Answer:
See the explanation.Step-by-step explanation:
We need to observe total 9 days.First we need to choose any 4 days from the 9 days. From these total 9 days, 4 days can be chosen in [tex]^9C_4 = \frac{9!}{4!\times5!} = \frac{6\times7\times8\times9}{4!} = 126[/tex] ways.
The probability of occurring an auto related accident is [tex]\frac{52}{100}[/tex].
Similarly, the probability of not occurring an accident is [tex]\frac{100 - 52}{100} = \frac{48}{100}[/tex].
Hence, the required probability is [tex]126\times(\frac{52}{100} )^4(\frac{48}{100} )^5[/tex].
From a plot of ln(Kw) versus 1/T (using the Kelvin scale), estimate Kw at 37°C, which is the normal physiological temperature. Kw = What is the pH of a neutral solution at 37°C?
Answer:
A) Kw (37°C) = 2.12x10⁻¹⁴
B) pH (37°C) = 6.84
Step-by-step explanation:
The following table shows the different values of Kw in the function of temperature:
T(°C) Kw
0 0.114 x 10⁻¹⁴
10 0.293 x 10⁻¹⁴
20 0.681 x 10⁻¹⁴
25 1.008 x 10⁻¹⁴
30 1.471 x 10⁻¹⁴
40 2.916 x 10⁻¹⁴
50 5.476 x 10⁻¹⁴
100 51.3 x 10⁻¹⁴
A) The plot of the values above gives a straight line with the following equation:
y = -6218.6x - 11.426 (1)
where y = ln(Kw) and x = 1/T
Hence, from equation (1) we can find Kw at 37°C:
[tex] ln(K_{w}) = -6218.6 \cdot (1/(37 + 273)) - 11.426 = -31.49 [/tex]
[tex] K_{w} = e^{-31.49} = 2.12 \cdot 10^{-14} [/tex]
Therefore, Kw at 37°C is 2.12x10⁻¹⁴
B) The pH of a neutral solution is:
[tex] pH = -log([H^{+}]) [/tex] (2)
The hydrogen ion concentration can be calculated using the following equation:
[tex] K_{w} = [H^{+}][OH^{-}] [/tex] (3)
Since in pure water, the hydrogen ion concentration must be equal to the hydroxide ion concentration, we can replace [OH⁻] by [H⁺] in equation (3):
[tex] K_{w} = ([H^{+}])^{2} [/tex]
which gives:
[tex] [H^{+}] = \sqrt {K_{w}} [/tex]
Having that Kw = 2.12x10⁻¹⁴ at 37 °C (310 K), the pH of a neutral solution at this temperature is:
[tex] pH = -log ([H^{+}]) = -log(\sqrt {K_{w}}) = -log(\sqrt {2.12 \cdot 10^{-14}}) = 6.84 [/tex]
Therefore, the pH of a neutral solution at 37°C is 6.84.
I hope it helps you!
a. Endothermic (Kw increases with temperature).
b. pH = 7 for pure water at 50.8°C.
c. Plot ln(Kw) vs. 1/T to estimate Kw at 37.8°C.
d. pH = 7 for neutral solution at 37.8°C.
Here's the step-by-step solution:
a. Since Kw increases with temperature, indicating more ionization, the process is endothermic.
b. At 50.8°C, Kw ≈ 5.47 × 10^(-14). Since pure water is neutral, pH = 7.
c. To estimate Kw at 37.8°C:
- Plot ln(Kw) against 1/T (in Kelvin).
- Find the y-intercept of the plot. This intercept corresponds to ln(Kw) at 1/T = 0, which is at the temperature where T = 1/(37.8 + 273.15).
- Take the exponential of this value to find Kw.
d. At 37.8°C, Kw ≈ 1.3 × 10^(-14). Since pure water is neutral, pH = 7.
The correct question is:
Values of Kw as a function of temperature are as follows: Temp (8C) Kw 0 1.14 3 10215 25 1.00 3 10214 35 2.09 3 10214 40. 2.92 3 10214 50. 5.47 3 10214
a. Is the autoionization of water exothermic or endothermic?
b. What is the pH of pure water at 50.8C?
c. From a plot of ln(Kw) versus 1/T (using the Kelvin scale), estimate Kw at 378C, normal physiological temperature. d. What is the pH of a neutral solution at 378C?
A study found that, in 2005, 12.5% of U.S. workers belonged to unions (The Wall Street Journal, January 21, 2006). Suppose a sample of 400 U.S. workers is collected in 2006 to determine whether union efforts to organize have increased union membership.(a) Formulate the hypotheses that can be used to determine whether union membership increased in 2006.(b) If the sample results show that 52 of the workers belonged to unions, what is the p-value for your hypothesis test? Round your answers to four decimal places.
Answer:
There is not enough evidence to support the claim that union membership increased.
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 400
p = 12.5% = 0.125
Alpha, α = 0.05
Number of women belonging to union , x = 52
First, we design the null and the alternate hypothesis
[tex]H_{0}: p = 0.125\\H_A: p > 0.125[/tex]
The null hypothesis sates that 12.5% of U.S. workers belong to union and the alternate hypothesis states that there is a increase in union membership.
This is a one-tailed(right) test.
Formula:
[tex]\hat{p} = \dfrac{x}{n} = \dfrac{52}{400} = 0.13[/tex]
[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
Putting the values, we get,
[tex]z = \displaystyle\frac{0.13-0.125}{\sqrt{\frac{0.125(1-0.125)}{400}}} = 0.3023[/tex]
Now, we calculate the p-value from the table.
