Answer:
q=171
Step-by-step explanation:
Hello there!
In this explanation k is constant and it means that q is some constant amount bigger than r.
q=kr
76=k×20
76÷20=k
k= 3.8
q=3.8r
q=3.8×45
q=171
Tory plans to enter the path at mile marker 1 and hike for 4 miles. Tory is a beginner hiker, so he wants to choose the path that is less steep on average.
Which path should he choose? Why?
Describe how you could draw a diagram for a problem about finding the total length for two strings, 15 inches long and 7 inches long
[tex] \frac{10.5}{1.79} [/tex]
[tex] \frac{16}{2.19} [/tex]
How would I find out the unit rate for these two equations?
PLEASE HELLPPPP!!!!!!!!!!!!
Figure BMHF is rotated how many degrees clockwise?
Which best describes a triangle with side lengths of 6 inches 8 inches and 9 inches?
What is the area of a figure whose vertices are (-2,1), (3,1), (2,-3), and (-3,-3)?
the volume of water in a bowl is given by V=1/3pih^2(60-h), where h is the depth of the water in centimeters. If the depth is increasing at the rate of 3cm/sec when the water is 10 cm deep, how fast is the volume increasing at that instant
tanθ=sinθ/cosθ
True or False
Marcus drew a quadrilateral with only 1 pair of parallel sides. Which quadrilateral did Marcus draw?
Find the area of the shaded regions. Give your answer as a completely simplified exact value in terms of π (no approximations).
Answer: 40 pi
Step-by-step explanation:
write a number sentence to compare 7.17 and 7.170
Answer:
They are equal
Step-by-step explanation:
What is the standard form of the equation of a circle given by x^2 + y^2 - 18x + 8y + 5 = 0
Answer:
x-9,y+4,92
Step-by-step explanation:
At a pie-eating contest, Isla ate 1⁄4 of a pie before time was called. Alexa only finished 1⁄8 of a pie. How much more pie did Isla eat than Alexa?
The area of a rectangular patio is 5 and 5/8 square yards, and its length is 1 and 1/2 yards. What is the patio's width in yards?
If a quantity you calculated has units of (kg*m^2)/(s^2*c), what is that quantity?
Final answer:
The quantity with units of (kg*m²)/(s²*c) is closely related to the energy concept in physics, specifically within the framework of Einstein's mass-energy equivalence formula, E = mc², albeit with an added factor of the speed of light in the denominator.
Explanation:
The quantity with units of (kg*m^2)/(s^2*c) is related to the concept of mass-energy equivalence, as seen in Albert Einstein's famous equation, E = mc².
In this equation, E represents energy in joules, m represents mass in kilograms, and c represents the speed of light in meters per second.
The units of c² are m²/s², making the units of energy (when mass is given in kilograms) equivalent to kg*m²/s², or joules.
However, the specific units you mentioned, (kg*m²)/(s²*c), seem to incorporate an additional factor of the speed of light in the denominator, suggesting a possible deviation or specific application within the broader context of relativistic physics or energy conversions where the factor of c is explicitly considered.
The average number of mosquitoes in a stagnant pond is 70 per square meter with a standard deviation of 8 per square meter. if one square meter of the pond is chosen at random for a mosquito count, find the probability that the average of the count is more than 68 mosquitoes per square meter. assume that the variable is normally distributed.
The probability that the average mosquito count in a randomly chosen square meter of the pond is more than 68 mosquitoes is approximately 59.87%.
To find the probability that the average count of mosquitoes in a randomly chosen square meter of the pond is more than 68 mosquitoes per square meter, we can use the normal distribution properties.
Given:
Mean (">"): 70 mosquitoes per square meterStandard deviation (σ): 8 mosquitoes per square meterWe need to find P(X > 68). To do this, we convert the raw score to a Z-score using the formula:
Z = (X - μ) / σ
Substitute the given values:
Z = (68 - 70) / 8 = -2 / 8 = -0.25
Next, we need to find the probability corresponding to Z > -0.25 using the standard normal distribution table.
Looking up Z = -0.25, we find the cumulative probability (P(Z < -0.25)) is approximately 0.4013.
To find P(Z > -0.25):
P(Z > -0.25) = 1 - P(Z < -0.25) = 1 - 0.4013 = 0.5987
Therefore, the probability that the average mosquito count in a randomly chosen square meter is more than 68 mosquitoes per square meter is approximately 0.5987 or 59.87%.
Sally has x books that weigh 2 pounds each and 8 books that weigh 3 pounds each. the total weight of her books is 62 pounds. which equation below could be used to find how many 2-pound books sally has?
Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.
a2 − 2a − 224 = 0
8, –28
16, –14
–16, 14
–8, 28
A professor teaching a discrete math course gives a multiple-choice quiz that has ten questions, each with four possible responses: a, b, c,
d. what is the minimum number of students that must be in the professor's class in order to guarantee that at least three answer sheets must be identical? (assume that no answers are left blank.)3
The correct minimum number of students required to guarantee that at least three answer sheets must be identical is 5.
To solve this problem, we can use the pigeonhole principle. The pigeonhole principle states that if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon.
In this scenario, each question can be thought of as a pigeonhole, with the possible answers (a, b, c, d) as the pigeons. Since there are four possible answers for each question, we have four pigeonholes for each of the ten questions.
Now, let's consider the number of students (pigeons) and the number of unique answer sheets (pigeonholes). We want to find the minimum number of students such that at least three of them have identical answer sheets.
For two students, there are [tex]\(4^{10}\)[/tex] possible combinations of answers for the ten questions. For three students, the number of possible combinations of answers is [tex]\(4^{10} \times 4^{10}\)[/tex]. In general, for [tex]\(n\)[/tex] students, the number of possible combinations is [tex]\(4^{10} \times 4^{10} \times \ldots \times 4^{10}\) (\(n\) times)[/tex].
We need to find the smallest [tex]\(n\)[/tex] such that the number of combinations exceeds the number of ways to distribute the answer sheets without having three identical ones.
The number of ways to distribute the answer sheets without having three identical is given by the sum of the combinations of choosing 1, 2, ...,[tex]\(n-1\)[/tex] students out of [tex]\(n\)[/tex] to have unique answer sheets, and then multiplying by [tex]\(4^{10}\)[/tex] for each unique combination. This is because we are allowing for the possibility that some students have the same answer sheets, but not more than two.
The sum is given by:
[tex]\[ \binom{n}{1} \times 4^{10} + \binom{n}{2} \times 4^{10} + \ldots + \binom{n}{n-1} \times 4^{10} \][/tex]
We want to find the smallest [tex]\(n\)[/tex] such that:
[tex]\[ \binom{n}{1} \times 4^{10} + \binom{n}{2} \times 4^{10} + \ldots + \binom{n}{n-1} \times 4^{10} < 4^{10n} \][/tex]
For [tex]\(n = 4\)[/tex], we have:
[tex]\[ \binom{4}{1} \times 4^{10} + \binom{4}{2} \times 4^{10} + \binom{4}{3} \times 4^{10} = 4 \times 4^{10} + 6 \times 4^{10} + 4 \times 4^{10} = 14 \times 4^{10} \][/tex]
For [tex]\(n = 5\)[/tex], we have:
[tex]\[ \binom{5}{1} \times 4^{10} + \binom{5}{2} \times 4^{10} + \binom{5}{3} \times 4^{10} + \binom{5}{4} \times 4^{10} \][/tex]
[tex]\[ = 5 \times 4^{10} + 10 \times 4^{10} + 10 \times 4^{10} + 5 \times 4^{10} = 30 \times 4^{10} \][/tex]
Now, we compare [tex]\(30 \times 4^{10}\) with \(4^{10 \times 5}\)[/tex]. Since [tex]\(4^{10 \times 5}\)[/tex] is much larger than[tex]\(30 \times 4^{10}\)[/tex], we can see that for [tex]\(n = 5\)[/tex], the number of possible combinations exceeds the number of ways to distribute the answer sheets without having three identical ones.
Therefore, the minimum number of students required to guarantee that at least three answer sheets must be identical is 5.
T-shirts at a clothing store cost $19 each Rory bought several T-shirts at this price and paid $57 how many t-shirts did Rory buy
nine times five less than twice x
[tex]nine times fivelesstwicex[/tex]
what is the tenth rule for A(n)=-9+5(n-1)
Explain how you can tell when a formula represents a length an area or a volume
Area and volume are both used to determine the amount of space.
While area is used for two-dimensional shapes, volume is used for three-dimensional shapes.
The difference between the given parameters are:
Length is often represented as the difference between points.Area is represented as the product of two dimensions or the square of a dimensionVolume is represented as the product of three dimensions or the cube of a dimensionFor instance:
[tex]\mathbf{Length = b -a}[/tex]
[tex]\mathbf{Area = ab}[/tex]
[tex]\mathbf{Volume= abc}[/tex]
Read more about lengths, areas and volumes at:
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Can someone please help me
Leslie is a biologist. She is going to randomly select one animal from her lab to study. There are 5 salamanders, 3 crayfish, and 12 minnows in her lab. What is \text{P(salamander}
The sample space contains three types of animals; 5 salamanders, 3 crayfish, and 12 minnows.
So population of sample space would be, n(S) = Salamanders + Crayfish + Minnows.
n(S) = 5 + 3 + 12 = 20.
