Answer: f
= x = − 3 x + 1 = x = 1 4 g = x 2 + 2 x − 5f( x ) = f ( x )
Step-by-step explanation:
Which would be the next step? Does anyone know? Please, help.
Answer:
∠DAB = ∠DBA
Then AD=DB from above statement
can you guys help me please as much as you can ^^
Answer:
Step-by-step explanation:
m<1 = 50
m<2 = 22
m<3 = 108
m<4 = 50
m<5 = 65
m<6 = 50
m<7 = 43
m<8 = 65
m<9 = 75
m<10 = 18
m<11 = 25
m<12 = 40
m<13 = 115
m<14 = 65
m15 = 115
Suppose we are testing the null hypothesis H_o: µ = 16 against the alternative H_a: µ > 16 from a normal population with known standard deviation σ=4. A sample of size 324 is taken. We use the usual z statistic as our test statistic. Using the sample, a z value of 2.34 is calculated. (Remember z has a standard normal distribution.)
a) What is the p value for this test? ______
b) Would the null value have been rejected if this was a 2% level test?
OY
ON
c) Would the null value have been rejected if this was a 1% level test?
OY
ON
d) What was the value of x calculated from our sample? _______
a) The p- value is 0.0094.
b) True
c) Yes
To calculate the p-value for the test, we can use the standard normal distribution table or a statistical calculator.
a) The p-value is the probability of obtaining a test statistic as extreme as the observed value or more extreme if the null hypothesis is true. Since we are testing the alternative hypothesis H_a: µ > 16, the p-value is the probability of getting a z-value greater than 2.34.
Using a standard normal distribution table, the p-value corresponding to a z-value of 2.34 is 0.0094.
b) If this was a 2% level test, the null hypothesis would be rejected if the p-value is less than the significance level of 0.02.
Since the p-value (0.0094) is less than the significance level, the null hypothesis would have been rejected at the 2% level.
c) If this was a 1% level test, the null hypothesis would be rejected if the p-value is less than the significance level of 0.01.
Since the p-value (0.0094) is greater than the significance level, the null hypothesis would not have been rejected at the 1% level.
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In a survey of 3203 adults, 1447 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results. A 99% confidence interval for the population proportion is ??
Answer:
(0.4291, 0.4743)
Step-by-step explanation:
Using the relation :
p ± Zcritical * Sqrt[(p(1-p)) / n]
P = x / n =. 1447 / 3203 = 0.4517
1 - p = 0.5483
Zcritical at 99% = 2.575
Sqrt[(p(1-p)) / n] = sqrt(0.4517(0.5483)) / 3203) = 0.008793
p ± Zcritical * 0.008793
Lower boundary = 0.4517 - (2.575 * 0.008793) = 0.4291
Upper boundary = 0.4517 + (2.575 * 0.008793) = 0.4743
(0.4291, 0.4743)
Tuition for one year at Mount Tusk University is $13,400. Elijah has saved
$1,800 for this year's tuition. His parents have agreed to pay $5,000 this
year, and he earned a football scholarship for $3,500. Elijah hopes to earn
the rest of the money through his work-study job tutoring in the math
center. How much money does Elijah need to earn tutoring? (Enter the
quantity only)
Your answer
Answer:
The answer is 3,100
Step-by-step explanation:
If you subtract the 8,500 from his parents and his football scholarship you get 4,900 and then take the 1,800 from his saved money, You get what he still needs, which is 3,100.
Suppose you place a $35 bet on a horse with a 2:7 odds against winnning. Determine the winning payout for this horse.
The winning payout for the horse is $45
To determine the winning payout for the horse, you need to use the following formula:
Odds against winning: B / (A + B)
Betting amount: X Payout: X + (X * B / A)where A is the denominator of the odds and B is the numerator of the odds.
Here, the odds against winning are 2:7.
So, the denominator (A) is 7 and the numerator (B) is 2.
The betting amount is $35.
Plugging these values into the formula:
Payout = 35 + (35 * 2 / 7)
Payout = $45.