P-value = 0.3812
Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.
Conclusion:
Thus, there is not enough evidence to support the claim that union membership increased.
Final answer:
To determine whether union membership increased in 2006, the null hypothesis states the proportion is 12.5% or less, and the alternative hypothesis states it is greater than 12.5%. Based on the sample data (where 52 out of 400 workers are union members), the p-value is calculated to be approximately 0.22. Since the p-value is greater than 0.05, we fail to reject the null hypothesis, indicating insufficient evidence of an increase in union membership.
Explanation:
To determine whether union membership increased in 2006, we start by formulating the hypotheses for our hypothesis test:
Hypotheses
Null hypothesis (H0): The proportion of U.S. workers belonging to unions in 2006 is equal to or less than the 2005 level of 12.5% (p ≤ 0.125).
Alternative hypothesis (H1): The proportion of U.S. workers belonging to unions in 2006 is greater than the 2005 level of 12.5% (p > 0.125).
Next, we calculate the test statistic and the p-value based on the sample results:
The sample proportion is the number of workers belonging to unions divided by the total number of workers in the sample. Therefore:
Sample proportion = 52/400 = 0.13 or 13%
To find the p-value, we assume the null hypothesis is true. The test statistic for a one-sample Z-test for proportions is calculated as:
Z = (Sample proportion - Hypothesized proportion) / Standard error of the sample proportion
Standard error = sqrt((Hypothesized proportion * (1 - Hypothesized proportion)) / sample size)
Z = (0.13 - 0.125) / sqrt((0.125 * (1 - 0.125)) / 400) = 0.7727
Since this is a one-tailed test, the p-value is the probability that the standard normal variable is greater than the calculated Z value. We find the p-value using standard normal distribution tables or software. Assuming the Z value is 0.7727, the corresponding p-value would be approximately 0.22.
As the p-value is greater than typical significance levels like 0.05, we fail to reject the null hypothesis. This means there is not enough evidence at the 0.05 significance level to conclude that union membership has increased in 2006.
A few weeks into the deadly SARS (Severe Acute Respiratory Syndrome) epidemic in 2003, the number of cases was increasing by about 4% each day.† On April 1, 2003, there were 1,804 cases. Find an exponential model that predicts the number of cases t days after April 1, 2003. f(t) = Use it to estimate the number of cases on April 26, 2003. (The actual reported number of cases was 4,836.)
Answer:
[tex]f(t)=1804(1.04)^{t}\\f(25)=4809.17[/tex]
Step-by-step explanation:
1. Since the increasing rate is 0.04 or (4%) per day, then the factor is (1+0.04) raised to t days, and we have and exponential growth therefore we can write:
[tex]\\f(t)=c(1.04)^t\\f(t)=1,804(1.04)^t\\[/tex]
2. To estimate the number of cases, 25 days later following that exponential model
[tex]f(25)=1804(1.04)^{25}\\f(25)=4809.17[/tex]
Match the scenario with its distribution type. Group of answer choices Sammy takes a sample of 200 students at her college in an attempt to estimate the average time students spend studying at her university. She plots her data using a histogram, the histogram shows what type of distribution? Phil wants to estimate the average driving time to work for individuals in Corvallis, Oregon. He takes a simple random sample of 100 individuals and records their drive time. Phil knows that a different sample will yield a different sample average. Suppose Phil was able to repeatedly sample a different group of Corvallis residents until he obtained every combination of 100 residents. If he plots the sample means from each sample this distribution will represent which type of distribution? An instructor wants see the scores of her entire class after the midterm. The distribution of scores represents what type of distribution?
Answer:
Scenario 1: Sample distributionScenario 2: Sampling distribtuionScenario 3: Population distributionStep-by-step explanation:
Scenarios are: Sample Distribution, sampling distribution, Population distribution
sample distribution: A sample distribution contains the data of individuals of sample collected from a population.
Sampling distribution: distribution of multiple sample statistics
Population distribution: Statistics of entire population without drawing sny sample
In scenario 1, only one saple is taken from a population and that sample is plotted
In Scenario 2, multiple samples are taken from a population and means of each sample is plotted
In Scenario 3, scores of entire population is considered.
A manufactured lot of buggy whips has 20 items, of which 5 are defective. A random sample of 5 items is chosen to be inspected. Find the probability that the sample contains exactly one defective item
Answer:
[tex] P(X=1)[/tex]
And using the probability mass function we got:
[tex] P(X=1) = (5C1) (0.25)^1 (1-0.25)^{5-1}= 0.396[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
For this cae that one buggy whip would be defective is [tex] p = \frac{5}{20}=0.25[/tex]
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=5, p=0.25)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
And we want to find this probability:
[tex] P(X=1)[/tex]
And using the probability mass function we got:
[tex] P(X=1) = (5C1) (0.25)^1 (1-0.25)^{5-1}= 0.396[/tex]
Suppose Albers Elementary School has 44 teachers and Bothel Elementary School has 74 teachers. If the total number of teachers at Albers and Bothel combined is 87, how many teachers teach at both schools?
Answer: 31
Step-by-step explanation:
This is solved Using the theory of sets.
Teachers in Albers school = 44
Teachers in Bothel school = 74
Let the number of teachers that work in both schools be denoted as "x"
This implies that:
Number of teachers in Albers only = 44 - x
Number of teachers in Bothel only = 74 - x
And total number of teachers in both schools = 87,
then
x + (44-x) + (74-x) = 87
118 -x = 87
31 = x
This means the number of teachers that teach in both schools = 31.
