Answer:
9
Step-by-step explanation:
Plug in 3 for x, and the equation is 6 + 3, which is equal to 9.
Answer:
6 + x = 9
Step-by-step explanation:
If x equals 3
6 plus x (when x equals 3)
Thus, 6 plus 3
= 9
9 will be your answer
Which statements are true about the x-intercepts of quadratic functions?
1) An x-intercept of a quadratic function is also called a zero of the function.
2)An x-intercept is located at the point on the function where the value of x is 0.
3)All quadratic functions have exactly one x-intercept.
4)A quadratic function can have up to two x-intercepts.
5)If x = a is an x-intercept of a quadratic function, then (x + a) is a factor of the function.
Answer:
A quadratic function can have up to two x-intercepts.
An x-intercept of a quadratic function is also called a zero of the function.
Step-by-step explanation:
None needed.
The true statement is
An x-intercept of a quadratic function is also called a zero of the function.
A quadratic function can have up to two x-intercepts.
What is a quadratic function?A quadratic polynomial function has one or more variables and a variable with a maximum exponent of two. It is also known as the polynomial of degree 2 because the second-degree component in a quadratic function is the biggest degree term. A quadratic function must contain at least one term of the second degree. It possesses algebraic qualities.
we have,
The x-intercepts of quadratic functions,
The standard form of quadratic equation is
ax² + bx + c = 0
The zeroes of the quadratic functions are the values of x, which is also known as the x-intercept.
Because the degree of the polynomial is 2, there will be at least two x-intercept values.
The x-intercept is the x position at which the value of y is zero.
Let x = a be an x-intercept, and the function factor is (x - a).
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A sanitation department is interested in estimating the mean amount of garbage per bin for all bins in the city. In a random sample of 46 bins, the sample mean amount was 49.9 pounds and the sample standard deviation was 3.641 pounds. Construct a 95.7% confidence interval for the expected amount of garbage per bin for all bins in the city. Answer to 3 decimals (a) What is the lower limit of the 95.7% interval
Answer:
The 95.7% confidence interval for the expected amount of garbage per bin for all bins in the city
(48.937 , 50.863)
Step-by-step explanation:
Explanation:-
Given data random sample of 46 bins, the sample mean amount was 49.9 pounds and the sample standard deviation was 3.641
The sample size 'n' =46
mean of the sample x⁻ = 49.9
Standard deviation of the sample S = 3.641
Confidence intervals:-
The 95.7% confidence interval for the expected amount of garbage per bin for all bins in the city
[tex](x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha } \frac{S}{\sqrt{n} })[/tex]
Degrees of freedom = n-1 = 46-1 =45
The tabulated value t₀.₉₆ = 1.794 ( from t-table)
[tex](49.9 - 1.794 \frac{3.641}{\sqrt{46} } ,49.9+ 1.794 \frac{3.641}{\sqrt{46} })[/tex]
(49.9 -0.9630 , 49.9+0.9630)
(48.937 , 50.863)
Conclusion:-
The 95.7% confidence interval for the expected amount of garbage per bin for all bins in the city
(48.937 , 50.863)
Help me solve this problem
Answer: -2x+1
Step-by-step explanation:
Slope formula. -3-3/2- -1 = -6/3=-2=m
Y=mx+b use(-1,3) -1=x 3=y
3=-2(-1)+b
3=2+b
Subtract 2
1=b
Complete the sentence below. The standard deviation of the sampling distribution of x overbarx, denoted sigma Subscript x overbarσx, is called the _____ _____ of the _____. The standard deviation of the sampling distribution of x overbarx, denoted sigma Subscript x overbarσx, is called the standard distribution of the sample.
Answer:
The standard deviation of the sampling distribution of x overbarx, denoted sigma Subscript x overbarσx, is called the standard error of the mean . The standard deviation of the sampling distribution of x overbarx, denoted sigma Subscript x overbarσx, is called the standard distribution of the sample.
Step-by-step explanation:
The standard error (SE) of a statistic is the standard deviation of its sampling distribution or an estimate of that standard deviation. It is called the standard error of the mean (SEM) if the parameter or the statistic is the mean.
Which of the following life processes is not necessary for an individual organism to survive, but is necessary for the survival of the species?
A. digestion
B. respiration
C. regulation
D. reproduction
pls help me
Answer:
D. reproduction
Step-by-step explanation:
If there is no reproduction involved, then the species will go extinct
Answer:
D. reproduction
Step-by-step explanation:
How many centimeters is s ?