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A line that includes the points (-2, c) and (-1, 10) has a slope of 2. What is
the value of c?
Answer:
8
Step-by-step explanation:
m = (Y2-Y1) ÷ (X2- X1) 2 = (10-c) ÷ (-1-(-2)) 2 = (10-c) ÷( 1)2= 10-cc = 10-2c= 8(-2,c) and (-1,10)
10-c
-1--2
10-c
1
We need the fraction to be 2, or 2/1. The bottom number is 1, so we technically already have the answer. But we still need to plug in a number for c. To get 2, we need to subtract 8 from 10.
So c is 8.
---
hope it helps
sorry my work was a mess
Shirley is saving money to buy a computer.
• The computer she will buy costs $1,200.00
• She has already saved $300.00
Shirley will save $60.00 per week until she has enough money to buy
the computer.
Answer:
It will take her 15 weeks to have enough money for the computer.
Step-by-step explanation:
$1200 - $300 = $900
900 divided by 60 gives you 15
Therefor, it'll take Shirley 15 weeks to save $1200.
A bakery made 55 boxes of rolls. Each box holds 12 rolls. How many rolls were made in all?
Make an equation to represent the problem. Drag numbers and symbols to the lines.
55
12
+
X
If 25% of a number is 65 and 60% of the same number is 156, find 35% of that number.
Answer:
35% of that number is 91.
Step-by-step explanation:
we need to first find 25% of what number equals 65.
65×100÷25 = 260
Then, if you use the same method again, 60% of 260 would be 156.
We know "that" number is 260, and all we need to do is find 35% of it.
35% of 260 = 91
Hope this helped :)
. A random variable X has pdf fX(x) = 2e −2x , x ≥ 0.
(a) Use Chebyshev’s inequality to obtain an upper bound for P(X /∈ (µX − 1, µX + 1))
(b) Use Chebyshev’s inequality to obtain a lower bound for P(X ∈ (µX − 3, µX + 3))
(a) The upper bound for P(X ∈ (µX − 1, µX + 1)) using Chebyshev's inequality is 0.75.
(b) The lower bound for P(X ∈ (µX − 3, µX + 3)) using Chebyshev's inequality is 0.55.
(a) The upper bound for \(P(X \notin (\mu_X - 1, \mu_X + 1))\) using Chebyshev's inequality can be found as follows:
Chebyshev's inequality states that for any random variable \(X\) with mean \(\mu_X\) and standard deviation \(\sigma_X\), the probability that \(X\) deviates from its mean by more than \(k\) standard deviations is at most \(1/k^2\).
In this case, we have the random variable \(X\) with the probability density function (pdf) \(f_X(x) = 2e^{-2x}\) for \(x \geq 0\). The mean \(\mu_X\) of this distribution can be calculated as \(\mu_X = \int_0^\infty xf_X(x) dx\). By integrating, we find \(\mu_X = \frac{1}{2}\).
To calculate the standard deviation \(\sigma_X\), we need to find the variance first. The variance \(\text{Var}(X)\) is given by \(\text{Var}(X) = E[X^2] - (E[X])^2\). Evaluating the integral, we find \(E[X^2] = \frac{3}{4}\).
Thus, the variance is \(\text{Var}(X) = \frac{3}{4} - \left(\frac{1}{2}\right)^2 = \frac{1}{4}\). Taking the square root of the variance gives us the standard deviation \(\sigma_X = \frac{1}{2}\).
Now, applying Chebyshev's inequality with \(k = 1\), we have \(P(X \notin (\mu_X - 1, \mu_X + 1)) \leq \frac{1}{1^2} = 1\).
Therefore, the upper bound for \(P(X \notin (\mu_X - 1, \mu_X + 1))\) is 1.
Chebyshev's inequality is a probabilistic bound that gives us an estimate of how likely a random variable is to deviate from its mean by a certain number of standard deviations. In this case, we used Chebyshev's inequality to find an upper bound for the probability that \(X\) falls outside the interval \((\mu_X - 1, \mu_X + 1)\).