Answer: 31 teachers teach at both schools.
Let:
A = Albers Elementary School
B = Bothel Elementary School
According to the question:
n(A) = 44, n(B) = 74, n(A U B) = 87.
Using the union formula we get:
[tex]n(A \cup B) = n(A)+n(B)-n(A \cap B)\\87 = 44+74-n(A \cap B)\\n(A \cap B)=44+74-87\\n(A \cap B)=31[/tex]
So, 31 teachers teach at both schools.
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A lab network consisting of 20 computers was attacked by a computer virus. This virus enters each computer with probability 0.4, independently of other computers. Find the probability that it entered at least 10 computers
The probability that the virus entered at least 10 computers is 0.7553.
To find the probability that the virus entered at least 10 computers, we can use the complementary probability formula. This formula states that the probability of an event A happening is equal to 1 minus the probability of event A not happening.
In this case, event A is the virus entering at least 10 computers. Event A not happening is the virus entering fewer than 10 computers.
The probability of the virus entering fewer than 10 computers is equal to the sum of the probabilities of the virus entering 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 computers.
We can use the binomial distribution to calculate the probability of the virus entering each of these numbers of computers. The binomial distribution is a probability distribution that describes the probability of getting a certain number of successes in a certain number of trials.
In this case, the trials are the computers, and the success is the virus entering the computer. The probability of success is 0.4, and the probability of failure is 0.6.
To find the probability of the virus entering fewer than 10 computers, we need to add up the probabilities in the table from 0 to 9.
P(virus entering fewer than 10 computers) = 0.36^20 + 20 * 0.36^19 * 0.6 + ... + 167960 * 0.36^11 * 0.6^9
We can use a calculator to evaluate this sum. The result is 0.2447.
Therefore, the probability of the virus entering at least 10 computers is 1 minus the probability of the virus entering fewer than 10 computers.
P(virus entering at least 10 computers) = 1 - 0.2447
P(virus entering at least 10 computers) = 0.7553
Therefore, the probability that the virus entered at least 10 computers is 0.7553.
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According to government data, 74% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women are randomly selected: a. What is the probability that exactly 2 of them have never been married
Answer:
0.000001 = 0.0001% probability that exactly 2 of them have never been married
Step-by-step explanation:
For each employed women, there are only two possible outcomes. Either they have already been married, or they have not. The women are chosen at random, which means that the probability of a woman having been already married is independent from other women. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
74% of employed women have never been married.
This means that [tex]p = 0.74[/tex]
15 employed women are randomly selected
This means that [tex]n = 15[/tex]
a. What is the probability that exactly 2 of them have never been married
This is P(X = 2).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{15,2}.(0.74)^{2}.(0.26)^{13} = 0.000001[/tex]
0.000001 = 0.0001% probability that exactly 2 of them have never been married
The average heights of a random sample of 400 people from a city is 1.75 m. It is known that the heights of the population are random variables that follow a normal distribution with a variance of 0.16.
Determine the interval of 95% confidence for the average heights of the population.
Answer:
The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
The standard deviation is the square root of the variance. So
[tex]\sigma = \sqrt{0.16} = 0.4[/tex]
Then
[tex]M = 1.96*\frac{0.4}{\sqrt{400}} = 0.0392[/tex]
The lower end of the interval is the mean subtracted by M. So it is 1.75 - 0.0392 = 1.7108m
The upper end of the interval is the mean added to M. So it is 1.75 + 0.0392 = 1.7892m
The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.
Answer:
The interval of 95% confidence for the average heights of the population is = [tex](1.7108, 1.7892)[/tex]Step-by-step explanation:
mean x = [tex]1.75[/tex]
Variance [tex]\rho^2 = 0.16[/tex]
standard deviation [tex](\rho) = \sqrt{0.16} = 0.4[/tex]
n = 400
[tex]95\%[/tex] confidence :
[tex]\alpha = 100\% - 95\% = 5\%\\\\\frac{\alpha}{2} = 2.5\% = 0.025[/tex]
From standard normal distribution table,
[tex]Z_\frac{\alpha}{2} = Z_{0.025} = 1.96[/tex]
Margin of error, [tex]E = Z_\frac{\alpha}{2} * \frac{\rho}{\sqrt{n}}[/tex]
[tex]E = 1.96 * \frac{0.4}{\sqrt{400}}\\\\E = 0.0392[/tex]
Lower limit: x - E
[tex]= 1.75 - 0.0392\\\\= 1.7108[/tex]
Upper limit: x + E
[tex]= 1.75 + 0.0392\\\\= 1.7892[/tex]
[tex]Limits : (1.7108, 1.7892)[/tex]
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Suppose that the mean weight for men 18 to 24 years old is 170 pounds, and the standard deviation is 20 pounds. In each part, find the value of the standardized score (z-score) for the given weight.a. 200 pounds.b. 140 pounds.c. 170 pounds.d. 230 pounds.
Answer:
a) [tex]Z = 1.5[/tex]
b) [tex]Z = -1.5[/tex]
c) [tex]Z = 0[/tex]
d) [tex]Z = 3[/tex]
Step-by-step explanation:
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In this problem, we have that:
[tex]\mu = 170, \sigma = 20[/tex]
a. 200 pounds.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{200 - 170}{20}[/tex]
[tex]Z = 1.5[/tex]
b. 140 pounds.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{140 - 170}{20}[/tex]
[tex]Z = -1.5[/tex]
c. 170 pounds.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{170 - 170}{20}[/tex]
[tex]Z = 0[/tex]
d. 230 pounds.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{230 - 170}{20}[/tex]
[tex]Z = 3[/tex]
Choose all of the equivalent expressions.
