Answer:
f(t) = -16t²+64t
t = number of secondsf(t) = height (in feet)The x-coordinate(t) of the vertex can be found using formula [tex]\frac{-b}{2a}[/tex], where:
a = -16b = 64Therefore the t-value of the vertex is:
[tex]\frac{-64}{2(-16)} =\frac{-64}{-32} =2[/tex]
The baseball will reach its maximum height after two seconds.
A teacher calculates for the test grades in
Class A, mean = 32 and sd = 4
Class B, mean = 32 and sd = 8
a. If the teacher was going to guess what any student in his/her class would earn, what is the best score
to guess?
b. Which of the classes has more consistency in their scores? Why?
Answer:
a. best score to guess would be 33
b. Standard deviation simplifies the square root of the mean so makes it closer to 1 has more consistency as the mean of 32 when squared is sqrt 32 is Class A as class a = 4 and is closer to 5.65685425
as 5.65685425^2 = 32
Step-by-step explanation:
If you are comparing two normally-distributed variables on the same measurement scale then yes, you can regard the standard deviation as an indicator of how reliable the mean is--the smaller the standard deviation, the better able you are to "zero in" on the actual population mean.
a. proofs;
We find 32/6 = 5.333 and 32/5 = 6.4 and 6.4 is closer to both sd 4 and 8 than 5.33 is. As 6.4 it is closer to 6
But when we use 33/6 = 5.5 and therefore shows close range 6
therefore the two sd proves it is slightly high 32 score average for both classes A + B when joined and high 32 = 33 mean when classes A+B are joined or you could say 32/8 = 4 is class B becomes lower tests scores as 32/4 = 8 of class A that has higher test scores.
In this exercise we have to use probability and statistics to organize the students' grades, so we have:
A) best score is 33
B) Class A
In the first part of the exercise we have to analyze the grades of each class, like this:
A)Class A: 32/4
Class B: 32/8
Dividing each of them we have:
[tex]32/4=8 \\32/8=4[/tex]
B) With the information given above, we can say that the best class is A.
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Joe is four years older than Tim. Ten years ago, Joe was twice as old as Tim. Find their ages now?
Answer:
Joe: 18 years old
Tim: 14 years old
Given a set of data that is skewed-left, there is at least _____ % of the data within 2 standard deviations.
Answer:
75
Step-by-step explanation:
For non-normal distributions, we use Chebyshev's Theorem.
Chebyshev Theorem
The Chebyshev Theorem states that:
At least 75% of the measures are within 2 standard deviations of the mean.
At least 89% of the measures are within 3 standard deviations of the mean.
An in general terms, the percentage of measures within k standard deviations of the mean is given by [tex]100(1 - \frac{1}{k^{2}})[/tex].
In this question:
Within 2 standard deviations of the mean, so 75%.
An isosceles right triangle has a hypotenuse that measures 4√2 cm. What is the area of the triangle?
Answer:
It would be B. 16 centimeters^2
Step-by-step explanation:
Which of these are related functions. Plato
Answer:
◦•●◉these are related functions
Find the missing length on this triangle
Answer:
Step-by-step explanation:
This is a geometric means problem where 60, the side common to both the triangles, is the geometric mean. Set it up like this:
[tex]\frac{36}{60}=\frac{60}{x}[/tex] and cross multiply to get
36x = 3600 so
x = 100
Bill works for a large food service company. In one hour he can make 19 sandwiches or he can make 40 salads. Bill works 7 hours per day. If Bill needs to make 30 sandwiches then how many salads can he make
Answer:
[tex]x=216 salads[/tex]
Step-by-step explanation:
One Hour:
Salad=40
Sandwich=19
Total work time[tex]T=7[/tex]
Generally
Time to make 30 sandwiches is
[tex]T_s=\frac{30}{19}[/tex]
[tex]T-s=1.6hours[/tex]
Therefore
Bill has 7-1.6 hours to make salads and can make x about of salads in
[tex]x=(7-1.6)*40[/tex]
[tex]x=5.4*40[/tex]
[tex]x=216 salads[/tex]
Select the correct statement about what data scientists do during the Data Preparation stage.
a. During the Data Preparation stage, data scientists define the variables to be used in the model.
b. During the Data Preparation stage, data scientists determine the timing of events.
c. During the Data Preparation stage, data scientists aggregate the data and merge them from different sources.
d. During the Data Preparation stage, data scientists identify missing data.
e. All of the above statements are correct.
