+h instead of x:

then, f(x+h)=8(x+h)³+1

finally, = f(x+h)-f(x)/h

=(8(x+h)³+1))-(8x³+1)/h

=8(-x³+(h+x)³)/h

Hence, The different quotient for f(x)=8x³+1 is 8(-x³+(h+x)³)/h.

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Answer:

4

2x2=4

Step-by-step explanation:

Brainliest Plzzz!!!

Answer:

4

Step-by-step explanation:

i learned this in 1st grade

Answer:

$119.79

Step-by-step explanation:

im like 95% sure this is right

Answer:

In three years Bernie total saving account would be $119.79

Step-by-step explanation:

The answer varies each year. In one year 10% of $90 is $9. Bernie total saving account would be $99 in the first year. In two years it would be $108.90( 10% of $99 = $9.9, $9.9 + $99 = $108.90). In three years Bernie total saving account would be $119.79( 10% of 108.90 = 10.89, $10.89 + $108.9 = $119.79) and so on.

Answer:

both the second and third one

Answer:

6/15 divided by 6

Step-by-step explanation:

\(\frac{2}{5} / 6\\= \frac{6}{15} / 6\\\)

the probability of the event that  a sample of 5 cans will have a meaningful amount of at least 12.13 oz is 0.0078

We are provided with the information that cans of coke are filled in the account of having normal distribution, with the provided means which is  12.00 oz and the standard deviation of 0.12 oz and the sample size is

n= 5.

let X be the sample mean, so the probability will be :

P(X\(\geq\) 12.13) = 1- P(X<12.13)

               =1-  P(X-(μ/(σ/\(\sqrt{n}\))) < (12.13 -12.00)/(0.12/\(\sqrt{5}\)))

            = 1- P(z<2.42)

            = 1-0.9922

           =0.0078

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Cans of regular Coke are labeled as containing 12 oz. Statistics students weighted the content of 5 randomly chosen cans, and found the mean weight to be 12.13. Assume that cans of Coke are filled so that the actual amounts are normally distributed with a mean of 12.00 oz and a standard deviation of 0.12 oz. Find the probability that a sample of 5 cans will have a mean amount of at least 12.13 oz.

If two fair dice are rolled, the conditional probability that at least one is a 6 given that the two rolls give different numbers is 1/6

Two fair dice are rolled

Total number of outcomes = 6 × 6

= 36

The probability is the ratio of  number of favorable outcomes to total number of outcomes

The probability = Number of favorable outcomes / Total number of outcomes

Consider the two events E and F

P(E/F) = P(E∩F) / P(F)

P(E∩F) = 5/36

P(F) = 30/36

Substitute the values in the equation

P(E/F) = (5/36) / (30/36)

= (5/36) × (30/36)

= 1/6

Therefore, the conditional probability is 1/6

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~Denki~

Answer:

the length of "AC" should be the same as it was before it was reflected over the y-axis.

(there is no picture so i cant get as specific as intended)

hope this helps :D

Answer:

15% of $67.50 is $10.125 lets round it up to $10.13

Tip: $10.13

Total meal cost: $77.63

Brainly plz :))

Answer:

the waiter 10.13

your parents will pay 77.63

Step-by-step explanation:

67. 50 x 0.15 = 10.13

67.50 + 10.13 = 77.63

hope this helps

After looking carefully the six cards, I

Answer:24%

Step-by-step explanation:

When you multiply a fraction by 100 you will get the percent

12/50 x 100=24

The probability of choosing any 5 students will be 5% so for all the options the probability will be same.

Probability is the extent to which an event is likely to occur, measured by the ratio of the favourable cases to the whole number of cases possible.

Here total samples are [90]

favorable outcomes should be [5]

So the probability to select 5 random students from 90 will be

\(P=\dfrac{5}{90}=0.05=0.05\times 100=5\%\)

Hence the probability of choosing any 5 students will be 5% so for all the options the probability will be same.

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Answer:

21 or 12

Step-by-step explanation:

Suppose, the number of one of the houses is “xy”, whose numerical value = (10x + y).

So, the number of the other house is “yx”, whose numerical value = (10y + x).

Let, “xy” > “yx”.

Now, difference of house numbers = {(10x + y) - (10y + x)} = {9 * (x - y)}; so the difference is a multiple of 9.