By calculating the mean and standard deviation of the random variable \(X\), we were able to apply Chebyshev's inequality and determine that the probability is bounded above by 1. This means that it is guaranteed that \(X\) will be within the interval \((\mu_X - 1, \mu_X + 1)\) at least 0% of the time.
(b) The lower bound for \(P(X \in (\mu_X - 3, \mu_X + 3))\) using Chebyshev's inequality can be obtained as follows:
By the same reasoning as in part (a), we have the mean \(\mu_X = \frac{1}{2}\) and the standard deviation \(\sigma_X = \frac{1}{2}\) for the random variable \(X\) with pdf \(f_X(x) = 2e^{-2x}\) for \(x \geq 0\).
Applying Chebyshev's inequality with \(k = 3\), we have \(P(X \notin (\mu_X - 3, \mu_X + 3)) \leq \frac{1}{3^2} = \frac{1}{9}\).
To find the lower bound
for \(P(X \in (\mu_X - 3, \mu_X + 3))\), we subtract the upper bound from 1: \(P(X \in (\mu_X - 3, \mu_X + 3)) \geq 1 - \frac{1}{9} = \frac{8}{9}\).
Therefore, the lower bound for \(P(X \in (\mu_X - 3, \mu_X + 3))\) is \(\frac{8}{9}\).
Chebyshev's inequality allows us to establish a lower bound for the probability that a random variable falls within a certain range around its mean. In this case, we used Chebyshev's inequality to find a lower bound for the probability that \(X\) falls within the interval \((\mu_X - 3, \mu_X + 3)\).
By calculating the mean and standard deviation of the random variable \(X\), we applied Chebyshev's inequality with \(k = 3\) to obtain an upper bound for the probability of being outside the interval.
Subtracting this upper bound from 1 gives us the lower bound for the desired probability, which is \(\frac{8}{9}\). This means that at least 88.9% of the time, \(X\) will fall within the interval \((\mu_X - 3, \mu_X + 3)\).
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MATH PLEASE HELP!! HELP BADLY NEEDED....
Select the correct answer.
What are the asymptote and the y-intercept of the function shown in the graph?
f(x) = 3(0.2)^x + 2
A. asymptote: y = -2
y-intercept: (0,5)
B. asymptote: y = 2
y-intercept: (0,5)
C. asymptote: y = 2
y-intercept: (0,4)
D. asymptote: y = -2
y-intercept: (0,3)
Answer:
B
Step-by-step explanation:
The function reaches the y-axis at the point (0,5).
The asymptote is the line that the function follows but never quite reaches. In this case, the function follows the path of y = 2. However, it never exactly fits the line.
The y-intercept is (0,5) and the asymptote is y = 2. The answer, then, is B.
Good luck ^^
The equation of the asymptote is y = 2 and the coordinate of the y-intercept will be (0, 5). Then the correct option is B.
What is asymptote?An asymptote is a line that constantly reaches a given curve, but does not touch at any infinite distance.
The equation of the function is given below.
[tex]\rm f(x) = 3(0.2)^x + 2[/tex]
The asymptote of the function is given as by substituting x as infinity, then the equation of the asymptote will be
[tex]\rm y = 3(0.2)^{\infty} + 2\\\\y = 2[/tex]
Then the y-intercept of the function will be given by substituting y = 0, then the y-intercept will be
y = 3(0.2)⁰ + 2
y = 3 + 2
y = 5
The coordinate of the y-intercept will be (0, 5).
Then the correct option is B.
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Mikel is trying to save for college expenses. He has a job, but can only afford to put $20 per month aside. He has 4 years until he will need to pay for college. How much will he have saved by the end of 4 years?
Answer:
960
20 times 12 moths to get 240 then multiply that by 4 years and get 960 .
Step-by-step explanation:
sorry if wrong ]
have a good day /night
may i pleae have a branlliest .
20 x 12=240 x 4 =960
Answer:
960
Step-by-step explanation:
what two number have an absolute value of 12?
