Final answer:
To factor a composite number into primes, divide it by its smallest divisor that is not 1, then continue dividing the quotient until all factors are prime. An example is the number 60, which factors into 2 x 2 x 3 x 5, or 2² x 3 x 5 when using exponents for the repeated factor of 2.
Explanation:
The subject of the question is to select a composite number and break down its factors until all the factors are prime. As an example, let's choose the composite number 60. Here's how you can factor it into primes:
First, note that 60 is an even number, so it is divisible by 2. Start by dividing 60 by 2 to get 30.
Now, 30 is still even, so we can divide by 2 again to get 15.
15 is divisible by 3, so when we divide it by 3, we get 5, which is a prime number.
So, the prime factorization of 60 is 2 x 2 x 3 x 5, often written using exponents for any repeated factors as 22 x 3 x 5.
Through factorization, we have converted the composite number into a product of prime factors. Each factor multiplication is a step that requires one to find two numbers that multiply to the number we are factoring and continuing this process until we reach numbers that are prime.
find the zero of picture is included
Answer:
x = -12/6 = -2
x = -18/6 = -3
Step-by-step explanation:
The zeros in an equation is the same thing as the x's.
3x²+15x+18 = 0 (Where a = 3, b = 15, c = 18)
One thing you could do to find the answer is use the quadratic formula:
x = (-b ± √(b²-4ac))/2a =
(-15 ± √15² - (4*3*18))/2*3 =
(-15 ± √225 - 216)/6 =
(-15 ± √9)/6 =
(-15 ± 3) / 6.
To find the first x we need to do (-15 + 3) / 6 and to find the second x we need to do (-15-3) / 6.
x = -12/6 = -2
x = -18/6 = -3
To find the zeros of the quadratic equation 3x^2 + 15x + 18 = 0, use the quadratic formula to obtain the values of x.
To find the zeros of the quadratic equation 3x^2 + 15x + 18 = 0, we can use the quadratic formula.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
Plugging in the values from our equation, we get:
x = (-15 ± sqrt(15^2 - 4*3*18)) / (2*3)
After evaluating the expression, we find that the zeros of the equation are x = -3 and x = -2.
The probable question may be:
find the zero of 3x^2+15x+18=0 using any method.
The function 3(2)t models the number of leaves on a plant, where t represents the number of weeks since it was planted. Which statement is the best interpretation of one of the values in this function? A) The number of leaves increases by 6 each week. B) The initial number of leaves on the plant was 2. C) The initial number of leaves on the plant was 3. D) The number of leaves on the plant increases by 3% each week.
Answer:
C) The initial number of leaves on the plant was 3.
Step-by-step explanation:
The plant has initially 3 leaves and is multiplied by 2 every week. Therefore, the answer is C.
A baseball is hit, and rises to a height above the ground, f(t), in feet, at time t, in seconds, can be estimated by the formula: f(t)=-10t^(2)+40t+8
Answer:
The maximum height reached by the baseball is 48 feet.Step-by-step explanation:
The given function is
[tex]f(t)=-10t^{2}+40t+8[/tex]
To find the maximum height reached by the ball, we need to find the vertex of the parabolla represented by the given quadratic equation.
Remember that a vertex has coordinates (h,k), where [tex]h=-\frac{b}{2a}[/tex] and [tex]k=f(h)[/tex].
According to the given function, we have
[tex]a=-10[/tex] and [tex]b=40[/tex]
[tex]h=-\frac{40}{2(-10)}=\frac{40}{20}=2[/tex]
[tex]k=f(2)=-10(2)^{2} +40(2)+8=-10(4)+80+8=-40+80+8=48[/tex]
So, the vertex has coordinates of (2,48), which means the maximum height reached by the baseball is 48 feet.
austin swam 1.4 laps each minute. Ashley is Austin's older sister. she trains in the same pool and can swim 9 laps in 6 minutes. is ashley a faster swimmer than Austain?
Answer:
Since Ashley swam more laps per minute, she's a faster swimmer than Austin.
Step-by-step explanation:
Austin:
She swam 1.4 laps per minute
Ashley:
9 laps in 6 minutes
So, per minute
9/6 = 1.5 laps per minute
Since Ashley swam more laps per minute, she's a faster swimmer than Austin.