Answer:
e. All of the above statements are correct.
Option e is correct. All of the above statements are correct.
What is Data science?Data science is an interdisciplinary academic field that uses statistics, scientific computing, scientific methods, processes, algorithms and systems to extract or extrapolate knowledge and insights from noisy, structured and unstructured data
Data Scientist makes value out of data, he is expert in various tools and technologies like machine learning, deep learning, artificial intelligence and he solve business problems by presenting a model to predict business future.
During data preparation, data scientists and DBAs aggregate the data and merge them from different sources. During data preparation, data scientists and DBAs define the variables to be used in the model.
Hence, All of the above statements are correct, Option e is correct.
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g Let the joint probability density function of random variables X and Y. (a) Calculate the marginal probability densities of X and Y . (b) Calculate the expected values of X and Y . be given by
Answer: hello your question is incomplete attached below is the complete question
answer:
a) Fx(X) = 0 ≤ x ≤ 2, Fy(Y) = y - y^3/4
b) E(X) = 32/20 , E(Y) = 64/60
Step-by-step explanation:
a) Marginal probability density
Fx(X) = [tex]\int\limits^x_0 {\frac{xy}{2} } \, dy[/tex]
∴ probability density of X = 0 ≤ x ≤ 2
Fy(Y) = [tex]\int\limits^2_y {\frac{xy}{2} } \, dx[/tex]
∴ probability density of Y = y - y^3/4
b) Determine the expected values of X and Y
E(X) = 32/20
E(Y) = 64 /60
attached below is the detailed solution
A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 444 gram setting. It is believed that the machine is underfilling the bags. A 41 bag sample had a mean of 440 grams with a variance of 441. Assume the population is normally distributed. A level of significance of 0.05 will be used. Specify the type of hypothesis test.
Answer:
The null hypothesis is [tex]H_0: \mu = 444[/tex]
The alternative hypothesis is [tex]H_1: \mu < 444[/tex]
The p-value of the test is 0.1148 > 0.05, which means that the sample does not give enough evidence to conclude that the machine is underfilling the bags.
Step-by-step explanation:
A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 444 gram setting.
At the null hypothesis, we test if the machine works correctly, that is, the mean is of 444. So
[tex]H_0: \mu = 444[/tex]
At the alternative hypothesis, we test if they are underfilling, that is, if the mean is of less than 444. So
[tex]H_1: \mu < 444[/tex]
The test statistic is:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.
444 is tested at the null hypothesis:
This means that [tex]\mu = 444[/tex]
A 41 bag sample had a mean of 440 grams with a variance of 441.
This means that [tex]n = 41, X = 440, s = \sqrt{441} = 21[/tex]
Value of the test statistic:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{440 - 444}{\frac{21}{\sqrt{41}}}[/tex]
[tex]t = -1.22[/tex]
P-value of the test:
Right-tailed test(test if the mean is less than a value), with 41 - 1 = 40 df and t = -1.22.
Using a t-distribution calculator, this p-value is of 0.1148
The p-value of the test is 0.1148 > 0.05, which means that the sample does not give enough evidence to conclude that the machine is underfilling the bags.
Help I’ll mark you!!
Answer:
A.
Step-by-step explanation:
Each mark is worth two. We are inbetween the first mark and 0 on the left. Half of two is one. and since we are in the left quadrant we know it to be negative. Looking down, we see that we are exactly one mark down. As a mark is two, ans that we are going down, this will be a negative two. That leaves us with the answer of (-1, -2)
Answer:
A. (-1,-2)
Step-by-step explanation:
just trust me...I promise it right
According to records from a large public university, 88% of students who graduate from the university successfully find employment in their chosen field within three months of graduation. What is the probability that of nine randomly selected students who have graduated from this university, at least six of them find employment in their chosen field within three months
Answer:
0.9842 = 98.42% probability that at least six of them find employment in their chosen field within three months.