Answer:
The absolute value of 12, is 12...
Step-by-step explanation:
Answer:
-12
Step-by-step explanation:
Jasmine scored a 85 on the last math test. The class average was a 76 with a standard deviation of 4.5. So, XN(76,4,5) Jasmine's Z-score is This tells you that 85 is standard deviations to the left or right) of the mean, 14. The number of problems on all math exams are normal distributed. What is the probability a randomly selected math exam has fewer than 15 questions if the mean is 20 questions with a standard deviation of 2.5? Use the empirical rule. Enter your answer as a percent rounded to two decimal places if necessary. A city has around 890 thousand people. There are 123 parks in this city. What is the number of parks per capita in this city? Write your answer in scientific notation.
Jasmine's Z-score is 2. The number of parks per capita in the city is [tex]1.38 * 10^-4[/tex] in scientific notation.
To calculate Jasmine's Z-score, we can use the formula:
Z = (X - μ) / σ
where X is the individual score (85), μ is the mean (76), and σ is the standard deviation (4.5).
Z = (85 - 76) / 4.5
Z = 9 / 4.5
Z = 2
Since Jasmine's Z-score is 2, this tells us that her score of 85 is 2 standard deviations to the right of the mean.
Now let's calculate the probability of randomly selecting a math exam with fewer than 15 questions using the mean of 20 and a standard deviation of 2.5.
To apply the empirical rule, we need to determine how many standard deviations 15 is away from the mean.
Z = (X - μ) / σ
Z = (15 - 20) / 2.5
Z = -5 / 2.5
Z = -2
Since 15 is 2 standard deviations to the left of the mean, we can use the empirical rule to estimate the probability.
According to the empirical rule:
The data is within one standard deviation of the mean for about 68% of the time.
The data is within 2 standard deviations of the mean for about 95% of the time.
99.7% of the data are contained within a 3 standard deviation range around the mean.
Since 15 is beyond 2 standard deviations to the left, the probability of randomly selecting a math exam with fewer than 15 questions would be very close to 0. In this case, we can assume it's effectively 0%.
Now let's calculate the number of parks per capita in the city with 890,000 people and 123 parks.
Number of parks per capita = Number of parks / Population
Number of parks per capita = 123 / 890,000
To write the answer in scientific notation, we can express 890,000 as 8.9 x 10^5:
Number of parks per capita =[tex]123 / (8.9 * 10^5)[/tex]
Calculating the result:
Number of parks per capita =[tex]1.38 * 10^-4[/tex]
Therefore, the number of parks per capita in the city is[tex]1.38 * 10^-4[/tex] in scientific notation.
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An airline has a policy of booking as many as 11 persons on an airplane that can seat only 10. (Past studies have revealed that only 86.0% of the booked passengers actually arrive for the flight.) Find the probability that if the airline books 11 persons, not enough seats will be available. Is it unlikely for such an overbooking to occur? The probability that not enough seats will be available is (Round to four decimal places as needed.) Is it unlikely for such an overbooking to occur? A. It is unlikely for such an overbooking to occur, because the probability of the overbooking is less than or equal to than 0.05. B. It is unlikely for such an overbooking to occur, because the probability of the overbooking is greater than 0.05. OC. It is not unlikely for such an overbooking to occur, because the probability of the overbooking is less than or equal to than 0.05. OD. It is not unlikely for such an overbooking to occur, because the probability of the overbooking is greater than 0.05.
The probability that there won't be enough seats available if the airline books 11 persons is 0.3274. It is not unlikely for such an overbooking to occur because the probability of the overbooking is greater than 0.05.
To find the probability that there won't be enough seats available, we need to calculate the probability that more than 10 persons show up out of the 11 booked. This can be done using the binomial distribution.
The probability of a person showing up for the flight is given as 86.0%, which means the probability of not showing up is 14.0%. Since the events of individuals showing up or not showing up are independent, we can use the binomial distribution to calculate the probability.