Now, the smallest positive number, divisible by 9, that ends with 2 is 72 = (9 * 8).

So, (x - y) = 8.

Since, 0 ≤ x, y ≤ 9; so, there are two solutions in hand:

I) x = 8, y = 0……………………..this makes the house numbers: “80” and “08”.

II) x = 9, y = 1…………………..this makes the house numbers: “91” and “19”.

So, the lowest possible numbers of the houses are “80” and “8” (not “91” and “19”).

A single digit number can not have its reverse number. hence 10,20,30,40,50,60,70,80 and 90 can not be the house number

Reverse of these numbers is same, hence 11,22,33,44,55,66,77,88 and 99 can not be the house number

For other reverse numbers the difference will be in multiples of 9 like 9,18,27,36,45,54,63,72,81

As the difference between house numbers ends in two, the lowest possible numbers of our house are 19 and 91

When 0 (zero) also considred as part of house number, the lowest possible numbers of our house are 08 and 80

Answer:

3

Step-by-step explanation:

From inspection, we can see that the dilated triangle (red) has side lengths that are 3 times bigger than the side lengths of the original triangle (blue).

Therefore, the scale factor for the dilation is 3 (about the origin)

On solving the provided question, we cans ay that the percentage of people that vote in favour is required by the random variable supplied in b, making that option the best one.

A percentage in mathematics is a figure or ratio that is stated as a fraction of 100. The abbreviations "pct.," "pct," and "pc" are also occasionally used. It is frequently denoted using the percent symbol "%," though.

We estimate the parameter based on the sample values when the parameter is unknown. A point estimate is used to estimate the unknown parameter and is a function of sample values. Additionally, we are aware that statistic refers to the function of sample values. If a confidence interval is described as having a lower endpoint equal to the point estimate minus the margin of error, then

The proportion is used to calculate the percentage of people who possess a given quality. Since the random variables in letters a, c, and d are numerical, the confidence interval for the mean will be applied to them.

Additionally, the percentage of people that vote in favour is required by the random variable supplied in b, making that option the best one.

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Answer:

\(3a\geq 300\)

Step-by-step explanation:

Answer: first number = 24

second number = 19

third number = 57

Explanation:

Let the second number be x. From the information given,

the first number is 5 more than the second. This means that

the first number is x + 5

The third number is 3 times the second. This means that

the second number is 3x

If the sum of the three numbers is 100, it means that

x + 5 + x + 3x = 100

5x + 5 = 100

5x = 100 - 5 = 95

x = 95/5

x = 19

The first number = 19 + 5 = 24

The second number = 19

the third number = 3 * 19 = 57

To calculate the covariance and correlation coefficient of U and V, we need to first find the expected value and variance of U and V.

The covariance of two variables is a gauge of their relationship. In other words, it is a measure of how much the values of two variables tend to fluctuate together. Positive covariance shows that the values of the two variables tend to rise or decrease together, whereas negative covariance indicates that the values of one variable tend to increase when the values of the other variable drop. In probability theory and statistics, covariance is frequently employed to determine the association between two variables. It is determined by multiplying the variances of the two variables from their respective means and dividing the result by the total number of observations. This metric can be helpful for seeing trends and patterns in data and aiding analysts in forecasting.

How to solve?

The expected value of U is $E[U] = E[2 - X_1 - 2X_2] = 2 - m - 2m = -2m. Similarly, the expected value of V is $E[V] = E[3 + 3X_2 - X_3] = 3 + 3m - m = 4m.

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To calculate the value of covariance and correlation coefficient of U and V, we have to find their expected value along with variance.

A measure of the link between two random variables and how much they fluctuate together is called covariance. Alternately, we may say that it establishes the relationship between the two variables' changes, i.e., that a change in one variable is equivalent to a change in the other.

When one variable changes, covariance shows the link between the other two variables. Both variables are said to have positive covariance if a rise in one causes an increase in the other. Reduced values of one variable also result in reduced values of the other.

for expected value of U ;

= $E[U]

= E[2 - X_1 - 2X_2]

= 2 - m - 2m = -2m.

Similarly, the expected value of V;

= $E[V]

= E[3 + 3X_2 - X_3]

= 3 + 3m - m

= 4m.

now we can easily calculate the covariance and correlation coefficient of U,V.

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