Answer:
t
Step-by-step explanation:
Daniel believes that people perform better in the barbell curl, on average, if they are encouraged by a coach. He recruited 29 subjects to participate in an experiment and randomly assigned them into two groups. Daniel gave one group verbal encouragement during the exercise and was quiet the exercise for the other group. He recorded the total number of barbell curls each subject was able to complete before setting the bar down.
(a) Explain the purpose of random assignment in this experiment.
(b) Compare these distributions,
(c) State the hypotheses Daniel should use to test his belief about receiving encouragement during exercise. Make sure to define any parameters you use.
(d) Identify the significance test Daniel should use to analyze the results of his experiment and show that the conditions for this test are met.
(e) The P-value for Daniel's test is 0.107. What conclusion should Daniel make at the a=0.05 significance level?
Answer:
a) to remove bias in the study
b) The number of barbell curls performed by group who was encourged was higher than the group who was not encouraged
c) H0: mean of barbell curls for group encouraged= mean of barbell curls for group not encouraged
Ha: mean of barbell curls for group encouraged≠ mean of barbell curls for group not encouraged
d) two sample t-test.
Requirement has been met as there are two separate groups whose means have to be compared. Median of box plot is equal to mean. Standard Deviation is equal to range divided by four. Sample size will have to be assumed
e) Encouragement affects the number of barbell curl performed by an individua
Step-by-step explanation:
a) random assignemnt removes bias in any study.
b) box plot is attached
c) null hypthesis and alternate hypothesis would compare means of the two distributions
d) As we are comparing two groups, we will use two sample t-test.
The median of box plot is equal to mean for both groups.
The rnge/4 is equal to standard deviation.
mean of encouraged group= 30
mean of not encoouraged group=19
SD of encouraged group= (75-5)/4= 15
SD of not encouraged group= (60-0)/4= 15
Assume number of inidivudals in encouraged group =15
assume number of individuals in not encouraged group = 14
e) test statistic= (mean1-mean2)/√(SD1²/(N1-1)+ SD2²/(N2-1))
= (30-19)/(√(15²/14+ 15²/13)
= 1.9734
Since test statistic is greater than p-value, null hypothesis is rejected.
Encouragement affects the number of barbell curl performed by an individual
Email marketing, the online counterpart of direct mail marketing, can be a highly personalized, very targeted means of communication. With its very low cost and wide reach, you feel that email marketing must play a major role in Caffè Gustoso's online advertising campaign.
The owners want to send as many emails as possible to potential customers. However before you send out the first email, you have to choose an email permission strategy.
Select an option from the choices below:
a. Opt-in: Request the consumer's permission before sending any Caffé Gustoso email correspondence.
b. Unsolicited: Send unsolicited Caffe Gustoso email to a large number of potential customer.
c. Opt-out: Allow consumers to quest to be removed from have been contacted
Answer:
c. Opt-out: Allow consumers to quest to be removed from have been contacted.
Step-by-step explanation:
The Caffe Gutoso is considering a marketing strategy to promote its business. The owner want to send emails to its potential customers. Before sending the email the sender must insert email permission strategy as send an email from Caffe Gutoso and place and Opt-out option for those customers who are not willing to receive any email from the Caffe.
mean of 153 176 249 150 620 242
Answer: 265
Step-by-step explanation:
Add up all the numbers and divide this value by the total number of terms,in this case 6 terms.
(153+ 176+ 249+ 150+ 620+ 242)/6 =265
Find the value of y+5 given that -3y-3=6
Simplify
A particular fruit's weights are normally distributed, with a mean of 426 grams and a standard deviation of 37 grams. If you pick 9 fruit at random, what is the probability that their mean weight will be between 413 grams and 464 grams. Round to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 2 decimal places are accepted.
Answer:
Probability that their mean weight will be between 413 grams and 464 grams is 0.8521.
Step-by-step explanation:
We are given that a particular fruit's weights are normally distributed, with a mean of 426 grams and a standard deviation of 37 grams.
Also, you pick 9 fruit at random.
Let [tex]\bar X[/tex] = sample mean weight
The z-score probability distribution for sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean weight = 426 grams
[tex]\sigma[/tex] = population standard deviation = 37 grams
n = sample of fruits = 9
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.
