Answer:
The proportions of the diameters that are greater than 25.4 millimeters is 5%.
Step-by-step explanation:
Given;
mean of the normal distribution, m = 25.1 millimeters
standard deviation, d = 0.08 millimeter
1 standard deviation above the mean = m + d = 25.1 + 0.08 = 25.18
2 standard deviation above mean = m + 2d = 25.1 + 2(0.08) = 25.26
3 standard deviation above the mean = m + 3d = 25.1 + 3(0.08) = 25.34
4 standard deviation above the mean = m + 4d = 25.1 + 4(0.08) = 25.42
To obtain a diameter greater than 25.4, we select data after 4 standard deviation above the mean.
Data within 4 standard deviation above the mean is 95%
Data outside 4 standard deviation above the mean is 5%
Therefore, the proportions of the diameters that are greater than 25.4 millimeters is 5%.
g(x) = f(x+1) using f(x)= x to the power of 2
Answer:
g(x) = x² + 2x + 1
General Formulas and Concepts:
Algebra I
Terms/Coefficients
ExpandingFunctions
Function NotationStep-by-step explanation:
Step 1: Define
Identify
g(x) = f(x + 1)
f(x) = x²
Step 2: Find
Substitute in x [Function f(x)]: f(x + 1) = (x + 1)²Expand: f(x + 1) = x² + 2x + 1Redefine: g(x) = x² + 2x + 1Select the expression that has a value of 13.
9 + 3 x (2 ÷ 3) + 6
(9 + 3) x 2 ÷ 3 + 6
9 − (3 x 2) ÷ 3 + 6
(9 + 3 x 2) ÷ 3 + 6
Answer:
9 − (3 x 2) ÷ 3 + 6 is the answer
a) __m=10km 25m =___km
b) __m=__km__m=1.5 km
Example :
a) 7250m= 7km 250m = 7.250km
Please help me
Answer:
a) 10,025 m = 10km 25m = 10.025 km
b) 1,500 m = 1 km 500 m = 1.5 km
Answer:
a) 10025m = 10km 25m = 10.025km
b) 1500m = 1km 500m = 1.5km
Step-by-step explanation:
Concept:
Here, we need to know the idea of unit conversion.
Unit conversion is the conversion between different units of measurement for the same quantity.
1 km = 1000 m
Solve:
a)
10km 25m = 10×1000 + 25 = 10025 m10km 25m = 10 + 25/1000 = 10.025 kmb)
1.5km = 1 + 0.5 × 1000 = 1km 500m1.5km = 1.5 × 1000 = 1500mHope this helps!! :)
Please let me know if you have any questions
The scores on an entrance exam to a university are known to have an approximately normal distribution with mean 65% and standard deviation 7.1%. Using the normalcdf function on your graphing calculator, what percentage of students would score 70 or better on this entrance exam?
A. 28.4%
B. 18.9%
C. 24.1%
D. 22.3%
Answer:
The correct answer is - C. 24.1%
Step-by-step explanation:
Given:
mean μ = 65%
standard deviation δ = 7.1 %
solution:
Prob( X>70) = 1 - Prob(x<70)
= P (x-μ/δ ≥ 70 -65/7.1)
= 1 - Prob( (70-65)/7.1)
= 1 - Prob ( z < 0.7042553)
= 0.24065
the percentage of students scoring 70 or more in the exam
= 24.065*100
= 24.1%
Graph the linear equation by find
2 = 4x + y
2=4x+y
y=2-4x
Александр Мазепов
Step-by-step explanation:
Step 1: Solve for y
[tex]2 = 4x + y[/tex]
[tex]2 - 4x = 4x + y - 4x[/tex]
[tex]y = 2 - 4x[/tex]
Step 2: Solve for x
[tex]2 - 2 - 4x = 0 - 2[/tex]
[tex]-4x / -4 = -2 / -4[/tex]
[tex]x = 1/2[/tex]
Step 3: Solve for y
[tex]y = 2 - 4(0)[/tex]
[tex]y = 2[/tex]
Step 4: Graph the equation
Graph the x-intercept, (1/2, 0), the y-intercept, (0, 2) and draw a line between them. Look at the attached picture for the graph:
I want to know how to solve this equation
Answer:
your answer will be Option D
Step-by-step explanation:
log 10
The manager of a juice bottling factory is considering installing a new juice bottling machine which she hopes will reduce the amount of variation in the volumes of juice dispensed into 8-fluid-ounce bottles. Random samples of 10 bottles filled by the old machine and 9 bottles filled by the new machine yielded the following volumes of juice (in fluid ounces).