Using the binomial distribution formula, we can calculate the probability of 11 or more persons showing up out of 11 bookings. This gives us a probability of 0.3274.
To determine if it is unlikely for such an overbooking to occur, we compare the probability to a significance level of 0.05. If the probability is less than or equal to 0.05, we can consider it unlikely. However, in this case, the probability of 0.3274 is greater than 0.05, indicating that it is not unlikely for such an overbooking to occur.
Therefore, the correct answer is OD. It is not unlikely for such an overbooking to occur, because the probability of the overbooking is greater than 0.05.
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maths equation : solve (2 - x)² < 4/25
Answer:
8/5 < x < 12/5
Step-by-step explanation:
(2 - x)² < 4/25
2 - x < ±2/5
2 - x < 2/5
-x < -8/5
x > 8/5
2 - x > -2/5
-x > -12/5
x < 12/5
Therefore, your answer is 8/5 < x < 12/5
If the point M(2,2) is reflected over the y axis, what will be the coordinates of the resulting point, M’?
Answer:
-8,5
Step-by-step explanation:
8,5 because the "m" is 5 squares down and 8 squares to the right.
selects a piece of candy and eats it (so it is NOT replaced!) Then selects a piece of candy and eats it. Find the probability of each event
Question:
There are 30 candies in a box, all identically shaped. 5 are filled with coconut, 10 with caramel, and 15 are solid chocolate.
You randomly select a piece of candy and eat it (so it is NOT replaced!), then select a second piece. Find the probability of each event
(a) The probability of selecting two solid chocolates in a row.
(b) The probability of selecting a caramel and then a coconut candy.
Answer:
[tex](a)[/tex] [tex]P(Chocolates) = \frac{7}{29}[/tex]
[tex](b)[/tex] [tex]P(Caramel\ and\ Coconut) = \frac{5}{87}[/tex]
Step-by-step explanation:
Given
[tex]Coconut = 5[/tex]
[tex]Caramel = 10[/tex]
[tex]Chocolate = 15[/tex]
[tex]Total = 30[/tex]
For probabilities without replacement, 1 is subtracted after the first selection.
So, we have:
Solving (a): Two solid chocolates
This is calculated as:
[tex]P(Chocolates) = P(First\ Chocolate) * P(Second\ Chocolate)[/tex]
[tex]P(Chocolates) = \frac{n(Chocolate)}{Total} * \frac{n(Chocolate) - 1}{Total - 1}[/tex]
[tex]P(Chocolates) = \frac{15}{30} * \frac{15 - 1}{30 - 1}[/tex]
[tex]P(Chocolates) = \frac{15}{30} * \frac{14}{29}[/tex]
[tex]P(Chocolates) = \frac{1}{2} * \frac{14}{29}[/tex]
[tex]P(Chocolates) = \frac{7}{29}[/tex]
Solving (a): Caramel and Coconut
This is calculated as:
[tex]P(Caramel\ and\ Coconut) = P(Caramel) * P(Coconut)[/tex]
[tex]P(Caramel\ and\ Coconut) = \frac{n(Caramel)}{Total} * \frac{n(Coconut)}{Total - 1}[/tex]
[tex]P(Caramel\ and\ Coconut) = \frac{10}{30} * \frac{5}{30- 1}[/tex]
[tex]P(Caramel\ and\ Coconut) = \frac{10}{30} * \frac{5}{29}[/tex]
[tex]P(Caramel\ and\ Coconut) = \frac{1}{3} * \frac{5}{29}[/tex]
[tex]P(Caramel\ and\ Coconut) = \frac{5}{87}[/tex]
What is the value of the output when the
input is 18?
Answer:
Any number less than greater than or equal to 18
Step-by-step explanation:
Joshua has 3.95 pounds of candy. He is placing the candy into 5 equal size bags. How much candy will be in each bag?
Find the value of the expression below
when x =3/4
4x² + 8x - 5
Answer:
4x² + 8x - 5
= 4(3/4)² + 8(3/4) - 5
= 4 × 9/16 + 24/4 - 5
= 36/16 + 24/4 - 5
= 9/4 + 6 - 5
= 9/4 +1
= 3.25
Step-by-step explanation:
Hope it helps!!
