Answer:
The 95 percent confidence interval for the true mean metal thickness is between 0.2903 mm and 0.2907 mm
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 the margin of error 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.
[tex]M = 1.96\frac{0.000586}{\sqrt{59}} = 0.0002[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 0.2905 - 0.0002 = 0.2903 mm
The upper end of the interval is the sample mean added to M. So it is 0.2905 + 0.0002 = 0.2907 mm
The 95 percent confidence interval for the true mean metal thickness is between 0.2903 mm and 0.2907 mm
Lee watches TV for 2 hours per day. During that time, the TV consumes 150 watts per hour. Electricity costs (12 cents)/(1 kilowatt-hour). How much does Lee's TV cost to operate for a month of 30 days?
Answer:
$1.08
Step-by-step explanation:
30 days × (2 hrs/day) × (150 W) × (1 kW / 1000 W) × (0.12 $/kWh) = $1.08
Which of the following is false? Correlation coefficient and the slope always have the same sign (positive or negative). If the correlation coefficient is 1, then the slope must be 1 as well. If the correlation between two variables is close to 0.01, then there is a very weak linear relation between them. Correlation measures the strength of linear association between two numerical variables.
Answer:
If the correlation coefficient is 1, then the slope must be 1 as well.
Step-by-step explanation:
Coefficient of correlation is used in statistics to determine the relationship between two variables. Correlation coefficient and slope always have same sign. It measures the strength of linear relation between two variables. The values of correlation coefficient ranges between 0 to 1. where 0 determines that there is no relationship between two variables.
If the correlation coefficient is 1, then the slope must be 1 as well.
The correlation coefficient (ρ) is a measure that determines the degree to which the movement of two different variables is associated.
Correlation coefficient and the slope both quantify the direction and strength of the relationship between two numeric variables. When the correlation (r) is negative, the regression slope (b) will be negative. When the correlation is positive, the regression slope will be positive.If the correlation between two variables is close to 0.01, then there is a very weak linear relation between them.
So, the false statement is:
If the correlation coefficient is 1, then the slope must be 1 as well.
Learn more:https://brainly.com/question/16557696
The mean and standard deviation of a random sample of n measurements are equal to 34.5 and 3.4, respectively.A. Find a 95 % confidence interval for μ if n=49.B. Find a 95% confidence interval for μ if n=196.C. Find the widths of the confidence intervals found in parts a and b.D. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient fixed?1. Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the confidence interval by a factor of 4.2. Quadrupling the sample size while holding the confidence coefficient fixed increases the width of the confidence interval by a factor of 2.3. Quadrupling the sample size while holding the confidence coefficient fixed increases the width of the on confidence interval by a factor of 4.4. Quadrupling the sample size while holding the confidence coefficient fixed does not affect the width of the confidence interval.5. Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the confidence interval by a factor of 2.
Answer:
a. The 95% confidence interval for the mean is (33.52, 35.48).
b. The 95% confidence interval for the mean is (34.02, 34.98).
c. n=49 ⇒ Width = 1.95
n=196 ⇒ Width = 0.96
Note: it should be a factor of 2 between the widths, but the different degrees of freedom affects the critical value for each interval, as the sample size is different. It the population standard deviation had been used, the factor would have been exactly 2.
d. 5. Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the confidence interval by a factor of 2.
Step-by-step explanation:
a. We have to calculate a 95% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
The sample mean is M=34.5.
The sample size is N=49.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{49}}=\dfrac{3.4}{7}=0.486[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=49-1=48[/tex]
The t-value for a 95% confidence interval and 48 degrees of freedom is t=2.011.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=2.011 \cdot 0.486=0.98[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M-t \cdot s_M = 34.5-0.98=33.52\\\\UL=M+t \cdot s_M = 34.5+0.98=35.48[/tex]
The 95% confidence interval for the mean is (33.52, 35.48).
b. We have to calculate a 95% confidence interval for the mean.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{196}}=\dfrac{3.4}{14}=0.243[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=196-1=195[/tex]
The t-value for a 95% confidence interval and 195 degrees of freedom is t=1.972.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=1.972 \cdot 0.243=0.48[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M-t \cdot s_M = 34.5-0.48=34.02\\\\UL=M+t \cdot s_M = 34.5+0.48=34.98[/tex]
The 95% confidence interval for the mean is (34.02, 34.98).
c. The width of the intervals is:
[tex]n=49\rightarrow UL-LL=33.52-35.48=1.95\\\\n=196\rightarrow UL-LL=34.02-34.98=0.96[/tex]
d. The width of the intervals is decreased by a factor of √4=2 when the sample size is quadrupled, while the others factors are fixed.