Answer:
4 cm
Step-by-step explanation:
The area of the net is 6 times the area of one square, so one square has an area of ...
(96 cm^2)/6 = 16 cm^2
The side length (s) is the square root of this value:
s = √(16 cm^2)
s = 4 cm
A cereal company is putting 1 11 of 3 33 prizes in each box of cereal. The prizes are evenly distributed so the probability of winning any given prize is always 1 / 3 1/31, slash, 3. Adam wonders how many boxes he should expect to buy to get all 3 33 prizes. He carried out 32 3232 trials of a simulation and his results are shown below. Each dot represents how many boxes it took to get all 3 33 prizes in that trial. 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13# of boxes purchased Use his results to estimate the probability that it takes 9 99 or more boxes to get all 3 33 prizes. Give your answer as either a fraction or a decimal. P ( 9 or more boxes ) ≈ P(9 or more boxes)≈P, left parenthesis, 9, start text, space, o, r, space, m, o, r, e, space, b, o, x, e, s, end text, right parenthesis, approximately equals
There is 18.75% probability that it takes 9 or more boxes to get all 3 prizes.
The calculation is as follows:No. of Boxes | frequency
3 | 12
4 | 4
5 | 5
6 | 5
7 | 0
8 | 0
9 | 2
10 | 2
11 | 0
12 | 2
Here 9 or more boxes represent that it is equal to 9 boxes or more than 9 boxes.
P(x ≥ 9) = No. of boxes equal or more than 9 / total no. of boxes
And, The total no. of boxes is 32
From the above data, we have to count the number of boxes equal to or more than 9. (6 times) i.e.
[tex]P(x $\geq$ 9) = 6\div 32\\\\P(x $\leq$ 9) = 3\div 16[/tex]
P(x ≥ 9) = 0.1875
P(x ≥ 9) = 18.75%
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Which expression is equivalent to 6(3x-4)
Answer:
18x+24
Step-by-step explanation:
For this, you have to multiply 6 to the numbers inside the parenthesis.
SO, 6 x 3x and 6 x 4
6 x 3x=18x
6 x 4=24
The answer will be 18x+24
7. There are seven clarinet players in the concert band. In how
many ways can they be seated in seven chairs at a concert?
Use the Fundamental Counting Principle.
A. 5,040
C. 840
B. 2,520
D. 210
Answer:
5,040
Step-by-step explanation:
Fundamental Counting Principle states that if there are m ways of doing a thing and there are n ways of doing other thing then there are total m*n ways of doing both things.
Example : If there are 5 path to reach a destination and 3 mode of transport ( bike, car, bicycle) . Then are 5 * 3 ways to reach the destination using the different mode of transport available.
______________________________________
Given no of clarinet players = 7
no of chairs = 7
First player can be seated on seven chairs in seven ways
( for illustration let the seat be a , b, c , d , e , f , g. he can sit on any of the chairs)
since one chair is occupied and 6 chair is available
second player can be seated on 6 chairs in 6 ways
for illustration let the seat be a be occupied then b, c , d , e , f , g are the chairs on which second player can sit.
\
Similarly 3rd, 4th, 5th, 6th and 7th player can be seat on chairs in
5, 4 , 3, 2, and 1 way respectively.
___________________________________________
now using fundamental Counting Principle
since 7 players can sit on chair in
7 , 6, 5 , 4, 3 , 2, 1 ways then together they can be seated in
7 * 6 * 5 * 4 * 3 * 2 * 1 ways = 5, 040 ways
In a study, researchers wanted to measure the effect of alcohol on the hippocampal region, the portion of the brain responsible for long-term memory storage, in adolescents. The researchers randomly selected 22 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of 9.02 cm cubed. An analysis of the sample data revealed that the hippocampal volume is approximately normal with no outliers and x overbarequals8.02 cm cubed and sequals0.7 cm cubed. Conduct the appropriate test at the alphaequals0.01 level of significance. State the null and alternative hypotheses. Upper H 0: mu equals 9.02 Upper H 1: mu less than 9.02 (Type integers or decimals. Do not round.) Identify the t-statistic.
The t-statistic is calculated as: -6.70
What is the test statistic?
The appropriate test to conduct at the alpha equals 0.01 level of significance is a one-tailed t-test.
The null hypothesis is that the mean hippocampal volume of adolescents with alcohol use disorders is equal to 9.02 cm cubed, while the alternative hypothesis is that the mean hippocampal volume is less than 9.02 cm cubed.
