Answer:
60%
Step-by-step explanation:
Percent means out of 100
Changing 3/5 to a denominator of 100
3/5*20/20
60/100
The percent is 60 %
Urban Community College is planning to offer courses in Finite Math, Applied Calculus, and Computer Methods. Each section of Finite Math has 40 students and earns the college $40,000 in revenue. Each section of Applied Calculus has 40 students and earns the college $60,000, while each section of Computer Methods has 10 students and earns the college $26,000. Assuming the college wishes to offer a total of seven sections, accommodate 220 students, and bring in $292,000 in revenues, how many sections of each course should it offer?
Finite Math section(s)
Applied Calculus section(s)
Computer Methods section(s)
Answer:
meh
Step-by-step explanation:
What’s the correct answer for this question?
Answer:
B.
Step-by-step explanation:
A cone with a radius of 1-unit will be obtained by rotating the 3-D figure.
3. Bob the Builder wants to earn an annual rate of 10% on his investments,
how much (to the
nearest cent) should he pay for a note that will be worth $3,000 in 9 months?
Answer:
He should pay $2,790.7.
Step-by-step explanation:
This is a simple interest problem.
The simple interest formula is given by:
[tex]E = P*I*t[/tex]
In which E is the amount of interest earned, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time, in years.
After t years, the total amount of money is:
[tex]T = E + P[/tex]
In this question:
Rate of 10%, so I = 0.1.
9 months, so [tex]t = \frac{9}{12} = 0.75[/tex]
How much should he pay for a note that will be worth $3,000 in 9 months?
We have to find P for which T = 3000. So
[tex]T = E + P[/tex]
[tex]3000 = E + P[/tex]
[tex]E = 3000 - P[/tex]
Then
[tex]E = P*I*t[/tex]
[tex]3000 - P = P*0.1*0.75[/tex]
[tex]1.075P = 3000[/tex]
[tex]P = \frac{3000}{1.075}[/tex]
[tex]P = 2790.7[/tex]
He should pay $2,790.7.
Tony rode his bicycle 3 7/10 miles to school. What is this distance written as a decimal?
Answer:
the distance written as a decimal is 3.7
37/10 = 3,7
Achievements :)
The accompanying data are the times to failure (in millions per cycle) of high-speed turbine engine bearings made out of two different compounds. These were taken from "Analysis of Single Classification Experiments Based on Censored Samples from the Two-parameter Weibull Distribution" by J.I. McCool (The Journal of Statistical Planning and Inference, 1979) Compound 1 3.03 5.53 5.60 9.30 9.92 12.51 12.95 15.21 16.04 16.84 Compound 2 3.19 4.26 4.47 4.53 4.67 4.69 5.78 6.79 9.37 12.75 (a) Find the 0.84 quantile of the Compound 1 failure times (b) Give the coordinates of the two lower-left points that would appear on a normal plot of the compound 1 data (c) Make back-to-back stem-and-leaf plots for comparing the life length properties of bearings made from Compounds 1 and 2 (d) Make (to scale) side-by-side boxplots for comparing the life lengths for the two compounds. Mark numbers on the plots indicating the locations of their main features (e) Compute the sample means and standard deviations of the two sets of lifetimes (f) Describe what your answers to parts (c), (d), and (e) above indicate about the life lengths of these turbine bearings.
Find the given attachments
Solve for X 10(x-1) = 8x-2
Answer:
X = 4 x
over 5 ( x − 1 ) − 1 5 ( x − 1 )
Answer:
x = 4
Step-by-step explanation:
10(x-1) = 8x-2
Distribute
10x-10 = 8x-2
Subtract 8 from each side
10x-10-8x = 8x-2-8x
2x-10 = -2
Add 10 to each side
2x-10+10 = -2+10
2x = 8
Divide each side by 2
2x/2 = 8/2
x = 4
Researchers want to know about the true proportion of adults with at least a high school education. 1000 adults are surveyed, and 710 of them have at least a high school education. Create a 95% confidence interval for the true population proportion of adults with at least a high school education. Interpret this interval in context of the problem.