So, probability that the mean weight of 9 fruits picked at random will be between 413 grams and 464 grams is given by = P(413 grams < [tex]\bar X[/tex] < 464 grams) = P([tex]\bar X[/tex] < 464 grams) - P([tex]\bar X[/tex] [tex]\leq[/tex] 413 grams)
P([tex]\bar X[/tex] < 464 grams) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{464-426}{\frac{37}{\sqrt{9} } }[/tex] ) = P(Z < 3.08) = 0.99896
P([tex]\bar X[/tex] [tex]\leq[/tex] 413 grams) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{413-426}{\frac{37}{\sqrt{9} } }[/tex] ) = P(Z [tex]\leq[/tex] -1.05) = 1 - P(Z < 1.05)
= 1 - 0.85314 = 0.14686
{Now, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 3.08 and x = 1.05 in the z table which has an area of 0.99896 and 0.85314 respectively.}
Therefore, P(413 grams < [tex]\bar X[/tex] < 464 grams) = 0.99896 - 0.14686 = 0.8521
Hence, the probability that their mean weight will be between 413 grams and 464 grams is 0.8521.
The typical lifespan for various mammal species in captivity (L) in years has been related to averageadult size (M) in kilograms according to the regression equation seen below. A typical adult Meerkat weighs about 0.9 kilograms. What is the predicted lifespan of a Meerkat incaptivity, according to this equation? * 2 points 11.6 years 2.4 years 14.1 years 2.6 years 28.8 years
Complete Question:
The typical lifespan for various mammal species in captivity (L) in years has been related to average adult size (M) in kilograms according to the regression equation seen below.
[tex]ln L = 2.468 + 0.2 (lnM)[/tex]
A typical adult Meerkat weighs about 0.9 kilograms. What is the predicted lifespan of a Meerkat incaptivity, according to this equation? * 2 points 11.6 years 2.4 years 14.1 years 2.6 years 28.8 years
Answer:
Option A) L = 11.6 years
Step-by-step explanation:
From the given equation:
[tex]ln L = 2.468 + 0.2 (lnM)[/tex]..........(1)
Average adult size, M = 0.9 kg
Putting the value of M into the regression equation in (1)
[tex]ln L = 2.468 + 0.2 (ln0.9)\\ln L = 2.468 + (-0.02107)\\ln L = 2.447\\L = e^{2.447} \\L = 11.55 years[/tex]
The question is an illustration of regressions.
The lifespan of a typical adult Meerkat that weighs about 0.9 kilograms is (a) 11.6 years
The regression equation is given as:
[tex]\ln(L) = 2.468 + 0.2\ln(M)[/tex]
From the question, we have:
M = 0.9
Substitute 0.9 for M in [tex]\ln(L) = 2.468 + 0.2\ln(M)[/tex]
[tex]\ln(L) = 2.468 + 0.2\ln(0.9)[/tex]
Take natural logarithm of 0.9
[tex]\ln(L) = 2.468 + 0.2\times -0.1054[/tex]
[tex]\ln(L) = 2.468 -0.02108[/tex]
[tex]\ln(L) = 2.44692[/tex]
Take exponents of both sides
[tex]L = e^{2.44692}[/tex]
[tex]L = 11.553[/tex]
Approximate
[tex]L = 11.6[/tex]
Hence, the lifespan is (a) 11.6 years
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Solve (x - 5)^2 = 3
Answer:
c
Step-by-step explanation:
√(x-5)²=±√3
x-5=±√3
x=5±√3
To solve the quadratic equation (x - 5)² = 3, take the square root of both sides to give x - 5 = ± √3, then add 5 to each side to give x = 5 ± √3. The two possible solutions are x = 5 + √3 and x = 5 - √3.
Explanation:The question is asking to solve the equation (x - 5)² = 3. This is a quadratic equation in the form (x-a)² = b. To solve it, you should start by taking the square root on both sides.
Step 1: √[(x - 5)²] = √[3]
This leads to: x - 5 = ± √3
Step 2: Followed by adding 5 to both sides
x = 5 ± √3
So there are two possible solutions: x = 5 + √3 and x = 5 - √3.
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Which theorem proves that angle EHF = Angle EGF?
A. The measure of an inscribed angle is half the measure of its intercepted arc.
B. Two inscribed angles that intercept the same arc are congruent.
C. An angle inscribed in a semicircle is a right angle.
D. The opposite angles of a quadrilateral inscribed in a circle are supplementary
Answer:B) Two inscribed angles that intercept the same arc are congruent.