Step-by-step explanation:
For each student, there are only two possible outcomes. Either they found employment, or they did not. The probability of a student finding employment is independent of any other student, which means that the binomial probability distribution is used 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.
88% of students who graduate from the university successfully find employment in their chosen field within three months of graduation.
This means that [tex]p = 0.88[/tex]
Nine randomly selected students
This means that [tex]n = 9[/tex]
What is the probability that of nine randomly selected students who have graduated from this university, at least six of them find employment in their chosen field within three months?
This is:
[tex]P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 6) = C_{9,6}.(0.88)^{6}.(0.12)^{3} = 0.0674[/tex]
[tex]P(X = 7) = C_{9,7}.(0.88)^{7}.(0.12)^{2} = 0.2119[/tex]
[tex]P(X = 8) = C_{9,8}.(0.88)^{8}.(0.12)^{1} = 0.3884[/tex]
[tex]P(X = 9) = C_{9,9}.(0.88)^{9}.(0.12)^{0} = 0.3165[/tex]
Then
[tex]P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) = 0.0674 + 0.2119 + 0.3884 + 0.3165 = 0.9842[/tex]
0.9842 = 98.42% probability that at least six of them find employment in their chosen field within three months.
The director of research and development is testing a new medicine. She wants to know if there is evidence at the 0.02 level that the medicine relieves pain in more than 384 seconds. For a sample of 41 patients, the mean time in which the medicine relieved pain was 387 seconds. Assume the population standard deviation is 23. Find the P-value of the test statistic.
Answer:
The p-value of the test statistic is 0.2019.
Step-by-step explanation:
Test if there is evidence at the 0.02 level that the medicine relieves pain in more than 384 seconds.
At the null hypothesis, we test if it relieves pain in at most 384 seconds, that is:
[tex]H_0: \mu \leq 384[/tex]
At the alternative hypothesis, we test if it relieves pain in more than 384 seconds, that is:
[tex]H_1: \mu > 384[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
384 is tested at the null hypothesis:
This means that [tex]\mu = 384[/tex]
For a sample of 41 patients, the mean time in which the medicine relieved pain was 387 seconds. Assume the population standard deviation is 23.
This means that [tex]n = 41, X = 387, \sigma = 23[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{387 - 384}{\frac{23}{\sqrt{41}}}[/tex]
[tex]z = 0.835[/tex]
P-value of the test:
The p-value of the test is the probability of finding a sample mean above 387, which is 1 subtracted by the p-value of z = 0.835.
Looking at the z-table, z = 0.835 has a p-value of 0.7981.
1 - 0.7981 = 0.2019
The p-value of the test statistic is 0.2019.
Which of the following tables represent valid functions?
Answer:
Step-by-step explanation:
A relation may or may not represent a function.
Table (a), (c) and (d) represent a function
The tables represent a relation
For a relation to be a function, then:
The y values must have unique (or distinct) x-values.
From the list of tables, we have the following observations
All y values in table (a), have different corresponding x valuesy values 3 and 6 in table (b), point to the same x value (2)All y values in table (c), have different corresponding x valuesAll y values in table (d), have different corresponding x valuesHence, all the tables represent a valid function, except table (b)
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What is 2/11 as a decimal rounded to 3 decimal places?
Answer: The answer is 0.182
Hope this help :)
what is the ratio of the two values and what new value do they produce? $280 in 7m
what is the ratio of the two values and what new value do they produce? 105 miles in 2 hours
what is the ratio of the two values and what new value do they produce? $33 for 5lb
what is the ratio of the two values and what new value do they produce? 50 pages in 2 hours
Answer:
The ratio between two values A and B is just the quotient between these two values:
ratio = A/B
a) $280 in 7m
Here the ratio is:
$280/7m = $40/m
This also can be read as:
$40 per meter.
b) 105 miles in 2 hours
Here the ratio is:
105mi/2h = 52.5 mi/h
This also can be read as:
52.5 miles per hour
c) $33 for 5lb
The ratio is:
$33/5lb = $6.6/lb
This can be read as:
$6.6 per pound.
d) 50 pages in 2 hours
the ratio is:
(50 pages)/2h = 25 pages/h
this can be read as:
25 pages per hour.