Old machine: 8.2, 8.0, 7.9, 7.9, 8.5, 7.9, 8.1,8.1, 8.2, 7.9
New machine: 8.0, 8.1, 8.0, 8.1, 7.9, 8.0, 7.9, 8.0, 8.1
Required:
Use a 0.05 significance level to test the claim that the volumes of juice filled by the old machine vary more than the volumes of juice filled by the new machine
Answer:
Reject H0 and conclude that volume filled by old machine varies more than volume filled by new machine
Step-by-step explanation:
Given the data:
Old machine: 8.2, 8.0, 7.9, 7.9, 8.5, 7.9, 8.1,8.1, 8.2, 7.9
New machine: 8.0, 8.1, 8.0, 8.1, 7.9, 8.0, 7.9, 8.0, 8.1
To test if volume filled by old machine varies more than volume filled by new machine :
Hypothesis :
H0 : s1² = s2²
H1 : s1² > s2²
Using calculator :
Sample size, n and variance of each machine is :
Old machine :
s1² = 0.37889
n = 10
New machine :
s2² = 0.006111
n = 9
Using the Ftest :
Ftest statistic = larger sample variance / smaller sample variance
Ftest statistic = 0.37889 / 0.006111
Ftest statistic = 62.001
Decision region :
Reject H0 ; If Test statistic > Critical value
The FCritical value at 0.05
DFnumerator = 10 - 1 = 9
DFdenominator = 9 - 1 = 8
Fcritical(0.05, 9, 8) = 3.388
Since 62 > 3.388 ; Reject H0 and conclude that volume filled by old machine varies more than volume filled by new machine
PLZ ANSWER QUESTION IN PICTURE
Answer:
X-int = -5 and y-int = 6
Step-by-step explanation:
1.2x+6 = 0
1.2x= -6
X = -6/1.2
X = -5
Which expression is equivalent to the given expression?
Answer:
Option C, a³b
Step-by-step explanation:
(ab²)³/b⁵
= a³b⁶/b⁵
= a³b
Answered by GAUTHMATH
If p is true and ~ q is false, then p ~ q is _____ false.
a. sometimes
b. always
c. never
What is the value of the capacitance of a capacitor that stores 40
μ
C on each plate, when a potential difference of 10 V is applied to it?
We know
[tex]\boxed{\sf Q=CV}[/tex]
[tex]\\ \large\sf\longmapsto C=\dfrac{Q}{V}[/tex]
[tex]\\ \large\sf\longmapsto C=\dfrac{40}{10}[/tex]
[tex]\\ \large\sf\longmapsto C=4\mu F[/tex]
A company that manufactures and bottles apple juice uses a machine that automatically fills 32-ounce bottles. There is some variation, however, in the amount of liquid dispensed into the bottles. The amount dispensed has been observed to be approximately normally distributed with mean 32 ounces and standard deviation 1 ounce. Determine the proportion of bottles that will have more than 30 ounces dispensed into them. (Round your answer to four decimal places.)
Answer:
The proportion of bottles that will have more than 30 ounces dispensed into them is 0.9772 = 97.72%.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
The amount dispensed has been observed to be approximately normally distributed with mean 32 ounces and standard deviation 1 ounce.
This means that [tex]\mu = 32, \sigma = 1[/tex]
Determine the proportion of bottles that will have more than 30 ounces dispensed into them.
This is 1 subtracted by the p-value of Z when X = 30, so:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30 - 32}{1}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a p-value of 0.0228.
1 - 0.0228 = 0.9772
The proportion of bottles that will have more than 30 ounces dispensed into them is 0.9772 = 97.72%.
A rectangular floor is 20 feet long and 16 feet broad. if it is to be paved with squared marbles of same size,find the greatest length of each squared marbles.