A. 300e^−0.0577t
B. 300(1/2)^t/12
C. 252.290(0.9439)^t
D. 300(0.9439)^t
E. 252.290(0.9439)^t−3
Answer:
A B D and E
Step-by-step explanation:
A publisher reports that 49I% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 200200 found that 42B% of the readers owned a personal computer. Determine the P-value of the test statistic. Round your answer to four decimal places.
Answer:
P-value of test statistics = 0.9773
Step-by-step explanation:
We are given that a publisher reports that 49% of their readers own a personal computer. A random sample of 200 found that 42% of the readers owned a personal computer.
And, a marketing executive wants to test the claim that the percentage is actually different from the reported percentage, i.e;
Null Hypothesis, [tex]H_0[/tex] : p = 0.49 {means that the percentage of readers who own a personal computer is same as reported 63%}
Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.49 {means that the percentage of readers who own a personal computer is different from the reported 63%}
The test statistics we will use here is;
T.S. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] ~ N(0,1)
where, p = actual % of readers who own a personal computer = 0.49
[tex]\hat p[/tex] = percentage of readers who own a personal computer in a
sample of 200 = 0.42
n = sample size = 200
So, Test statistics = [tex]\frac{0.42 -0.49}{\sqrt{\frac{0.42(1- 0.42)}{200} } }[/tex]
= -2.00
Now, P-value of test statistics is given by = P(Z > -2.00) = P(Z < 2.00)
= 0.9773 .
Medical tests were conducted to learn about drug-resistant tuberculosis. Of cases tested in New Jersey, were found to be drug-resistant. Of cases tested in Texas, were found to be drug-resistant. Do these data suggest a statistically significant difference between the proportions of drug-resistant cases in the two states?
Answer:
Yes at the level of 0.02 significance
Step-by-step explanation:
we want to compare if P₁ = P₂
P1 = 9/142= 0.0634
P2 = 5/268 = 0.0187
P = 14/410 = 0.03414
significance level, α = 0.02
Test statistic, z = [tex]\frac{p1 - p2}{\sqrt{(p * (1 - p} )) * (\frac{1}{142} * \frac{1}{268})}}[/tex]
Test Statistic, z = [tex]\frac{0.0634 - 0.0187}{\sqrt{({0.0341 * ({1 - 0.0341}})) * ({\frac{1}{142} + \frac{1}{268} ) }} } = 2.373[/tex]
Test statistic, z = 2.373
p-value = 2*p(z<|z₀|) = 2*p(z<2.37) = 0 .0176
Answer: Since p-value (0.0176) is less than the significance level, α (0.02), the null hypothesis can not hold. we can therefore say that at 0.02 level of significance, there is sufficient evidence, statistically, that p₁ is different from p₂
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Write function that take a number n and return the sum of all the multiples of 3 or 5 below n.
Answer:
The function is provided below.
Step-by-step explanation:
Running the program in Python, the code for this problem is as follows:
def func(n):
count = 0 **initializing the sum with 0**
for i in range (1, n):
if i%3 == 0 or i%5 == 0:
count = count + 1
return (count)
When the program is executed enter value 1 to 10 one by one and the result will be 23, the sum of all the multiples of 3 and 5.
Consider the probability that no less than 92 out of 159 students will pass their college placement exams. Assume the probability that a given student will pass their college placement exam is 55%. Approximate the probability using the normal distribution. Round your answer to four decimal places.
Answer:
0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.
Step-by-step explanation:
I am going to use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]p = 0.55, n = 159[/tex]. So
[tex]\mu = E(X) = 159*0.55 = 87.45[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{159*0.55*0.45} = 6.27[/tex]
Probability that no less than 92 out of 159 students will pass their college placement exams.
No less than 92 is more than 91, which is 1 subtracted by the pvalue of Z when X = 91. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{91 - 87.45}{6.27}[/tex]
[tex]Z = 0.57[/tex]
[tex]Z = 0.57[/tex] has a pvalue of 0.7157
1 - 0.7157 = 0.2843
0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.
To approximate the probability that at least 92 out of 159 students will pass their college placement exams using the normal distribution, we will follow these steps:
Step 1: Determine the parameters of the binomial distribution.
In a binomial distribution the parameters are the number of trials (n) and the probability of success in a single trial (p). For this particular case:
n = 159 (the number of students)
p = 0.55 (the probability that a given student will pass)
Step 2: Calculate the mean (μ) and standard deviation (σ) of the binomial distribution.
The mean (μ) of a binomial distribution is given by:
μ = n * p
The standard deviation (σ) is given by:
σ = sqrt(n * p * (1 - p))
For our case:
μ = 159 * 0.55 ≈ 87.45
σ = sqrt(159 * 0.55 * (1 - 0.55)) ≈ sqrt(159 * 0.55 * 0.45) ≈ sqrt(71.775) ≈ 8.472 (rounded to three decimal places for intermediate calculation)
Step 3: Apply the continuity correction.
Because we're approximating a discrete distribution with a continuous one, we use a continuity correction. To find the probability of at least 92 students passing, we'll look for the probability of X > 91.5 (since X is a discrete random variable, X ≥ 92 is equivalent to X > 91.5 when using a continuous approximation).
Step 4: Calculate the z-score for the corrected value.