She wants to randomly select an animal and check if it is Salamander or not.
population of favourable outcomes would be, n(salamander) = 5.
Now the probability for selecting a salamander would be :-
[tex]P(salamander) = \frac{n(salamander)}{n(S)} \\\\ P(salamander) = \frac{5}{20} \\\\ P(salamander) = \frac{1}{4} \;or\; 0.25 \;or\; 25\%[/tex]
Concerned about graffiti, mayors of nine suburban communities instituted a citizen community watch program. community monthly incidents after monthly incidents before burr oak 2 7 elgin corners 3 2 elm grove 6 2 greenburg 4 3 huntley 5 11 north lyman 3 9 south lyman 3 5 pin oak 5 7 victorville 2 3 click here for the excel data file (a) choose the appropriate hypotheses to see whether the number of graffiti incidents declined. assume μd is the mean difference in graffiti incidents before and after.
Final answer:
The appropriate hypotheses to determine if graffiti incidents declined involve a null hypothesis where the mean difference is greater than or equal to zero and an alternative hypothesis where the mean difference is less than zero. The μd represents the mean difference in graffiti incidents before and after the institution of the community watch program.
Explanation:
The appropriate hypotheses to test whether the number of graffiti incidents declined in the suburban communities after the institution of a citizen community watch program are as follows:
H0: μd ≥ 0 (Null hypothesis: The mean difference in graffiti incidents before and after is greater than or equal to 0, indicating no decrease.)Ha: μd < 0 (Alternative hypothesis: The mean difference in graffiti incidents before and after is less than 0, indicating a decrease.)Here, μd (mu-sub-d) represents the mean difference between the number of incidents before the program and after the program was implemented. To test these hypotheses, we would typically conduct a paired t-test if the samples are normally distributed or a Wilcoxon signed-rank test if the distribution is not normal.
Assuming μd is the mean difference in graffiti incidents before and after, the test will determine if there is statistically significant evidence to support that the community watch program resulted in a reduction of graffiti incidents.
Circle 1 is centered at (-4, 5) and has a radius of 2 centimeters. Circle 2 is centered at (2, 1) and has a radius of 6 centimeters. What transformations can be applied to circle 1 to prove that the circles are similar? The circles are similar because you can translate circle 1 using the transformation rule ( x, x) and then dilate it using a scale factor of ( ).
we know that
Figures can be proven similar if one, or more, similarity transformations
(reflections, translations, rotations, dilations) can be found that map one
figure onto another.
In this problem to prove circle 1 and circle 2 are similar, a
translation and a scale factor (from a dilation) will be found to map one
circle onto another.
we have that
step 1
Move the center of the circle 1 onto the center of the circle 2
the transformation has the following rule
(x,y)--------> (x+6,y-4)
so
(-4,5)------> (-4+6,5-4)-----> (2,1)
so
center circle 1 is now equal to center circle 2
The circles are now concentric (they have the same center)
step 2
A dilation is needed to increase the size of circle 1 to coincide with circle 2
scale factor=radius circle 2/radius circle 1-----> 6/2----> 3
radius circle 1 will be=2*scale factor-----> 2*3-----> 6 cm
radius circle 1 is now equal to radius circle
2
A
translation, followed by a dilation will map one circle onto
the other, thus proving that the circles are similar
the answer is
The circles are similar because you can translate circle 1 using the transformation rule ( x+6,y-4) and then dilate it using a scale factor of (3 )A work cell is scheduled to build 120 digital light processor (dlp) assemblies each week. these assemblies are later installed into home theater projection systems. the work cell has 7.5 hours of productive work each day, six days per week. what is takt time for this cell? express your answer in minutes rounded to one decimal place.
41 packages are randomly selected from packages received by a parcel service. the sample has a mean weight of 20.6 pounds and a standard deviation of 3.2 pounds. what is the best pint estimate for a confidence interval estimating the true mean weight, μ, of all packages received by the parcel service? answer: _____pounds
The best point estimate for the true mean weight of all packages received by a parcel service, given a sample mean of 20.6 pounds and a standard deviation of 3.2 pounds, is 20.6 pounds.
Explanation:In this problem, we are given that a sample of 41 packages were randomly selected. This sample provided a mean weight of 20.6 pounds and a standard deviation of 3.2 pounds. It is asked what the best point estimate for a confidence interval estimating the true mean weight of all packages is.
The best estimate of the true mean weight (µ) of all packages would be the mean of the sample that was taken, assuming that the sample was randomly selected and is representative of the entire population. This is due to the fact that in statistics, the sample mean is the best point estimate of the population mean.
Therefore, the best point estimate for the true mean weight of all the packages received by the parcel service is 20.6 pounds.
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I cant figure this out! please help.
solve for a.
3a+11.5