Step-by-step explanation:
I just took the test
The credit remaining on a phone card (in dollars) is a linear function of the total calling time made with the card (in minutes). The remaining credit after 31 minutes of calls is $21.28, and the remaining credit after 69 minutes of calls is $16.72. What is the remaining credit after 76 minutes of calls?
Answer:
$15.88
Step-by-step explanation:
[tex]\frac{(21.28 - 16.72)}{(69-31)}[/tex]
$4.56 / 38 min
$0.12 / 1 min
76 - 69 = 7 minutes
7 x .12 = .84
16.72 - .84 = 15.88
Moose Drool Makes Grass More Appetizing Different species can interact in interesting ways. One type of grass produces the toxin ergovaline at levels about 1.0 part per million in order to keep grazing animals away. However, a recent study27 has found that the saliva from a moose counteracts these toxins and makes the grass more appetizing (for the moose). Scientists estimate that, after treatment with moose drool, mean level of the toxin ergovaline (in ppm) on the grass is 0.183. The standard error for this estimate is 0.016.
Give notation for the quantity being estimated, and define any parameters used.
Give notation for the quantity that gives the best estimate, and give its value.
Give a 95% confidence interval for the quantity being estimated. Interpret the interval in context.
3.68 Bisphenol A in Your Soup Cans Bisphenol A (BPA) is in the lining of most canned goods, and recent studies have shown a positive association between BPA exposure and behavior and health problems. How much does canned soup consumption increase urinary BPA concentration? That was the question addressed in a recent study34 in which consumption of canned soup over five days was associated with a more than 1000% increase in urinary BPA. In the study, 75 participants ate either canned soup or fresh soup for lunch for five days. On the fifth day, urinary BPA levels were measured. After a two-day break, the participants switched groups and repeated the process. The difference in BPA levels between the two treatments was measured for each participant. The study reports that a 95% confidence interval for the difference in means (canned minus fresh) is 19.6 to 25.5 μg/L.
Is this a randomized comparative experiment or a matched pairs experiment? Why might this type of experiment have been used?
What parameter are we estimating?
Interpret the confidence interval in terms of BPA concentrations.
If the study had included 500 participants instead of 75, would you expect the confidence interval to be wider or narrower?
3.70 Effect of Overeating for One Month: Average Long-Term Weight Gain Overeating for just four weeks can increase fat mass and weight over two years later, a Swedish study shows.35 Researchers recruited 18 healthy and normal-weight people with an average age of 26. For a four-week period, participants increased calorie intake by 70% (mostly by eating fast food) and limited daily activity to a maximum of 5000 steps per day (considered sedentary). Not surprisingly, weight and body fat of the participants went up significantly during the study and then decreased after the study ended. Participants are believed to have returned to the diet and lifestyle they had before the experiment. However, two and a half years after the experiment, the mean weight gain for participants was 6.8 lbs with a standard error of 1.2 lbs. A control group that did not binge had no change in weight.
What is the relevant parameter?
How could we find the actual exact value of the parameter?
Give a 95% confidence interval for the parameter and interpret it.
Give the margin of error and interpret it.
In the first study, moose drool reduces the toxin levels on grass. In the second study, canned soup consumption increases urinary BPA concentrations. In the third study, overeating for four weeks leads to long-term weight gain.
Explanation:Question 1:
The notation for the quantity being estimated is the mean level of the toxin ergovaline on the grass (in ppm). Scientists estimate the mean level after treatment with moose drool to be 0.183 ppm.
The quantity that gives the best estimate is the mean level of the toxin ergovaline on the grass after treatment with moose drool (0.183 ppm).
A 95% confidence interval for the quantity being estimated is (0.183 - 1.96 * 0.016, 0.183 + 1.96 * 0.016), which is approximately (0.152, 0.214) ppm. This means there is a 95% chance that the true mean level of the toxin ergovaline on the grass after treatment with moose drool is between 0.152 and 0.214 ppm.
Question 2:
This is a matched pairs experiment because each participant is subjected to both treatments (canned soup and fresh soup). It was used to eliminate individual differences and control for variables that might affect the urinary BPA concentrations.
The parameter being estimated is the difference in means (canned minus fresh) of urinary BPA concentrations.