Answer:
The 95% confidence interval for the true population proportion of adults with at least a high school education is (0.6819, 0.7381). This means that we are 95% sure that the true proportion of adults in the entire population surveyed with at least a high school education is (0.6819, 0.7381).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 1000, \pi = \frac{710}{1000} = 0.71[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.71 - 1.96\sqrt{\frac{0.71*0.29}{1000}} = 0.6819[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.71 + 1.96\sqrt{\frac{0.71*0.29}{1000}} = 0.7381[/tex]
The 95% confidence interval for the true population proportion of adults with at least a high school education is (0.6819, 0.7381). This means that we are 95% sure that the true proportion of adults in the entire population surveyed with at least a high school education is (0.6819, 0.7381).
I need help please!!!
Answer:
A.
Step-by-step explanation:
So you would have f(3) = 3 + [tex]\frac{2}{2}[/tex]
f(3) = 3 + 1
f(3) = 4
Answer:
Step-by-step explanation: f(x)
3+ a square root of 3+1 / x-1
3+ a square root (4/2)
3+ square root of 2. That's the answer
You need a 55% alcohol solution. On hand, you have a 525 mL of a 45% alcohol mixture. You also have 90% alcohol mixture. How much of the 90% mixture will you need to add to obtain the desired solution? You will need _____mL of the 90% solution to obtain _____mL of the desired 55% solution.
Answer:
Let's call the amount of 90% solution x.
We can write:
45% * 525 + 90%x = 55%(x + 525)
0.45 * 525 + 0.9x = 0.55(x + 525)
Solving for x we get x = 150 so the first blank is 150 and the second blank is 525 + 150 = 675.
It is known that 50% of adult workers have a high school diploma. If a random sample of 8 adult workers is selected, what is the probability that less than 6 of them have a high school diploma
Answer:
85.56% probability that less than 6 of them have a high school diploma
Step-by-step explanation:
For each adult, there are only two possible outcomes. Either they have a high school diploma, or they do not. The probability of an adult having a high school diploma is independent of other adults. So we use the binomial probability distribution 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.
50% of adult workers have a high school diploma.
This means that [tex]p = 0.5[/tex]
If a random sample of 8 adult workers is selected, what is the probability that less than 6 of them have a high school diploma
This is P(X < 6) when n = 8.
[tex]P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{8,0}.(0.5)^{0}.(0.5)^{8} = 0.0039[/tex]
[tex]P(X = 1) = C_{8,1}.(0.5)^{1}.(0.5)^{7} = 0.0313[/tex]
[tex]P(X = 2) = C_{8,2}.(0.5)^{2}.(0.5)^{6} = 0.1094[/tex]
[tex]P(X = 3) = C_{8,3}.(0.5)^{3}.(0.5)^{5} = 0.2188[/tex]
[tex]P(X = 4) = C_{8,4}.(0.5)^{4}.(0.5)^{4} = 0.2734[/tex]
[tex]P(X = 5) = C_{8,5}.(0.5)^{5}.(0.5)^{3} = 0.2188[/tex]
[tex]P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0039 + 0.0313 + 0.1094 + 0.2188 + 0.2734 + 0.2188 = 0.8556[/tex]
85.56% probability that less than 6 of them have a high school diploma
A phone charger requires 0.5 A at 5V. It is connected to a transformer with 100 % of efficiency whose primary contains 2200 turns and is connected to 220-V household outlet.
(a) How many turns should there be in the secondary?
(b) What is the current in the primary?
(c) What would be the output current and output voltage values if number of secondary turns (N2) doubled of its initial value?