Answer:
4 ft
Step-by-step explanation:
I guess, the meaning is the largest marbles, so that we can pave the whole floor without cutting any marbles and leaving empty spots.
so, 20×16 ft²
we can have marbles 1/2 ft long. and it all fits well : 40×32 marbles.
we can have them 1 ft long, and it all fits well : 20×16 marbles.
we can have them 2ft long, and it still fits well : 10×8 marbles.
and so on.
so, actually, we are looking for the greatest common divisor (GCD) of 20 and 16. and that gives us the maximum length of a single marble to fulfill the requirement.
let's go for the prime factors starting with 2
20/2 = 10
10/2 = 5
5/3 fits not work
5/5 = 1 done
so, 20 = 2²×3⁰×5¹
16/2 = 8
8/2 = 4
4/2 = 2
2/2 = 1 done
16 = 2⁴
so, for the GCD I can only use powers of 2 (the only prime factors both numbers have in common).
and we have to use the smaller power of 2, which is 2, so, the GCD is 2² = 4
=>
the maximum length of the squared marbles is 4 ft.
that would pave the floor with 5×4 marbles completely.
In the Data Analysis portion of the article the authors report that they completed a power analysis to determine the power of their study with the sample size utilized. They report a power of 90%. What does this mean
Answer:
Kindly check explanation
Step-by-step explanation:
The power of a test simply gives the probability of Rejecting the Null hypothesis, H0 in a statistical analysis given that the the alternative hypothesis, H1 for the study is true. Hence, the power of a test can be referred to as the probability of a true positive outcome in an experiment.
Using this definition, a power of 90% simply means that ; there is a 90% probability that the a Pvalue less Than the α - value of an experiment is obtained if there is truly a significant difference. Hence, a 90% chance of Rejecting the Null hypothesis if truly the alternative hypothesis is true.
You pay $1.25 per pound for x pounds of apples?
Answer:
$1.25x
Step-by-step explanation:
Given :
Cost per pound = $1.25
Number of pounds of apple = x
The total cost of apple = (cost per pound * number of apple in pounds)
Hence,
Total cost of x pounds of apple is :
($1.25 * x)
= $1.25x
Write 6/7 as a decimal rounded to the nearest hundredth
Answer:
0.01
Step-by-step explanation:
6/7% = 6÷7÷100 = 0.0085714286 round to the nearest hundredth = 0.01
help with q25 please. Thanks.
First, I'll make f(x) = sin(px) + cos(px) because this expression shows up quite a lot, and such a substitution makes life a bit easier for us.
Let's apply the first derivative of this f(x) function.
[tex]f(x) = \sin(px)+\cos(px)\\\\f'(x) = \frac{d}{dx}[f(x)]\\\\f'(x) = \frac{d}{dx}[\sin(px)+\cos(px)]\\\\f'(x) = \frac{d}{dx}[\sin(px)]+\frac{d}{dx}[\cos(px)]\\\\f'(x) = p\cos(px)-p\sin(px)\\\\ f'(x) = p(\cos(px)-\sin(px))\\\\[/tex]
Now apply the derivative to that to get the second derivative
[tex]f''(x) = \frac{d}{dx}[f'(x)]\\\\f''(x) = \frac{d}{dx}[p(\cos(px)-\sin(px))]\\\\ f''(x) = p*\left(\frac{d}{dx}[\cos(px)]-\frac{d}{dx}[\sin(px)]\right)\\\\ f''(x) = p*\left(-p\sin(px)-p\cos(px)\right)\\\\ f''(x) = -p^2*\left(\sin(px)+\cos(px)\right)\\\\ f''(x) = -p^2*f(x)\\\\[/tex]
We can see that f '' (x) is just a scalar multiple of f(x). That multiple of course being -p^2.
Keep in mind that we haven't actually found dy/dx yet, or its second derivative counterpart either.
-----------------------------------
Let's compute dy/dx. We'll use f(x) as defined earlier.
[tex]y = \ln\left(\sin(px)+\cos(px)\right)\\\\y = \ln\left(f(x)\right)\\\\\frac{dy}{dx} = \frac{d}{dx}\left[y\right]\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(f(x)\right)\right]\\\\\frac{dy}{dx} = \frac{1}{f(x)}*\frac{d}{dx}\left[f(x)\right]\\\\\frac{dy}{dx} = \frac{f'(x)}{f(x)}\\\\[/tex]
Use the chain rule here.