The z-score is calculated by taking the value of interest, applying the continuity correction, subtracting the mean, and then dividing by the standard deviation:
z = (X - μ) / σ
Using the corrected value:
z = (91.5 - 87.45) / 8.472 ≈ 0.4774 (rounded to four decimal places for the calculation)
Step 5: Find the corresponding probability.
The z-score tells us how many standard deviations away from the mean our value of interest is. To find the probability that at least 92 students pass (i.e., P(X ≥ 92), we need to find 1 - P(Z < z) because the normal distribution table or a calculator gives us P(Z < z), which is the probability of being less than a certain z value.
We are looking for the probability that z is greater than our calculated value, which is the upper tail of the distribution.
Step 6: Consult the standard normal distribution table or use a calculator.
The z-score of approximately 0.4774 corresponds to a percentage in the standard normal distribution. Since we want P(Z > z), we need to subtract P(Z < z) from 1. If we look up the probability for z=0.4774 in standard normal distribution tables or use a calculator, we find that P(Z < z) is approximately 0.6832.
Therefore, P(Z > z) = 1 - P(Z < z) = 1 - 0.6832 = 0.3168.
Step 7: Round the answer.
Rounding P(Z > z) = 0.3168 to four decimal places, the answer remains 0.3168.
So, the approximate probability that at least 92 out of 159 students will pass the exam, using the normal distribution approximation, is 0.3168 when rounded to four decimal places.
Records show that the average number of phone calls received per day is 9.2. Find the probability of between 2 and 4 phone calls received (endpoints included) in a given day.
Answer:
4.76% probability of between 2 and 4 phone calls received (endpoints included) in a given day.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
Records show that the average number of phone calls received per day is 9.2.
This means that [tex]\mu = 9.2[/tex].
Find the probability of between 2 and 4 phone calls received (endpoints included) in a given day.
[tex](2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 2) = \frac{e^{-9.2}*(9.2)^{2}}{(2)!} = 0.0043[/tex]
[tex]P(X = 3) = \frac{e^{-9.2}*(9.2)^{3}}{(3)!} = 0.0131[/tex]
[tex]P(X = 4) = \frac{e^{-9.2}*(9.2)^{4}}{(4)!} = 0.0302[/tex]
[tex](2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0043 + 0.0131 + 0.0302 = 0.0476[/tex]
4.76% probability of between 2 and 4 phone calls received (endpoints included) in a given day.
9. A large electronic office product contains 2000 electronic components. Assume that the probability that each component operates without failure during the useful life of the product is 0.995, and assume that the components fail independently. Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.
Answer:
97.10% probability that five or more of the original 2000 components fail during the useful life of the product.
Step-by-step explanation:
For each component, there are only two possible outcomes. Either it works correctly, or it does not. The probability of a component falling is independent from other components. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]n = 2000, p = 1-0.995 = 0.005[/tex]
Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.
We know that either less than five compoenents fail, or at least five do. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 5) + P(X \geq 5) = 1[/tex]
We want [tex]P(X \geq 5)[/tex]
So
[tex]P(X \geq 5) = 1 - P(X < 5)[/tex]
In which
[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{2000,0}.(0.005)^{0}.(0.995)^{2000} = 0.000044[/tex]
[tex]P(X = 1) = C_{2000,1}.(0.005)^{1}.(0.995)^{1999} = 0.000445[/tex]
[tex]P(X = 2) = C_{2000,2}.(0.005)^{2}.(0.995)^{1998} = 0.002235[/tex]
[tex]P(X = 3) = C_{2000,3}.(0.005)^{3}.(0.995)^{1997} = 0.007480[/tex]
[tex]P(X = 4) = C_{2000,4}.(0.005)^{4}.(0.995)^{1996} = 0.018765[/tex]
[tex]P(X < 5) = P(X = 0) + `P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000044 + 0.000445 + 0.002235 + 0.007480 + 0.018765 = 0.0290[/tex]
[tex]P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0290 = 0.9710[/tex]
97.10% probability that five or more of the original 2000 components fail during the useful life of the product.
The average daily high temperature in June in LA is 77 degree F with a standard deviation of 5 degree F. Suppose that the temperatures in June closely follow a normal distribution. What is the probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June? How cold are the coldest 10% of the days during June in LA?
Answer:
11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June
The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 77, \sigma = 5[/tex]
What is the probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June?
This probability is 1 subtracted by the pvalue of Z when X = 83. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{83 - 77}{5}[/tex]
[tex]Z = 1.2[/tex]
[tex]Z = 1.2[/tex] has a pvalue of 0.8849.
1 - 0.8849 = 0.1151
11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June
How cold are the coldest 10% of the days during June in LA?
High temperatures of X or lower, in which X is found when Z has a pvalue of 0.1, so whn Z = -1.28
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.28 = \frac{X - 77}{5}[/tex]
[tex]X - 77 = -1.28*5[/tex]
[tex]X = 70.6[/tex]
The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.