Thus:
H₀: μ = 9.02 cm³
Hₐ: μ < 9.02 cm³
The t-statistic is calculated from the formula:
t = (x - μ)/(σ/√n)
t = (8.02 - 9.02) / (0.7/√(22))
= -6.70.
Therefore, the t-statistic is -6.70
Help me find x and y plz and thank you!
Answer:
x=3, y=5
Step-by-step explanation:
To find x:
Because opposite sides of a parallelogram are congruent, we can set up the equation 2x+5=12x-25. Solving the equation will show that x=3.
To find y:
Using the same method as x, we can find that y=5.
Hope this helped!
The school estimates attendance at the varsity basketball games to be between 200 and 275, inclusive. What is the most the percent error could be? Round your answer to the nearest hundredth.
Answer:37.5 %
Step-by-step explanation:
Given
School attendance varies between 200 and 275
Most percentage error arises when attendance of 275 students is expected but only 200 students are present
Percentage error[tex]=\frac{275-200}{200}\times 100=37.5\ \%[/tex]
if you sell a stock for more money than you paid for it, you have a gross capital loss. true or false
Answer:False
Step-by-step explanation:
False
Whenever a stock is sold more money than what is paid for it is termed as Gain or profit
For example if an item is bought for [tex]\$100[/tex] and it is sell for [tex]\$120[/tex]
then there is a profit of
[tex]\Rightarrow \frac{120-100}{100}\times 100[/tex]
[tex]\Rightarrow 20\%[/tex]
or a gain of [tex]\$20[/tex]
Answer: The answer is False, i know this because i took this quiz on edge and it was correct as false <3 hope this helps
Step-by-step explanation: brainliest please :3
A rectangle is four times as long as it is wide. If it has an area of 36 square inches, what are its dimension?.
a. 6 by 6
C. 4 by 9
b. 3 by 12
d. 4 and 8
Please select the best answer from the choices provided
Answer:
B. 3 by 12
Step-by-step explanation:
The rectangle is four times as long as it is wide,
(3 times 4=12)
and has an area of 36
(3 times 12=36)
so this is the correct answer.
If the area of the given rectangle is 36 square inches, then the length of rectangle is 12 inches and the breadth of rectangle is 3 inches.
What is area?"Area is the quantity that expresses the extent of a region on the plane or on a curved surface."
What is rectangle?"A rectangle is a shape with four straight sides and four angles of 90 degrees (right angles)."
Area of rectangle = 36 square inches
Let the width of rectangle be y
Length of rectangle be 4y
Area of rectangle = length × breadth
y×4y = 36
[tex]4y^{2} = 36[/tex]
[tex]y^{2}=9\\y=\sqrt{9}[/tex]
[tex]y=3[/tex]
Hence, the width of the rectangle is 3 inches and the length of the rectangle is 12 inches.
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It has long been stated that the mean temperature of humans is 98.6 degrees F. However, two researchers currently involved in the subject thought that the mean temperature of humans is less than 98.6 degrees F. They measured the temperatures of 56 healthy adults 1 to 4 times daily for 3 days, obtaining 250 measurements. The sample data resulted in a sample mean of 98.2 degrees F and a sample standard deviation of 0.9 degrees F. Use the P-value approach to conduct a hypothesis test to judge whether the mean temperature of humans is less than 98.6 degrees F at the alpha = 0.01 level of significance.1. State the hypotheses.A. Upper H 0H0:▼▼ 98.6 FB. Upper H 1H1:▼▼ 98.6 F2. Find the test statistic.a. t0 = ?b. the P-value is:____.3. What can be concluded?A. RejectUpper H0 since the P-value is less than the significance level.B. Reject Upper H0 since the P-value is not less than the significance level.C. Do not reject Upper H0 since the P-value is less than the significance level.D. Do not reject Upper H0 since the P-value is not less than the significance level.
Answer:
Reject null hypothesis ([tex]H_0[/tex]) since the P-value is less than the significance level.
Step-by-step explanation:
We are given that it has long been stated that the mean temperature of humans is 98.6 degrees F. However, two researchers currently involved in the subject thought that the mean temperature of humans is less than 98.6 degrees F.
The sample data resulted in a sample mean of 98.2 degrees F and a sample standard deviation of 0.9 degrees F.