A group of 120 students were surveyed about their interest in a new International Studies program. Interest was measured in terms of high, medium, or low. 30 students responded high interest; 50 students responded medium interest; 40 students responded low interest. What is the relative frequency of students with high interest? A. 30% B. 36.4% C. 25% D. Cannot be determined. Group of answer choices
Answer:
Option C (25%) is the correct answer.
Step-by-step explanation:
Given:
Number of students,
= 120
Students responded high interest,
= 30
Students responded medium interest,
= 50
Students responded low interest,
= 40
Now,
The relative frequency will be:
= [tex]\frac{30}{120}[/tex]
= [tex]0.25[/tex]
or,
= [tex]25[/tex]%
Solve for x:
|3x-1|=4
Answer:
x = 5/3 x= -1
Step-by-step explanation:
|3x-1|=4
There are two solutions to the absolute value equation, one positive and one negative
3x-1 =4 and 3x-1=-4
Add 1 to each side
3x-1+1 = 4+1 3x-1+1 = -4+1
3x=5 3x = -3
Divide by 3
3x/3 = 5/3 3x/3 = -3/3
x = 5/3 x= -1
Suppose 50.7 liters of water came out of a faucet today. If 2.6 liters of water come out each minute, for how many minutes was the faucet on?
Point A lies outside of plane P, how many lines can be drawn parallel to plane P that pass through point A?
A. 0
B. 1
C. 2
D. an infinite number
Answer:
B. an infinite number
Step-by-step explanation:
Since point A lies outside of P, the number of lines that can be drawn parallel to P and passing through point A is only infinite. It is infinite because it is just one given point lying outside the plane. If there is more than one point then it will be otherwise.
Answer:
yeah
Step-by-step explanation:
What is the value of y?
9514 1404 393
Answer:
(d) 2
Step-by-step explanation:
The parallel lines divide the transversals proportionally, so we have ...
3y/3 = 2y/y
y = 2 . . . . . . . . . simplify (assuming y ≠ 0)
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According to The Wedding Report, Inc., the mean cost for a wedding in the United States is $28732 (as of November 2008). Suppose the cost for a wedding is normally distributed with a standard deviation of $1500, and that a wedding is selected at random. Use the appropriate Excel function to calculate each of the following. (Note - Part (e) can be done by hand.)
(a) Find the probability that the wedding costs less than $22000.
(b) Find the probability that the wedding costs more than $32000.
(c) Find the probability that the wedding costs between $25000 and $30000.
(d) Find Q1 (the 25th percentile) and Q3 (the 75th percentile).
(e) Find the IQR for the wedding costs.
(f) The top 10% of weddings cost more than how much?
Answer:
Following are the solution to the given points:
Step-by-step explanation:
[tex]X = \text{cost of wedding}\sim \text{Normal}\ (\mu = 28732, \sigma= 1500)\\\\[/tex]
For point a:
[tex]Probability\ = 0.00000359\\\\ \text{(Using Excel function:} =NORMDIST(22000,28732,1500,1)).[/tex]
For point b:
[tex]Probability \ = 0.014678\\\\\text{(Using Excel function:} =1-NORMDIST(32000,28732,1500,1))\\\\[/tex]
For point c:
[tex]Probability\ = 0.794614436 \\\\[/tex]
[tex]\text{(Using Excel function:} \\=NORMDIST (30000,28732,1500,1)-NORMDIST(25000,28732,1500,1))\\\\[/tex]
For point d:
[tex]Q_1 = 27720.26537 \\\\\text{(Using Excel function:} =NORMINV(0.25,28732,1500)) \\\\Q_3 = 29743.73463 \\\\\text{(Using Excel function:} =NORMINV(0.75,28732,1500)).[/tex]
For point e:
[tex]IQR = Q_3 - Q_1 = 29743.73463 - 27720.26537 = 2023.469251.[/tex]
For point f:
[tex]Top\ 10\% = 30654.32735 \\\\\text{(Using Excel function:} =NORMINV(0.9,28732,1500)).[/tex]
A random sample of n1 = 296 voters registered in the state of California showed that 146 voted in the last general election. A random sample of n2 = 215 registered voters in the state of Colorado showed that 127 voted in the most recent general election. Do these data indicate that the population proportion of voter turnout in Colorado is higher than that in California? Use a 5% level of significance.