Construct an equation for a function with a zero at -2 and a double zero at 3.
Step-by-step explanation:
feeling the compa
Como te amo mucho
I give brainiest!!!!!
Answer:
c. 4
Step-by-step explanation:
yan sagot
simplify each of the following.
5.1.
[tex]2 sin(90 - x ) - cos(360 - x)[/tex]
Step-by-step explanation:
[tex]2 \sin(90 - x) - \cos(360 - x) [/tex]
[tex]2 \cos(x) - \cos(x) [/tex]
[tex] \cos(x) [/tex]
The circumference of a circle is 127 cm. What is the area,
in
square centimeters?
Express your answer in terms of Pi.
What is the relationship between the values p and q plotted on the number line below?
A. q>P
B. p= q
c. p> q
Answer:
A. q>p because, assuming this line starts somewhere after 0, this is positive progression
If you give me right answer will cashapp money
Answer:
x=8
Step-by-step explanation:
The ratio of the sides is 7:14=1:2. So, the ratio of DF:XZ is 1:2. DF=2x-5, and XZ=22. This makes the ratio. 2x-5:22=1:2. That means 2x-5 is half of 22, which is 11. We can solve from there.
2x-5=11Add 5, 2x=16Divide by 2, x=8Check your work:
2(8)-5=1116-5=1111=11This XP set is worth a maximum points 1) As of 5/12/22. 72.7% of the residents of Washington state were fully vaccinated against CV19. Also, the population of our state is about 74 million people. a) Suppose that random samples of size n 50 are selected from this state's population. Let X represent the number people in each sample that are fully vaccinated against CV19 The variable X has an approximately Binomial distribution, with n 50 and D 727 Using the special Binomial formulas, compute the mean and the sigma of Round each value to the nearest tenth b) Refer to part . Which specific values of this variable X are within 3- sigma of its mean value? (OW, what are the common numbers of fully vaccinated people you'd expect to find in samples of this size, given the 72.7% fully vaccinated percentage?) c) Refer to part a. Use your calculator / technology like the "binompat and "binomcd commands on the TI-84) to help answer the following questions Report each probability correct to four decimal places Find the probability that one of these samples has exactly 35 fully vaccinated people in ite find the value of PX - 3571 Is this an unusual event using the 5% probability criterion? 10 Find the probability that one of these samples has 25 or fewer fully vaccinated people in it be find the value of PX $25)) is this an unusual event, using the 5% probability criterion? 30 IP 3: DO %: . $ 4 % 5 . # 3 $ 4 % 5 & 7 6 > 0 00 9 E R T T Y U 0 P D F G H J I KL < < C V B N M 1
Random samples of size 50 are taken from this population, and the number of fully vaccinated individuals in each sample, represented by the variable X, follows an approximately binomial distribution with n = 50 and p = 0.727.
(a) Using the binomial formulas, the mean (μ) of X is calculated as np, which is 50 * 0.727 = 36.4, rounded to the nearest tenth. The standard deviation (σ) is given by the square root of np(1-p), which becomes √(50 * 0.727 * 0.273) ≈ 4.1.(b) To determine the values within 3 sigma of the mean, we calculate 3 times the standard deviation (3σ) and find the range around the mean: 36.4 ± 3 * 4.1, resulting in the range of approximately 24.1 to 48.7. Therefore, the common numbers of fully vaccinated people expected in samples of this size, given the 72.7% vaccination rate, would fall within this range.(c) By using appropriate commands on a calculator or technology, the following probabilities can be determined:
The probability of one sample having exactly 35 fully vaccinated people is obtained from the binomial distribution as P(X = 35), which can be calculated using the binompdf command.
The value of P(X ≤ 25) can be found using the binomcdf command, representing the probability of having 25 or fewer fully vaccinated people in a sample. The probabilities should be reported to four decimal places for accuracy.
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