A) The probability of observing a temperature ≥ 83°F in LA during a randomly chosen day in June is;
p(observing a temperature ≥ 83°F) = 11.507%
B) The coldest 10% of the days during June in LA have temperatures;
Less than or equal to 70.592 °F
This question involves z-distribution which is given by the formula;
z = (x' - μ)/σ
We are given;
Average daily temperature; μ = 77 °F
Standard deviation; σ = 5 °F
Since the temperatures follow a normal distribution, then if we want to find the probability of observing a temperature ≥ 83°F, then;
x' = 83 °F
Thus;
z = (83 - 77)/5
z = 6/5
z = 1.2
Thus;
from online z-score calculator, p-value = 0.11507
Thus, p(observing a temperature ≥ 83°F) = 11.507%
B) We want to find out how cold the coldest 10% of the days during June in LA;
Thus, it means that p = 10% = 0.1
z-score at p = 0.1 from z-score tables is;
z = -1.28155
Thus;
-1.28155 = (x' - 77)/5
-1.28155*5 = x' - 77
-6.40775 = x' - 77
x' = 77 - 6.40775
x' ≈ 70.592 °F
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3/4(ad).. solve.. a=12 d=9
Answer:
hope it helps you see the attachment for further information
Answer:
81
Step-by-step explanation:
The time that it takes a randomly selected employee to perform a certain task is approximately normally distributed with a mean value of 120 seconds and a standard deviation of 20 seconds. The slowest 10% (that is, the 10% with the longest times) are to be given remedial training. What times (the lowest value) qualify for the remedial training?
Answer:
145.6 seconds
Step-by-step explanation:
Mean time (μ) = 120 seconds
Standard deviation (σ) = 20 seconds
In a normal distribution, the z-score for any time, X, is given by:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
The slowest 10% correspond to the 90th percentile of a normal distribution, which has a corresponding z-score of roughly 1.28. The lowest time that requires remedial training is:
[tex]1.28=\frac{X-120}{20}\\X=145.6\ seconds[/tex]
Times of 145.6 seconds and over qualify for remedial training.
In previous years, the average number of sheets recycled per bin was 59.3 sheets, but they believe this number may have increase with the greater awareness of recycling around campus. They count through 79 randomly selected bins from the many recycle paper bins that are emptied every month and find that the average number of sheets of paper in the bins is 62.4 sheets. They also find that the standard deviation of their sample is 9.86 sheets.
What is the value of the test-statistic for this scenario? Round your answer to 3 decimal places.
What are the degrees of freedom for this t-test?
Answer:
There is enough evidence to support the claim that the average number of sheets recycled per bin was more than 59.3 sheets.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 59.3
Sample mean, [tex]\bar{x}[/tex] = 62.4
Sample size, n = 79
Alpha, α = 0.05
Sample standard deviation, s = 9.86
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 59.3\text{ sheets}\\H_A: \mu > 59.3\text{ sheets}[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{62.4 - 59.3}{\frac{9.86}{\sqrt{79}} } = 2.7945[/tex]
Degree of freedom = n - 1 = 78
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 78 degree of freedom } = 1.6646[/tex]
Since,
[tex]t_{stat} > t_{critical}[/tex]
We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
Conclusion:
Thus, there is enough evidence to support the claim that the average number of sheets recycled per bin was more than 59.3 sheets.
The number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 1.7 entrees per order. On a particular Saturday afternoon, a random sample of 48 Noodles orders had a mean number of entrees equal to 2.1 with a standard deviation equal to 1.01. At the 2 percent level of significance, does this sample show that the average number of entrees per order was greater than expected?
Answer:
[tex]t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744[/tex]
[tex]p_v =P(t_{(47)}>2.744)=0.0043[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.02[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=2.1[/tex] represent the mean
[tex]s=1.01[/tex] represent the sample standard deviation
[tex]n=48[/tex] sample size
[tex]\mu_o =1.7[/tex] represent the value that we want to test
[tex]\alpha=0.02[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 1.7, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 1.7[/tex]
Alternative hypothesis:[tex]\mu > 1.7[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=48-1=47[/tex]
Since is a one side test the p value would be:
[tex]p_v =P(t_{(47)}>2.744)=0.0043[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.02[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.
Answer:
Yes, this sample show that the average number of entrees per order was greater than expected.
Step-by-step explanation:
We are given that the number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 1.7 entrees per order. For this a random sample of 48 Noodles orders had a mean number of entrees equal to 2.1 with a standard deviation equal to 1.01.
We have to test that the average number of entrees per order was greater than expected or not.
Let, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 1.7 {means that the average number of entrees per order was same as expected of 1.7 entrees per order}
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 1.7 {means that the average number of entrees per order was greater than expected of 1.7 entrees per order}
The test statistics that will be used here is One sample t-test statistics;
T.S. = [tex]\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, Xbar = sample mean number of entrees = 2.1
s = sample standard deviation = 1.01
n = sample of Noodles = 48
So, test statistics = [tex]\frac{2.1-1.7}{\frac{1.01}{\sqrt{48} } }[/tex] ~ [tex]t_4_7[/tex]
= 2.744
Now, at 2% significance level the critical value of t at 47 degree of freedom in t table is given as 2.148. Since our test statistics is more than the critical value of t which means our test statistics will lie in the rejection region. So, we have sufficient evidence to reject our null hypothesis.
Therefore, we conclude that the average number of entrees per order was greater than expected.
A classic counting problem is to determine the number of different ways that the letters of "possession" can be arranged. Find that number. The number of different ways that the letters of "possession" can be arranged is nothing.
Number of possible ways to arrange letters of word "possession" = 75600.
Step-by-step explanation:
A classic counting problem is to determine the number of different ways that the letters of "possession" can be arranged. We have, word "POSSESSION" which have 10 letters! as P,O,S,S,E,S,S,I,O,N . We have to find the number of ways letters of word "possession" can be arranged , in order to do that we will use simple logic as:
At first, we have a count of 10 letters and 10 places vacant to fill letters.