Answer:
The p-value of the test is 0.0139 < 0.05, which means that these data indicates that the population proportion of voter turnout in Colorado is higher than that in California.
Step-by-step explanation:
Before testing the hypothesis, we need to understand the central limit theorem and subtraction of normal variables.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
Subtraction between normal variables:
When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.
California:
Sample of 296 voters, 146 voted. This means that:
[tex]p_{Ca} = \frac{146}{296} = 0.4932[/tex]
[tex]s_{Ca} = \sqrt{\frac{0.4932*0.5068}{296}} = 0.0291[/tex]
Colorado:
Sample of 215 voters, 127 voted. This means that:
[tex]p_{Co} = \frac{127}{215} = 0.5907[/tex]
[tex]s_{Co} = \sqrt{\frac{0.5907*0.4093}{215}} = 0.0335[/tex]
Test if the population proportion of voter turnout in Colorado is higher than that in California:
At the null hypothesis, we test if it is not higher, that is, the subtraction of the proportions is at most 0. So
[tex]H_0: p_{Co} - p_{Ca} \leq 0[/tex]
At the alternative hypothesis, we test if it is higher, that is, the subtraction of the proportions is greater than 0. So
[tex]H_1: p_{Co} - p_{Ca} > 0[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{s}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, and s is the standard error.
0 is tested at the null hypothesis:
This means that [tex]\mu = 0[/tex]
From the two samples:
[tex]X = p_{Co} - p_{Ca} = 0.5907 - 0.4932 = 0.0975[/tex]
[tex]s = \sqrt{s_{Co}^2+s_{Ca}^2} = \sqrt{0.0291^2+0.0335^2} = 0.0444[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{s}[/tex]
[tex]z = \frac{0.0975 - 0}{0.0444}[/tex]
[tex]z = 2.2[/tex]
P-value of the test and decision:
The p-value of the test is the probability of finding a difference above 0.0975, which is 1 subtracted by the p-value of z = 2.2.
Looking at the z-table, z = 2.2 has a p-value of 0.9861.
1 - 0.9861 = 0.0139.
The p-value of the test is 0.0139 < 0.05, which means that these data indicates that the population proportion of voter turnout in Colorado is higher than that in California.
A game consists of tossing three coins. If all three coins land on heads, then the player wins $75. If all three coins land on tails, then the player wins $45. Otherwise, the player wins nothing. On average, how much should a player expect to win each game
Answer:
On average, a player should expect to win $15.
Step-by-step explanation:
The expected value in an event with outcomes:
x₁, x₂, ..., xₙ
Each with probability:
p₁, ..., pₙ
is given by:
Ev = x₁*p₁ + ... +xₙ*pₙ
In this case we have 3 outcomes:
player wins $75 = x₁
player wins $45 = x₂
player does not win = x₃
Let's find the probabilities of these events.
player wins $75)
Here we must have the 3 coins landing on heads, so there is only one possible outcome to win $75
While the total number of outcomes for tossing 3 coins, is the product between the number of outcomes for each individual event (where the individual events are tossing each individual coin, each one with 2 outcomes)
Then the number total of outcomes is:
C = 2*2*2 = 8
Then the probability of winning $75 is the quotient between the number of outcomes to win (only one) and the total number of outcomes (8)
p₁ = 1/8
Win $45:
This happens if the 3 coins land on tails, so is exactly equal to the case above, and the probability is the same:
p₂ = 1/8
Not wining:
Remember that:
p₁ + p₂ + ... + pₙ = 1
Then for this case, we must have:
p₁ + p₂ + p₃ = 1
1/8 + 1/8 + p₃ = 1
p₃ = 1 - 1/8 - 1/8
p₃ = 6/8
Then the expected value will be:
Ev = $75*1/8 + $45*1/8 + $0*6/8 = $15
On average, a player should expect to win $15.