Cheryl bought 3.4 pounds of coffee that cost $6.95 per pound . How many did she spend on coffee
Answer:
23.63
Step-by-step explanation:
multiply the cost by the pounds
Answer:
$23.63
Step-by-step explanation:
3.4 X 6.95 = 23.63
A toy falls from a window 80 feet above the ground. How long does it take the toy to hit the ground?
Answer:
2.24 s
Step-by-step explanation:
Given:
Δy = 80 ft
v₀ = 0 ft/s
a = 32 ft/s²
Find: t
Δy = v₀ t + ½ at²
80 ft = (0 ft/s) t + ½ (32 ft/s²) t²
t = 2.24 s
The curvature of a plane parametric curve x = f(t), y = g(t) is $ \kappa = \dfrac{|\dot{x} \ddot{y} - \dot{y} \ddot{x}|}{[\dot{x}^2 + \dot{y}^2]^{3/2}}$ where the dots indicate derivatives with respect to t. Use the above formula to find the curvature. x = 6et cos(t), y = 6et sin(t)
Answer:
The curvature is modelled by [tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex].
Step-by-step explanation:
The equation of the curvature is:
[tex]\kappa = \frac{|\dot {x}\cdot \ddot {y}-\dot{y}\cdot \ddot{x}|}{[\dot{x}^{2}+\dot{y}^{2}]^{\frac{3}{2} }}[/tex]
The parametric componentes of the curve are:
[tex]x = 6\cdot e^{t} \cdot \cos t[/tex] and [tex]y = 6\cdot e^{t}\cdot \sin t[/tex]
The first and second derivative associated to each component are determined by differentiation rules:
First derivative
[tex]\dot{x} = 6\cdot e^{t}\cdot \cos t - 6\cdot e^{t}\cdot \sin t[/tex] and [tex]\dot {y} = 6\cdot e^{t}\cdot \sin t + 6\cdot e^{t} \cdot \cos t[/tex]
[tex]\dot x = 6\cdot e^{t} \cdot (\cos t - \sin t)[/tex] and [tex]\dot {y} = 6\cdot e^{t}\cdot (\sin t + \cos t)[/tex]
Second derivative
[tex]\ddot{x} = 6\cdot e^{t}\cdot (\cos t-\sin t)+6\cdot e^{t} \cdot (-\sin t -\cos t)[/tex]
[tex]\ddot x = -12\cdot e^{t}\cdot \sin t[/tex]
[tex]\ddot {y} = 6\cdot e^{t}\cdot (\sin t + \cos t) + 6\cdot e^{t}\cdot (\cos t - \sin t)[/tex]
[tex]\ddot{y} = 12\cdot e^{t}\cdot \cos t[/tex]
Now, each term is replaced in the the curvature equation:
[tex]\kappa = \frac{|6\cdot e^{t}\cdot (\cos t - \sin t)\cdot 12\cdot e^{t}\cdot \cos t-6\cdot e^{t}\cdot (\sin t + \cos t)\cdot (-12\cdot e^{t}\cdot \sin t)|}{\left\{\left[6\cdot e^{t}\cdot (\cos t - \sin t)\right]^{2}+\right[6\cdot e^{t}\cdot (\sin t + \cos t)\left]^{2}\right\}^{\frac{3}{2}}} }[/tex]
And the resulting expression is simplified by algebraic and trigonometric means:
[tex]\kappa = \frac{72\cdot e^{2\cdot t}\cdot \cos^{2}t-72\cdot e^{2\cdot t}\cdot \sin t\cdot \cos t + 72\cdot e^{2\cdot t}\cdot \sin^{2}t+72\cdot e^{2\cdot t}\cdot \sin t \cdot \cos t}{[36\cdot e^{2\cdot t}\cdot (\cos^{2}t -2\cdot \cos t \cdot \sin t +\sin^{2}t)+36\cdot e^{2\cdot t}\cdot (\sin^{2}t+2\cdot \cos t \cdot \sin t +\cos^{2} t)]^{\frac{3}{2} }}[/tex]
[tex]\kappa = \frac{72\cdot e^{2\cdot t}}{[72\cdot e^{2\cdot t}]^{\frac{3}{2} } }[/tex]
[tex]\kappa = [72\cdot e^{2\cdot t}]^{-\frac{1}{2} }[/tex]
[tex]\kappa = 72^{-\frac{1}{2} }\cdot e^{-t}[/tex]
[tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex]
The curvature is modelled by [tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex].