There's no need to plug in the expressions f(x) or f ' (x) as you'll see in the last section below.
Now use the quotient rule to find the second derivative of y
[tex]\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]\\\\\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{f'(x)}{f(x)}\right]\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-f'(x)*f'(x)}{(f(x))^2}\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2}\\\\[/tex]
If you need a refresher on the quotient rule, then
[tex]\frac{d}{dx}\left[\frac{P}{Q}\right] = \frac{P'*Q - P*Q'}{Q^2}\\\\[/tex]
where P and Q are functions of x.
-----------------------------------
This then means
[tex]\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} + \left(\frac{f'(x)}{f(x)}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} +\frac{(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2+(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\[/tex]
Note the cancellation of -(f ' (x))^2 with (f ' (x))^2
------------------------------------
Let's then replace f '' (x) with -p^2*f(x)
This allows us to form ( f(x) )^2 in the numerator to cancel out with the denominator.
[tex]\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*f(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*(f(x))^2}{(f(x))^2} + p^2\\\\-p^2 + p^2\\\\0\\\\[/tex]
So this concludes the proof that [tex]\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2 = 0\\\\[/tex] when [tex]y = \ln\left(\sin(px)+\cos(px)\right)\\\\[/tex]
Side note: This is an example of showing that the given y function is a solution to the given second order linear differential equation.
How many women must be randomly selected to estimate the mean weight of women in one age group? We want 90% confidence that the sample mean is within 3.7 lbs of the populations mean, and population standard deviation is known to be 28 lbs.
Answer:
155 women must be randomly selected.
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 - 0.9}{2} = 0.05[/tex]
Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.05 = 0.95[/tex], so Z = 1.645.
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.
The population standard deviation is known to be 28 lbs.
This means that [tex]\sigma = 28[/tex]
We want 90% confidence that the sample mean is within 3.7 lbs of the populations mean. How many women must be sampled?
This is n for which M = 3.7. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]3.7 = 1.645\frac{28}{\sqrt{n}}[/tex]
[tex]3.7\sqrt{n} = 1.645*28[/tex]
[tex]\sqrt{n} = \frac{1.645*28}{3.7}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.645*28}{3.7})^2[/tex]
[tex]n = 154.97[/tex]
Rounding up:
155 women must be randomly selected.
An F test for the two coefficients of promotional expenditures and district potential is performed. The hypotheses are H0: 1 = 4 = 0 versus Ha: at least one of the j is not 0. The F statistic for this test is 1.482 with 2 and 21 degrees of freedom. What can we say about the P-value for this test?
Answer:
Pvalue > 0.10
Step-by-step explanation:
Given the hypothesis :
H0 : β1 = β4 = 0
H1 : Atleast one of βj is not 0
F statistic = 1.482 ;
Degree of freedom = 2 and 21 ;
DFnumerator = 2
DFdenominator = 21
Using the Pvalue calculator from Fstatistic ;
Pvalue(1.482, 2, 21) = 0.24999 = 0.25
Hence, Pvalue for the test is 0.25
Pvalue > 0.10
Find the missing numerator: 3 1/3 = x/6
[tex]\sf\huge\underline\color{pink}{༄Answer:}[/tex]
[tex]\tt3 \frac{1}{3} = \frac{x}{6} \\ = \tt \frac{10}{3} = \frac{x}{6} \\ = \tt \frac{x}{6} = \frac{10}{3} \\ = \tt6 \frac{x}{6} = 6( \frac{10}{3} ) \\ = \tt\large\boxed{\tt{\color{pink}{x = 20}}}[/tex]
[tex]\color{pink}{==========================}[/tex]
#CarryOnLearning
Ray’s weight increased by 11% in the last two years. If he gained 16.5 pounds, what was his weight two years ago?
I need answering ASAP please
Answer:
The choose (D) 1/3
I hope I helped you^_^
Find the value of x.
Answer:
x=3
Step-by-step explanation:
Find the radius and use Pythagoras on the right side
According to Fidelity Investment Vision Magazine, the average weekly allowance of children varies directly as their grade level. In a recent year, the average allowance of a 9th-grade student was 9.66 dollars per week. What was the average weekly allowance of a 5 th-grade student?