The 95% confidence interval for the difference in means is (19.6, 25.5) μg/L. This means that we are 95% confident that the true difference in mean urinary BPA concentrations between canned soup and fresh soup is between 19.6 and 25.5 μg/L.
If the study had included 500 participants instead of 75, we would expect the confidence interval to be narrower because a larger sample size reduces the standard error, leading to a more precise estimate.
Question 3:
The relevant parameter is the mean weight gain for participants two and a half years after the experiment.
The actual exact value of the parameter cannot be found unless we measure the weight gain for all individuals in the population.
A 95% confidence interval for the parameter is (6.8 - 1.96 * 1.2, 6.8 + 1.96 * 1.2), which is approximately (4.44, 9.16) lbs. This means there is a 95% chance that the true mean weight gain for participants two and a half years after the experiment is between 4.44 and 9.16 lbs.
The margin of error is 1.96 * 1.2 lbs, which is approximately 2.35 lbs. This means that the estimated mean weight gain may vary by up to 2.35 lbs from the true mean weight gain.
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What is the Simplify expression of (15+39)÷6
Answer:
9
Step-by-step explanation:
[tex] \frac{(15 + 39)}{6} \\ \frac{54}{6 } \\ ans = 9 \\ [/tex]
What is the value and is it minimum or max
Answer:
minimum and value of 4
Step-by-step explanation:
[tex]\frac{dx}{dy} = 2x-8[/tex]
When [tex]\frac{dx}{dy}=0[/tex] we will be able to get the critical points.
[tex]2x-8 =0[/tex][tex]x=4[/tex]
The graph is a quadratic graph with a 'U' shape, thus it has a minimum critical point.
Find the common difference for the sequence shown.
56, 49, 42, 35, ...
-7
-12
7
Answer:
-7
Step-by-step explanation:
Given : sequence 56, 49, 42, 35, ...
To find : Find the common difference.
Solution : We have given that 56, 49, 42, 35, ...
By the common difference of an arithmetic sequence is:
d = a_{2} -a_{1}a2−a1
Where, a_{1}a1 is the first term and a_{1}a1 is the second term.
Then common difference = 49 -56
= -7 .
find the value of p+7 when p=19
Answer:
26
Step-by-step explanation:
p = 19
p + 7 = 19 + 7
= 26
So, the value of p + 7 is 26
Hope it helps and is useful :)
The slope of a pipe with a 1/4 inch of drop has a run of 1 foot. What is the run of a pipe that has a 3/4 inch drop?
Answer: 3 foot
Step-by-step explanation:
evaluate the expression when y=-6
y squared+7y+4
Answer:
In the equation of a straight line (when the equation is written as "y = mx + b"), the slope is the number "m" that is multiplied on the x, and "b" is the y-intercept (that is, the point where the line crosses the vertical y-axis). This useful form of the line equation is sensibly named the "slope-intercept form".
Step-by-step explanation:
This is just something to keep in mind
Answer:
-2
Step-by-step explanation:
Well, lt me hope the expression is complete amd correct.
Let's go on with the information you submitted.
y² + 7y + 4
When y= -6.
Let's substitute the value, y= -6 in the expression. We'll have to be careful with the signs.
y² + 7y + 4
-6² + 7(-6) + 4
36 - 42 + 4
40 - 42 =
-2
As a result of complaints by the staff about noise, the coffee and recreation area for student interns at OHaganBooks will now be in a 512 square foot rectangular area in the headquarter's basement against the southern wall. (The specified area was arrived at in complex negotiations between the student intern representative and management.) The construction of the partition will cost $16 per foot for the north wall and $4 per foot for the east and west walls. What are the dimensions of the cheapest recreation area that can be made?