Let [tex]\mu[/tex] = mean temperature of humans.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu\geq[/tex] 98.6°F {means that the mean temperature of humans is more than or equal to 98.6°F}
Alternate Hypothesis, [tex]H_A[/tex] : p < 98.6°F {means that the mean temperature of humans is less than 98.6°F}
The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean temperature = 98.2°F
[tex]\sigma[/tex] = sample standard deviation = 0.9°F
n = sample of healthy adults = 56
So, test statistics = [tex]\frac{98.2-98.6}{\frac{0.9}{\sqrt{56} } }[/tex] ~ [tex]t_5_5[/tex]
= -3.326
The value of t test statistics is -3.326.
Now, P-value of the test statistics is given by following formula;
P-value = P( [tex]t_5_5[/tex] < -3.326) = 0.00077 or 0.08%
Since, P-value of the test statistics is less than the level of significance as 0.08% < 1%, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that the mean temperature of humans is less than 98.6°F.
To determine if the mean human temperature is less than 98.6°F, we conducted a left-tailed t-test using the provided sample data. We calculated the t-test statistic and then used it to find the p-value. We then used the p-value to make our final decision on whether to reject or fail to reject our null hypothesis.
Explanation:In this scenario, we are performing a hypothesis test about the average human temperature. Let's formulate our hypothesis first:
Null hypothesis (H0): The average human temperature equals 98.6 F. Mathematically, H0: μ = 98.6 F.Alternative hypothesis (H1): The average human temperature is less than 98.6 F. Mathematically, H1: μ < 98.6 F.
We will conduct a left-tailed t-test because we are testing whether the average human temperature is less than a stated value.
Given the data: the sample size n = 250, the sample mean (x_bar) = 98.2 F, and the standard deviation (s) = 0.9 F.
To calculate the t-test statistic, use the formula: t0 = (x_bar - μ) / (s/√n)
For getting the p-value, you would use a statistical table or software with the above t statistic and degree of freedom (which is n-1 in this case).
In the end, if your p-value is less than the significance level (α = 0.01 in this case), we reject the null hypothesis, if not, we fail to reject the null hypothesis.
If the P-value is less than α, we would conclude that the research may be correct, and the average human temperature is indeed lower than 98.6 F (37.0 °C).
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A hot air balloon rising vertically is tracked by an observer located 4 km from the lift‑off point. At a certain moment, the angle between the observer's line of sight and the horizontal is π/5, and it is changing at a rate of 0.4 rad/min. How fast is the balloon rising at this moment? Let y be the height of the balloon (in km), t be time (in minutes), and θ the angle between the line‑of‑sight and the horizontal (in radians).
Answer:
1.22 km/min
Step-by-step explanation:
Let Q be baloon height at a time t. Our goal is to determine the speed of the baloon at the moment.
The dy / dt velocity of the baloon when = π/5?
So we can restate the question as follows:
Owing to the fact d / dt = 0.2 rad / min at some stage = π/5
from fig:
tanθ = y/4
Differentiating w.r.t "t"
sec2 θ * dθ/dt = 1/4(dy/dt)
=> dy/dt = (4/cos2 θ)dθ/dt
At the given moment θ = and dθ/dt = 0.2 rad/min.
dy/dt = (4/cos2)* (0.2)
= 1.22 km/min
And the velocity of the baloon currently is 1.22 km / min.
A file consisting of X packets is sent over the network to an end user. It is known that X is a random variable with PMF PX(x) = cx2 for x = 1, 2, 3, 4; and PX(x) = 0 otherwise. Each packet is received correctly with probability p, independently of the other packets. a) Find the constant c. b) Find P[X > 2]. c) Find the probability that the entire file is received correctly, i.e., all X packets are received correctly
Answer:
See explaination
Step-by-step explanation:
Please kindly check attachment for the step by step solution of the given problem.
From a point 31 meters from the base of a telephone pole the angle of elevation to a worker on the pole is 35 degrees. The angle of elevation to the top of the pole is 68 degrees. Find the distance from the worker to the top of the pole.
Answer:
51.02 metres
Step-by-step explanation:
The pictorial representation of the problem is attached.
We are required to find the distance from the worker to the top of the pole, |DA| in the diagram.
In Triangle ABC
[tex]Tan 68^\circ =\frac{|AB|}{|BC|}\\Tan 68^\circ =\frac{|AB|}{31}\\|AB|=31 X Tan 68^\circ=72.73m[/tex]
In Triangle BCD
[tex]Tan 35^\circ =\frac{|BD|}{|BC|}\\Tan 35^\circ =\frac{|BD|}{31}\\|BD|=31 X Tan 35^\circ=21.71m[/tex]
The distance, |DA| =|AB|-|BD|
=72.73-21.71
=51.02 metres
The distance from the worker to the top of the pole is 51.02 metres.