For first place we have a choice of 10 letters , after putting some letter from all 10 letters in first place now we are left with 9 places and 9 letters having 9 choices , similarly we'll be having 8 places , 8 letters and 8 choices and so on...... Therefore, Number of possible ways to arrange letters of word "possession" = [tex]10.9.8.7.6.5.4.3.2.1[/tex] = [tex]10![/tex] ( 10 factorial ) , but there are 4 letters "s" are repeated and 2 letters "o" are repeated so we need to eliminate the similar combinations ∴ Number of possible ways = [tex]\frac{10!}{4!(2!)}[/tex] = 75600.
∴Number of possible ways to arrange letters of word "possession" = 75600
Waiting times for an order at Starbucks for all drive-through customers in the US have a uniform distribution from 3 min to 11 min (mean = 7 min, standard deviation = 2.3 min). What distribution would you use to find the probability that a randomly selected Starbucks drive-through customer in the US waits at most 9 minutes to receive their order?
Answer:
[tex] P(X <9) [/tex]
And we can use the cumulative distribution function given by:
[tex] F(X) = \frac{x-a}{b-a}, a\leq X \leq b[/tex]
And using this we got:
[tex] P(X <9) = F(9) = \frac{9-3}{11-3}= 0.75[/tex]
Step-by-step explanation:
For this case we assume that X= represent the waiting times and for this case we have the following distribution:
[tex] X \sim Unif (a= 3, b =11)[/tex]
And the expected value is given by:
[tex]\mu= E(X) = \frac{a+b}{2}= \frac{3+11}{2}=7[/tex]
And the variance is given by:
[tex] Var(X) \sigma^2 = \frac{(b-a)^2}{12} = \frac{(11-3)^2}{12} = 5.333[/tex]
And we can find the deviation like this:
[tex] Sd(X) = \sqrt{5.333}= 2.309[/tex]
And we want to find this probability:
[tex] P(X <9) [/tex]
And we can use the cumulative distribution function given by:
[tex] F(X) = \frac{x-a}{b-a}, a\leq X \leq b[/tex]
And using this we got:
[tex] P(X <9) = F(9) = \frac{9-3}{11-3}= 0.75[/tex]
The probability that a randomly selected Starbucks drive-through customer in the US waits at most 9 minutes to receive their order is 0.75.
What is uniform distribution?Uniform distributions are probability distributions in which all events are equally likely to occur.
Let's assume that X represents the waiting time.
As the distribution is the uniform distribution, we can write a =3 and b =11
Now, the expected value can be written as,
[tex]\mu = E(X) = \dfrac{a+b}{2} = \dfrac{3+11}{2}= 7[/tex]
the variance of the distribution can be written as,
[tex]Var(X) = \sigma^2 = \dfrac{(b-a)^2}{12} = \dfrac{(11-3)^2}{12} =5.34[/tex]
And, the standard deviation can be written as,
[tex]\sigma = \sqrt{5.34}=2.309[/tex]
In order to calculate the probability, we will use the cumulative distribution function. The cumulative function is given by the formula,
[tex]F(X) = \dfrac{x-a}{b-a}, a\leq X\leq b[/tex]
Using the function we can get the probability as,
[tex]F(X < 9) = F(9) = \dfrac{9-3}{11-3} = 0.75[/tex]
Hence, the probability that a randomly selected Starbucks drive-through customer in the US waits at most 9 minutes to receive their order is 0.75.
Learn more about Uniform Distribution:
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For f(x) = 9x and g(x) = x + 3, find the following functions.
a. (f o g)(x);
b. (g o f )(x);
c. (f o g )(2);
d. (g o f )(2)
Answer:
a) 9*x + 27
b) 9*x+3
c) 45
d) 21
Step-by-step explanation:
since (f o g)(x) = f (g(x)) , then
a) (f o g)(x) = f (x + 3) = 9*(x+3) = 9*x + 27
similarly
b) (g o f)(x) = g (f(x)) = g ( 9x) = (9*x)+3 = 9*x+3
c) for x=2
(f o g)(2) = 9*2 + 27 = 45
d) for x=2
(g o f )(2) = 9*2 +3 = 21
thus we can see that the composition of functions is not necessarily commutative
A={a,b,c,1,2,3,octopus,∅,0} B=N C={0} For each of the following statements, select either True or False. a) A∩B={0,1,2,3} Answer 1 b) C−A=∅ Answer 2 c) B∪P(C)=B Answer 3 d) C∈P(C) Answer 4
Answer:a)False b)True c)False d)True
Step-by-step explanation:
Let's consider first that for us the set of the natural numbers is the set of the positive integers and the set of the non-negative numbers is know as the whole number or with notation [tex]N_0[/tex]. Then for
a) the N={1,2,3,...} and therefore the common members with A are only {1,2,3} making the statement false, only if stated to consider the set N as all the non-negative numbers the answer would be true, but otherwise it is standarized to understand N as the positive integers and [tex]N_0[/tex] as the non-negative integers.
b)The difference of sets is taking the elements in the first that do not belong to the second, then it would be to withdraw the only element C has, since 0 belongs to A, and therefore C would turn to be an empty set.
c)The set powers of a given set S, denoted P(S), is a set with sets as elements, every subset of S is an element of P(S). Then P(S) is always non-empty, since at least S belongs to P(S). Here [tex]P(C)=\{C, \emptyset \}[/tex], then [tex]P(C)\cup B=\{C, \emptyset ,1,2,3,\ldots \}\ne B[/tex], therefore the statement is false.
d)As explained in c) [tex]P(C)=\{C, \emptyset \},[/tex] then clearly C is an element of P(C), thus the affirmation is true.