An industrial psychologist consulting with a chain of music stores knows that the average number of complaints management receives each month throughout the industry is 4, but the variance is unknown. Nine of the chain's stores were randomly selected to record complaints for one month; they received 2, 4, 3, 5, 0, 2, 5, 1, and 5 complaints. Using the .05 significance level, is the number of complaints received by the chain different from the number of complaints received by music stores in general?
1. Use the five steps of hypothesis testing.
2. Sketch the distributions involved
3. Explain the logic of what you did to a person who is familiar with hypothesis testing, but knows nothing about t tests of any kind. Be sure to explain how this problem differs from a problem with a known population variance and a single sample.
Answer: See explanation
Step-by-step explanation:
1. Use the five steps of hypothesis testing.
Step 1: The aim of the research is to conduct the five steps of hypothesis testing.
Step 2:
Null hypothesis: H0 u= 4
Population mean: H1 u = 4
Alternate hypothesis: u ≠ 4
Population mean: u ≠ 4
Step 3 and step 4 are attached.
Step 5: Based on the calculation, the calculated value of t is less than the t critical value, therefore, the null hypothesis will be failed to be rejected.
2. Sketch the distributions involved
This has been attached.
3. Explain the logic of what you did to a person who is familiar with hypothesis testing, but knows nothing about t tests of any kind.
The distribution is "t".
The means is tested by using T-test.
Chi-square is used to test the single variance.
if the average of b and c is 8, and d=3b-4, what is the average of c and d in terms of b?
[tex] \underline{ \huge \mathcal{ Ànswér} } \huge: - [/tex]
Average of b and c is 8, that is
[tex]➢ \: \: \dfrac{b + c}{2} = 8[/tex]
[tex]➢ \: \: b + c = 16[/tex]
[tex]➢ \: \: c = 16 - b[/tex]
now let's solve for average of c and d :
[tex]➢ \: \: \dfrac{c + d}{2} [/tex]
[tex]➢ \: \: \dfrac{16 - b + 3b - 4}{2} [/tex]
[tex]➢ \: \: \dfrac{12 + 2b}{2} [/tex]
[tex]➢ \: \: \dfrac{2(6 + b)}{2} [/tex]
[tex]➢ \: \: b + 6[/tex]
Therefore, the average of c and d, in terms of b is : -
[tex] \large \boxed{ \boxed{b + 6}}[/tex]
[tex]\mathrm{✌TeeNForeveR✌}[/tex]
Answer:
b+6
Problem:
If the average of b and c is 8, and d=3b-4, what is the average of c and d in terms of b?
Step-by-step explanation:
We are given (b+c)/2=8 and d=3b-4.
We are asked to find (c+d)/2 in terms of variable, b.
We need to first solve (b+c)/2=8 for c.
Multiply both sides by 2: b+c=16.
Subtract b on both sides: c=16-b
Now let's plug in c=16-b and d=3b-4 into (c+d)/2:
([16-b]+[3b-4])/2
Combine like terms:
(12+2b)/2
Divide top and bottom by 2:
(6+1b)/1
Multiplicative identity property applied:
(6+b)/1
Anything divided by 1 is that anything:
(6+b)
6+b
b+6
Draw clearly the graph of the linear equation. y=1/2x, where x= (-4 -2, 0, 2, 4)
Answer:
(in attachment)
Step-by-step explanation:
you can find the points by inputting the x-values into the equation to solve for the y-values, then connecting the plotted points to create the line.
When x=-4
y=1/2(-4)
y=-2
(-4,-2)
Repeat for all values.
add:7ab,8ab,-10ab,-3ab
Answer:
2ab
Step-by-step explanation:
7ab+8ab+(-10ab)+(-3ab)=
=15ab-13ab= 2ab
Answer:
2ab
Step-by-step explanation:
7ab+8ab+-10ab+-3ab
Factor out ab
ab(7+8-10-3)
ab(2)
2ab
What is the area of the given triangle? Round to the nearest tenth
Answer:
28.0125 cm^2 rounded to 28.0 cm^2
Step-by-step explanation:
Area = a*b*sin(c)*1/2
Area = 7 * 13 * sin(38) * 1/2
Area = 91/2 * 0.61566...
Area = 28.0125...