The average weekly allowance of a 5th grade student as calculated using direct variation with the information provided by Fidelity Investment Vision Magazine is 5.367 dollars per week.
The question given is a direct variation problem:
Let:
• Average weekly allowance = [tex]a[/tex]
• Grade level = [tex]g[/tex]
If Average weekly allowance varies directly as grade level , then , then the direct variation between the variables can be expressed as :
[tex]a = k * g[/tex]
Where , [tex]k[/tex] = constant of proportionality
We can obtain the value of k from the given values of a and g
[tex]9.66 = k * 9\\9.66 = 9k\\k = 9.66/9[/tex]
Our equation becomes:
[tex]a = (9.66/9) * g[/tex]
[tex]a = (9.66/9) * 5\\a = 5.367[/tex] (rounded to 3 decimal places)
Hence, using proportional relationship, the average weekly allowance for a 5th grade student is [tex]5.367[/tex] per week
Learn more about direct variation here:
https://brainly.com/question/17257139
Out of a total of 10 college textbooks estimate the standard deviation of their ages if the oldest textbook is known to be 7.9 years old and the newest textbook is 1.3 years old.
Answer:
Given that the maximum age of the textbook is 7.9 years and the minimum age of the textbook is 1.3 years.
Using the range rule, the standard deviation is estimated as,
S≈maximum−minimum/4
=7.9−1.3/4
=1.65
The required value of the approximate standard deviation is 1.65.
The standard deviation of the data is 1.65.
What is Standard Deviation?Standard deviation is the measure of the deviation of the data from the mean.
The total college textbooks is 10
The oldest book is 7.9 years old
The newest book is 1.3 years old
The standard deviation of range is equal to one fourth of the difference of maximum to minimum.
The standard deviation = ( 7.9 - 1.3 ) /4 = 1.65
To know more about Standard Deviation
https://brainly.com/question/14747159
#SPJ5
The starting line up for a basketball team is to consist of two forwards and three guards. Two brothers are on the team. Matthew is a forward and Tony a guard. There are four forwards and six guards from which to choose the line up. If the starting players are chosen at random, what is the probability that the two brothers will end up in the starting line up
Answer:
0.25 = 25% probability that the two brothers will end up in the starting line up
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
The order in which the players are chosen is not important, which means that the combinations formula is used to solve this question.
Combinations formula:
[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]
Desired outcomes:
Matthew plus another forward from a set of 3.
Tony plus another two guards from a set of 5.
So
[tex]D = C_{3,1}C_{5,2} = \frac{3!}{1!2!} \times \frac{5!}{2!3!} = 3*10 = 30[/tex]
Total outcomes:
Two forwards from a set of 4.
Three guards from a set of 6.
So
[tex]T = C_{4,2}C_{6,3} = \frac{4!}{2!2!} \times \frac{6!}{3!3!} = 6*20 = 120[/tex]
What is the probability that the two brothers will end up in the starting line up?
[tex]p = \frac{D}{T} = \frac{30}{120} = 0.25[/tex]
0.25 = 25% probability that the two brothers will end up in the starting line up
can anybody help with this ?
Answer:(
fx).(gx)=D. -40x^3+25x^2+45
Step-by-step explanation:
Two numbers total 31 and have a difference of 11. Find the two numbers
Answer:
let 2 no. be x and y
x+y=31 ..... (1)
x-y=11 .........(2)
from (1)
x=31-y ..........(3)
putting (3) in (2)
31-y-y=11
-2y=11-31
-2y=-20
2y=20
y=10
Can someone help me find the answer?
Answer: B. This function has no intercept. I think B is the correct answer.
I need help with that, if you can, plz. I ty it I think is a not sure
Answer:
-5≤x <1
Step-by-step explanation:
sqrt( x+5) / sqrt(1-x)
The numerator must be greater than zero since it is a square root
sqrt(x+5) ≥0
Square each side
x+5≥0
x≥-5
The denominator must be greater than zero (the denominator cannot be zero)
sqrt(1-x)> 0
Square each side
1-x > 0
1>x
Putting these together
-5≤x <1