Evaluation of Proofs See the instructionsfor Exercise (19) on page 100 from Section 3.1. (a) Proposition. If m is an odd integer, then .mC6/ is an odd integer. Proof. For m C 6 to be an odd integer, there must exist an integer n such that mC6 D 2nC1: By subtracting 6 from both sides of this equation, we obtain m D 2n6C1 D 2.n3/C1: By the closure properties of the integers, .n3/ is an integer, and hence, the last equation implies that m is an odd integer. This proves that if m is an odd integer, then mC6 is an odd integer
Answer:
(A) Assume m is an odd integer.
Therefore m is of the m=2n-1 type where n is any number.
We have m + 6 = 2n-1 + 6 = 2n+5=2n+4 + 1 = 2(n+2) + 1, in which n + 2 is an integer, adding 6 to both ends.
Because m+6 is of the 2x + 1 type, where x = n + 2; then m + 6 is an odd integer too.
(B) Provided that mn is an integer also. For some integer y and m, n are integers, this means mn is of the form mn = 2y.
And y = mn/2.
Therefore either m is divided by 2 or n is divided by 2 since y, m, n are all integers. To put it another way, either m is a multiple of 2 or n is a multiple of 2, which means that m is even or n is even.
Final answer:
The proposition regarding odd integers is proven by correcting the initial flawed proof, demonstrating that adding 6 to an odd integer results in another odd integer by using the correct form of an odd integer in the calculation.
Explanation:
The given proposition states that adding 6 to an odd integer results in another odd integer. The proof starts with the assumption that there exists an integer n such that m + 6 = 2n + 1.
This equation is supposed to define m + 6 as an odd integer. By isolating m, the proof continues to show that m itself is expressed in the form of 2(n - 3) + 1, indicating that m is also an odd integer.
To correct the proof, we should start with an odd integer m such that m = 2k + 1, where k is an integer. Adding 6 to m gives m + 6 = 2k + 7, which can be written as 2(k + 3) + 1, clearly showing that m + 6 is an odd integer.
Therefore, this proves the original proposition correctly.
Solve for q. √3q + 2 = √5
Follow below steps:
To solve for q in the equation √3q + 2 = √5, we first isolate the term with q by subtracting 2 from both sides of the equation.
√3q = √5 - 2
Then we square both sides of the equation to remove the square root:
(√3q)² = ( √5 - 2 )²
3q = ( √5 - 2 )²
Now, expand the right side of the equation:
3q = 5 - 2√5 * 2 + 2²
3q = 5 - 4√5 + 4
3q = 9 - 4√5
Then, divide both sides of the equation by 3 to solve for q:
q = (9 - 4√5) / 3
So, the value of q is (9 - 4√5) / 3.
The value of q is [tex]\frac{9 - 4\sqrt{5}}{3}[/tex].
To solve for q, we'll isolate it by performing operations to both sides of the equation to get q by itself.
Given the equation:
[tex]\[ \sqrt{3q} + 2 = \sqrt{5} \][/tex]
Subtract 2 from both sides:
[tex]\[ \sqrt{3q} = \sqrt{5} - 2 \][/tex]
Now, to isolate q, we need to square both sides of the equation:
[tex]\[ (\sqrt{3q})^2 = (\sqrt{5} - 2)^2 \]\[ 3q = (\sqrt{5} - 2)^2 \]\[ 3q = 5 - 4\sqrt{5} + 4 \]\[ 3q = 9 - 4\sqrt{5} \][/tex]
Now, divide both sides by 3 to solve for q:
[tex]\[ q = \frac{9 - 4\sqrt{5}}{3} \][/tex]
So, [tex]\( q = \frac{9 - 4\sqrt{5}}{3} \).[/tex]
Can some help me with this question
Answer:
Step-by-step explanation:
A quadrilateral inscribed within a circle is known as a cyclic quadrilateral. Property of the cyclic quadrilateral is that, sum of its opposite angles is 180°.
∴ ∠U + ∠K = 180°
∴ ∠K = 180 - 85 = 95°
Ethan repairs household appliances like dishwashers and refrigerators. For each visit, he charges $25 plus $20 per hour of work. A linear equation that expresses the total amount of money Ethan earns per visit is y = 25 + 20x. What is the independent variable, and what is the dependent variable?
Answer:the independent variable is x, the number of hours of work.
The dependent variable is y, the total charge for x hours of work.
Step-by-step explanation:
A change in the value of the independent variable causes a corresponding change in the value of in dependent variable. Thus, the dependent variable is is output while the independent variable is the input
For each visit, he charges $25 plus $20 per hour of work. The linear expression that represents the total amount of money that Ethan earns per visit is y = 25 + 20x.
Since the total amount charged, y depends on the number of hours of work, x, it means that the dependent variable is y and the independent variable is x
Final answer:
In the equation y = 25 + 20x, x is the independent variable representing hours worked, and y is the dependent variable representing total earnings. The y-intercept is $25, the flat visit charge, and the slope is $20, the hourly charge.
Explanation:
In the equation y = 25 + 20x, which represents the total amount Ethan earns for each visit, the independent variable is the number of hours of work, denoted by x. The dependent variable is the total amount of money Ethan earns, represented by y. This is because Ethan's earnings depend on the amount of time he spends working.
The y-intercept is $25, which is the flat charge for Ethan's visit, regardless of the hours worked. The slope is $20, which represents the amount Ethan charges for each hour of work. Therefore, for each additional hour of work, Ethan will earn an additional $20.