CAN SOMEONE HELP ME ASAP
A. 5
B. 53‾√53
C. 10
D. 103√3
Answer:
n = 5
Step-by-step explanation:
Since this is a right triangle, we can use trig functions
tan theta = opp/ adj
tan 30 = n/ 5 sqrt(3)
5 sqrt(3) tan 30 = n
5 sqrt(3) * 1/ sqrt(3) = n
5 = n
The volume of a gas in a container varies inversely with the pressure on the gas. A container of helium has a volume of 370in3 under a pressure of 15psi (pounds per square inch). Write the equation that relates the volume, V, to the pressure, P. What would be the volume of this gas if the pressure was increased to 25psi?
Answer:
Step-by-step explanation:
When two variables vary inversely, it means that an increase in one would lead to a decrease in the other and vice versa. Since the volume of a gas, V in a container varies inversely with the pressure on the gas, P, if we introduce a constant of proportionality, k, the expression would be
V = k/P
If V = 370 in³ and P = 15psi, then
370 = k/15
k = 370 × 15 = 5550
The equation that relates the volume, V, to the pressure, P would be
V = 5550/P
if the pressure was increased to 25psi, the volume would be
V = 5550/25 = 222 in³
Answer:
v=5550/p
222
Step-by-step explanation:
You want to install a 1 1 yd wide walk around a circular swimming pool. The diameter of the pool is 23 yd. What is the area of the walk? Use 3.14 for pi π.
Complete Question:
You want to install a 1 yd wide walk around a circular swimming pool. The diameter of the pool is 23 yd. What is the area of the walk? Use 3.14 for pi π.
Answer:
75.36 square yard
Step-by-step explanation:
From the question,
The diameter of this circular pool inside is 23 yd.
This means that the radius = Diameter/2 = 23yd/2 = 11.5 yd.
The formula for the area of a circle =
A = πr²
A = π(11.5)²
A =3.14 × 11.5²
A = 415.265 yd²
This is the Area of the inner circle.
We were told in the question also that he wants to install a walk of 1 yard
Hence, the radius of outer circle =
radius of inner circle +length of the walk
11.5yard + 1 yard
= 12.5 yard
A = πr²
A = 3.14 × (12.5)²
A = 490.625yd²
Area of the walk = Area of the Outer circle - Area of the inner circle
= (490.625 - 415.265)yd = 75.36 yd²
Therefore, the area of the walk is 75.36 square yards.
A scooter runs 40 km using 1 litre of petrol tje distance covered by it using 15/4 litres of petrol is
Answer:
150 km
Step-by-step explanation:
1 liter ............ 40 km
15/4 liter .........x km
x = 15/4×40/1 = 600/4 = 150 km
An experiment consists of dealing 7 cards from a standard deck of 52 playing cards. What is the probability of being dealt exactly 4 clubs and 3 spades?
Answer: 0.00153
Step-by-step explanation:
Given: An experiment consists of dealing 7 cards from a standard deck of 52 playing cards.
Number of ways of dealing 7 cards from 52 cards = [tex]^{52}C_7[/tex]
Since there are 13 clubs and 13 spades.
Number of ways of getting exactly 4 clubs and 3 spades=[tex]^{13}C_4\times\ ^{13}C_3[/tex]
Now, the probability of being dealt exactly 4 clubs and 3 spades
[tex]=\dfrac{^{13}C_4\times\ ^{13}C_3}{^{52}C_7}\\\\\\=\dfrac{{\dfrac{13!}{4!(9!)}\times\dfrac{13!}{3!10!}}}{\dfrac{52!}{7!45!}}\\\\=\dfrac{715\times286}{133784560}\\\\=0.00152850224271\approx0.00153[/tex]
Hence, the probability of being dealt exactly 4 clubs and 3 spades = 0.00153
A 12 sided die is rolled the set of equally likely outcomes is 123 456-789-10 11 and 12 find the probability of rolling a number greater than three
Answer:
6
Step-by-step explanation:
nerd physics