Answer:
The dimensions are 32 ft by 16 ft
Step-by-step explanation:
Area of the coffee and recreation room=512 square foot
LB=512
L=512/B
Perimeter of the Room = Perimeter of north wall+Perimeter of east wall+ Perimeter of west wall =L+2B (West and East are opposite)
Cost of the Perimeter=16L+4(2B)
[tex]C(B)=16(\frac{512}{B})+8B\\C(B)=\frac{8192+8B^2}{B}[/tex]
To minimize cost, first, we take the derivative of C(B)
Using quotient rule
[tex]C^{'}(B)=\frac{8B^2-8192}{B^2}[/tex]
Setting the derivative equal to zero
[tex]8B^2-8192=0\\8B^2=8192\\B^2=1024\\B=32 ft[/tex]
[tex]L=\frac{512}{B}=\frac{512}{32}=16ft[/tex]
The dimension of the cheapest recreation area will be 32 ft by 16 ft
Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probabilities that the number of inquiries in a particular 50-millisecond stretch will be:
Answer: a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318
Step-by-step explanation:
Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probabilities that the number of inquiries in a particular 50-millisecond stretch will be:
Since we have given that
[tex]\lambda=0.1\ per\ millisecond=5\ per\ 50\ millisecond=5[/tex]
Using the poisson process, we get that
(a) less than or equal to 12
probability= [tex]P(X\leq 12)=\sum _{k=0}^{12}\dfrac{e^{-5}(-5)^k}{k!}=0.9980[/tex]
(b) equal to 13
probability= [tex]P(X=13)=\dfrac{e^{-5}(-5)^{13}}{13!}=0.0013[/tex]
(c) greater than 12
probability= [tex]P(X>12)=\sum _{k=13}^{50}\dfrac{e^{-5}.(-5)^k}{k!}=0.0020[/tex]
(d) equal to 20
probability= [tex]P(X=20)=\dfrac{e^{-5}(-5)^{20}}{20!}=0.00000026[/tex]
(e) between 10 and 15, inclusively
probability=[tex]P(10\leq X\leq 15)=\sum _{k=10}^{15}\dfrac{e^{-5}(-5)^k}{k!}=0.0318[/tex]
Hence, a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318
Using the Poisson distribution, the probability of x inquires in a particular 50-millisecond stretch will be:
[tex]P(X = x) = \frac{e^{-5}5^{x}}{(x)!}[/tex]
In this problem, we are given the mean during an interval, which means that the Poisson distribution is used.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
x is the number of successes
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
We are given an average rate of 0.1 inquires per millisecond.
50-millisecond interval, thus [tex]\mu = 50(0.1) = 5[/tex]
Then, the probability of x inquires in a particular 50-millisecond stretch will be:
[tex]P(X = x) = \frac{e^{-5}5^{x}}{(x)!}[/tex]
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basketball players score 19 times in one game he scored a total of 33.2 of each point to point shot in one of each free-throw how many to point shots did he make how many free throws
The answer is 17 free throws because I said so and that’s the correct answer
Answer:
my answer was gunna be his but that person answered first so ill just sound like i copied that person
Step-by-step explanation:
What is the answer?
A. Infinite Number of Solutions
B. (3, 3)
C. (3, -3)
D. (-3, 3)
Answer:
It's either B or A
Step-by-step explanation:
If I had to choose one I would pick B because I did the math; and logically, that answer makes the most sense.
If 3ab = c, then solve for a
To solve for an in the equation 3ab = c, divide both sides by 3b, yielding a = c / (3b), with the condition that b is not zero.
Explanation:To solve the equation 3ab = c for a, we need to isolate an on one side of the equation. Starting with the original equation:
3ab = c
We divide both sides of the equation by 3b to get:
a = c / (3b)
This is the solution for an in terms of b and c. It's important to note that b cannot be zero, as division by zero is undefined.
10/9 = 3/q
solve for q
Answer:
Q= 27/10
Step-by-step explanation:
Answer: [tex]q=\frac{27}{10}[/tex]
Step-by-step explanation:
[tex]\frac{10}{9}=\frac{3}{q}[/tex]
Multiply by q.
[tex]q*\frac{10}{9}=\frac{3}{q}*q[/tex]
[tex]\frac{10}{9}q=3[/tex]
Now multiply by the reciprocal of [tex]\frac{10}{9}[/tex] which is the inverted fraction. [tex]\frac{9}{10}[/tex]
[tex]\frac{9}{10}*\frac{10}{9}q=3*\frac{9}{10}[/tex]
[tex]q=\frac{27}{10}[/tex]
Find the sum of 526+213 by breaking apart the addenda into hundreds
What is the solution to the equation One-fourth x minus one-eighth = StartFraction 7 Over 8 EndFraction + one-half x?