ILL GIVE YOU BRAINLIST !!! *have to get it right * Find the slope of the line represented in the table.
Answer:
A. 5
Step-by-step explanation:
5 ÷ 1 = 5
10 ÷ 2 = 5
15 ÷ 3 = 5
Answer:
the slope should be 1/5.
Step-by-step explanation:
slope = change in y / change in x
2-1 = 1 = change in Y
10-5 = 5 = change in X
the slope should be 1/5.
apologies in advance if my answer is wrong or I explain badly.
-----------------------------------------------------------------------------------------------
hope this helps! <3
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Edit! someone else answered the question. their answer is probably correct, so please ignore my incorrect answer :'D
Compute the distance between (a,b,c) and (-2,5,6).
a.
(a + 2)^2 + (b - 5)^2 + (c - 6)^2
c.
a + b + c -(-2 + 5 + 6)
b.
sqrt((a+2)^2+(b-5)^2+(c-6)^2)
d.
((a + 2) + (b - 5) + (c - 6))^2
Answer:
b. sqrt(a+2)^2+(b-5)^2+(c-6)^2
Step-by-step explanation:
Use distance formula
sqrt (a-(-2)^2 + (b-5)^2 + (c-6) ^2
sqrt (a+2)^2 +(b+5)^2 + (c-6)^2
Answer:
B
Step-by-step explanation:
took it on edge, it's the right answer :)
what is the range of g(x)=3|x-1|-1?
The range of g(x) = 3|x-1|-1 is (-∞,∞).
What is range?The range is "set of all y-coordinates of the function's graph".
According to the question,
g(x) = 3 |x-1| - 1
To find range of linear polynomial 3|x-1|-1 put any value in right 'x' we can able to get any value 'y' in the set (-∞.∞).
Hence, the range of g(x) = 3|x-1|-1 is (-∞,∞).
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Cable Strength: A group of engineers developed a new design for a steel cable. They need to estimate the amount of weight the cable can hold. The weight limit will be reported on cable packaging. The engineers take a random sample of 45 cables and apply weights to each of them until they break. The mean breaking weight for the 45 cables is 768.2 lb. The standard deviation of the breaking weight for the sample is 15.1 lb. Using the sample information as given, construct a confidence interval for the mean breaking strength of the new steel cable.
Answer:
95% confidence interval for the mean breaking strength of the new steel cable is [763.65 lb , 772.75 lb].
Step-by-step explanation:
We are given that the engineers take a random sample of 45 cables and apply weights to each of them until they break. The mean breaking weight for the 45 cables is 768.2 lb. The standard deviation of the breaking weight for the sample is 15.1 lb.
Since, in the question it is not specified that how much confidence interval has be constructed; so we assume to be constructing of 95% confidence interval.
Firstly, the Pivotal quantity for 95% confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean breaking weight = 768.2 lb
s = sample standard deviation = 15.1 lb
n = sample of cables = 45
[tex]\mu[/tex] = population mean breaking strength
Here for constructing 95% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.
So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.02 < [tex]t_4_4[/tex] < 2.02) = 0.95 {As the critical value of t at 44 degree
of freedom are -2.02 & 2.02 with P = 2.5%}
P(-2.02 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.02) = 0.95
P( [tex]-2.02 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.02 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-2.02 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.02 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.02 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.02 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]768.2-2.02 \times {\frac{15.1}{\sqrt{45} } }[/tex] , [tex]768.2+2.02 \times {\frac{15.1}{\sqrt{45} } }[/tex] ]
= [763.65 lb , 772.75 lb]
Therefore, 95% confidence interval for the mean breaking strength of the new steel cable is [763.65 lb , 772.75 lb].
mDE = 115 and mbc =42. Find m
Plzzzzz help
45
55
95
Answer:
95
Step-by-step explanation:
yeah
All members of our painting team paint at the same rate. If 20 members can paint a 4800 square foot wall in 36 minutes, then how long would it take for 9 members to paint a 4200 square-foot wall, in minutes?
Answer:
70 minutes
Step-by-step explanation:
We assume that the time is proportional to the area and inversely proportional to the number of members.
Then the time for the second job will be the first job's time multiplied by ...
20/9 . . . the inverse of the ratio of members painting
4200/4800 = 7/8 . . . the ratio of area being painted
The time it would take is ...
(36 min)(20/9)(7/8) = 70 min
It will take the smaller team 70 minutes to paint a 4200 ft² wall.
A state Department of Transportation claims that the mean wait time for various services at its different locations is at most 6 minutes. A random sample of 34 services at different locations has a mean wait time of 10.3 minutes and a standard deviation of 8.0 minutes. Is there enough evidence to reject the claim at alpha equals 0.01? Find the critical value. Round your answer to the thousandths place.
To determine whether there is enough evidence to reject the claim of the state Department of Transportation, a hypothesis test needs to be performed. The null hypothesis (H0) is that the mean wait time is 6 minutes or less, while the alternative hypothesis (H1) is that the mean wait time is greater than 6 minutes. Using a one-sample t-test, the test statistic is calculated and compared to the critical value obtained from the t-distribution table or calculator. In this case, the test statistic is greater than the critical value, indicating that there is enough evidence to reject the null hypothesis.
Explanation:To determine whether there is enough evidence to reject the claim of the state Department of Transportation, we need to perform a hypothesis test.
Hypotheses:
Let:
μ be the mean wait time for various services at different locations.
The null hypothesis (H0) is that the mean wait time (μ) is 6 minutes or less:
H0: μ ≤ 6
The alternative hypothesis (H1) is that the mean wait time (μ) is greater than 6 minutes:
H1: μ > 6
Test statistic:
We will use a one-sample t-test since the population standard deviation is unknown.
The test statistic (t) is calculated using the formula:
t = (Xbar - μ)/(s/√(n))
Where:
Xbar is the sample mean - 10.3 minutes,
μ is the claimed mean - 6 minutes,
s is the sample standard deviation - 8.0 minutes, and
n is the sample size - 34.
Critical value:
To find the critical value, we need to determine the rejection region using the significance level (alpha) of 0.01 and the degrees of freedom (df) which is n - 1 = 33.
From the t-distribution table or calculator, the critical value for a one-tailed test at alpha = 0.01 and df = 33 is approximately 2.449.
Conclusion:
The test statistic (t) = (10.3 - 6)/(8.0/√(34)) = 3.720.
Since the test statistic (t) = 3.720 > 2.449 (the critical value), we have enough evidence to reject the null hypothesis.
Therefore, we can conclude that there is sufficient evidence to reject the claim of the state Department of Transportation at alpha = 0.01.
Use the function w = c + 9 to find the value of w when c = 2
Answer:
The value of w is 11 when c = 2
Step-by-step explanation:
Since c = 2, then we can just plug it into the given equation to find w.
[tex]w=(2)+9[/tex]
[tex]w=11[/tex]
What is the area of a circle with a radius of 7 cm? (Use 3.14 for x and round to the nearest tenth.)
38.5 cm
44.0 cm
150 cm?
153.9 cm?
Answer:
153.9 cm
Step-by-step explanation:
A = pi(r²)
= 3.14(7²)
= 3.14(49)
= 153.86
= 153.9 cm
Need help solving this equation -3(c-1)=33
Answer:
c= -10
Step-by-step explanation:
-3(c-1)= 33
-3(c) -3(-1)= 33 (expand)
-3c +3= 33
-3c= 33-3 (-3 on both sides)
-3c= 30 (simplify)
3c= -30 (divide by -1 throughout)
c= -30 ÷3
c= -10
Historically, a certain region has experienced 92 thunder days annually. (A "thunder day" is day on which at least one instance of thunder is audible to a normal human ear). Over the past fifteen years, the mean number of thunder days is 72 with a standard deviation of 38. Can you conclude that the mean number of thunder days is less than 92? Use the level of significance.
Answer:
We conclude that the mean number of thunder days is less than 92.
Step-by-step explanation:
We are given that Historically, a certain region has experienced 92 thunder days annually.
Over the past fifteen years, the mean number of thunder days is 72 with a standard deviation of 38.
Let [tex]\mu[/tex] = mean number of thunder days.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 92 days {means that the mean number of thunder days is more than or equal to 92}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 92 days {means that the mean number of thunder days is less than 92}
The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean number of thunder days = 72
s = sample standard deviation = 38
n = sample of years = 15
So, test statistics = [tex]\frac{72-92}{\frac{38}{\sqrt{15} } }[/tex] ~ [tex]t_1_4[/tex]
= -2.038
The value of z test statistics is -2.038.
Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical value of -1.761 at 14 degree of freedom for left-tailed test.
Since our test statistics is less than the critical value of t as -2.038 < -1.761, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that the mean number of thunder days is less than 92.