Answer:
The solution to the equation is
x = -4
Step-by-step explanation:
We want to find the solution to the equation
(1/4)x - 1/8 = 7/8 + (1/2)x
First, add -(1/2)x + 1/8 to both sides of the equation.
(1/4)x - 1/8 - (1/2)x + 1/8 = 7/8 + (1/2)x - (1/2)x + 1/8
[1/4 - 1/2]x = 7/8 + 1/8
x(1 - 2)/4 = (7 + 1)/8
-(1/4)x = 8/8
-(1/4)x = 1
Multiply both sides by -4
x = -4
Answer:
x=-4
Step-by-step explanation:
Mr. Gordon has 13 girls and 14 boys in his fourth period algebra class. One person is chosen at random.
What is the probability that the person chosen is a boy?
A. 1/14
B. 1/27
C. 13/14
D. 13/27
E. 14/27
Answer:
E
Step-by-step explanation:
It's E because you have to add 13 and 14 and as you can see there's 14 boys so it's E.
The probability that the person chosen is a boy is 14/27.
Number of girls = 13
Number of boys = 14
Total students = 27
What is probability?Probability is a measure of the likelihood of occurrence of an event.
P(event) = Favorable outcomes / total outcomes
Favourable outcomes = 14 because there are 14 boys
Total outcomes =27
So, the probability that the person chosen is a boy will be given by:
P(boy)=14/27
So, the probability that the person chosen is a boy is 14/27.
Hence, the probability that the person chosen is a boy is 14/27.
To get more about probability visit:
https://brainly.com/question/24756209
To get an estimate of consumer spending over a holiday season in 2009, 436 randomly sampled American adults were surveyed. Their spending for the six-day period after Thanksgiving, spanning the Black Friday weekend and Cyber Monday, averaged $84.71. A 95% confidence interval based on this sample is ($80.31, $89.11). Which of the following statements are true? Select all that apply. A 90% confidence interval would be narrower than the 95% confidence interval if we don't need to be as sure about our estimate. In order to decrease the margin of error of a 95% confidence interval to a third of what is is now, we would need to use a sample 3 times larger. This confidence interval is not valid since the distribution of spending in the sample data is right skewed. The margin of error is $4.4. We are 95% confident that the average spending of these 435 American adults over this holiday season is between $80.31 and $89.11. This confidence interval is valid since the sampling distribution of sample mean would be approximately normal with sample size of 436. 95% of random samples have a sample mean between $80.31 and $89.11. We are 95% confident that the average spending of all American adults over this holiday season is between $80.31 and $89.11.
Answer:
-A 90% confidence interval would be narrower than the 95% confidence interval if we don't need to be as sure about our estimate.
-This confidence interval is not valid since the distribution of spending in the sample data is right skewed.
-The margin of error is $4.4.
-This confidence interval is valid since the sampling distribution of sample mean would be approximately normal with sample size of 436.
-We are 95% confident that the average spending of all American adults over this holiday season is between $80.31 and $89.11.
Step-by-step explanation:
A 90% confidence interval would be narrower than the 95% confidence interval if we don't need to be as sure about our estimate.
TRUE. The 90% confidence is less strict in its probability of having the mean within the interval, so it is narrower than the 95% CI. It relies more in the information given by the sample.
In order to decrease the margin of error of a 95% confidence interval to a third of what is is now, we would need to use a sample 3 times larger.
FALSE. The margin of error is z*σ/(n^0.5). So to reduce it by two thirds, the sample size n needs to be 3^2=9 times larger.
This confidence interval is not valid since the distribution of spending in the sample data is right skewed.
FALSE. There is no information about the skewness in the sample.
The margin of error is $4.4.
TRUE. The margin of error is (89.11-80.31)/2=$4.4.
We are 95% confident that the average spending of these 435 American adults over this holiday season is between $80.31 and $89.11.
FALSE. The CI is related to the populations mean. We are 95% confident that the average spending of the population is between $80.31 and $89.11.
This confidence interval is valid since the sampling distribution of sample mean would be approximately normal with sample size of 436.
TRUE. This happens accordingly to the Central Limit Theorem.
95% of random samples have a sample mean between $80.31 and $89.11.
FALSE. The confidence interval refers to the population mean.
We are 95% confident that the average spending of all American adults over this holiday season is between $80.31 and $89.11.
TRUE. This is the conclusion that is looked for when constructing a confidence interval.