Answer:
a. 50 turns
b. 0.0114 A
c. 0.25 A, 10 V
Step-by-step explanation:
Given:-
- The required current ( Is ) = 0.5 A
- The required voltage ( Vs ) = 5 V
- Transformer is 100% efficient ( ideal )
- The number of turns in the primary coil, ( Np ) = 2200
- The Voltage generated by power station, ( Vp ) = 220 V
Find:-
a. The number of turns in the secondary coil of the transformer
b. The current supplied by the power station
c. The effect on output current and voltage when the number of turns of secondary coil are doubled.
Solution:-
- For ideal transformers that consists of a ferromagnetic core with two ends wounded by a conductive wire i.e primary and secondary.
- The power generated at the stations is sent to home via power lines and step-down before the enter our homes.
- A household receives a voltage of 220 V at one of it outlets. We are to charge a phone that requires 0.5 A and 5V for the process.
- The outlet and any electronic device is in junction with a smaller transformer.
- All transformers have two transformation ratios for current ( I ) and voltage ( V ) that is related to the ratio of number of turns in the primary and secondary.
Voltage Transformation = [tex]\frac{N_p}{N_s} = \frac{V_p}{V_s}[/tex]
Where,
Ns : The number of turns in secondary winding
- Plug in the values and evaluate ( Ns ):
[tex]N_s = N_p*\frac{V_s}{V_p} \\\\N_s = 2200*\frac{5}{220} \\\\N_s = 50[/tex]
Answer a: The number of turns in the secondary coil should be Ns = 50 turns.
- Similarly, the current transformation is related to the inverse relation to the number of turns in the respective coil.
Current Transformation = [tex]\frac{N_p}{N_s} = \frac{I_s}{I_p}[/tex]
Where,
Ip : The current in primary coil
- Plug in the values and evaluate ( Ip ):
[tex]I_p = \frac{N_s}{N_p}*I_s\\\\I_p = \frac{50}{2200}*0.5\\\\I_p = 0.0114[/tex]
Answer b: The current in the primary coil should be Ip = 0.0114 Amp.
- The number of turns in the secondary coil are doubled . From the transformation ratios we know that that voltage is proportional to the number of turns in the respective coils. So if the turns in the secondary are doubled then the output voltage is also doubled ( assuming all other design parameters remains the same ). Hence, the output voltage is = 2*5V = 10 V
- Similary, current transformation ratio suggests that the current is inversely proportional to the number of turns in the respective coils. So if the turns in the secondary are doubled then the output current is half of the required ( assuming all other design parameters remains the same ). Hence, the output current is = 0.5*0.5 A = 0.25 A
Solving for a Confidence Interval: Algebra 2 points possible (graded) In the problems on this page, we will continue building the confidence interval of asymptotical level 95% by solving for p as in the video. Recall that R1,…,Rn∼iidBer(p) for some unknown parameter p , and we estimate p using the estimator p^=R¯¯¯¯n=1n∑i=1nRi.
As in the method using a conservative bound, our starting point is the result of the central limit theorem:
In this second method, we solve for values of P that satisfy the inequality volves penat che non esito para polcomp R -P
To do this, we manipulate - ulate | " Vp(1-) 5 < 90/2 into an inequality involving a quadratic function App + Bp+C where A > 0, B, C la/2 into an inequality in depend on 13, 4a/2, and R. Which of the following is the correct inequality?
(We will use find the values of A, B, and C in the next problem.)
1. Ap^2 + Bp + C<0 where A >0.
2. Ap^2 + Bp+C>Owhere A >0.
Let P1 and P2 with 0
a. (P P2)
b. P
Answer:
Step-by-step explanation:
1) The given inequality is
[tex]|\sqrt{n} \frac{(\bar R_n-p)}{\sqrt{p(1-p)} } |<q_{\alpha /2}| \\\\ \to(\frac{(\sqrt{n} \bar R_n-p)}{\sqrt{p(1-p)} })<q^2_{\alpha /2}[/tex]
[tex]\to n( \bar R _n - p)^2<p(1-p)q^2_{\alpha /2}[/tex]
[tex]\to n\bar R +np^2-2nR_np<q^2_{\alpha /2 p- q^2_{\alpha /2}p^2[/tex]
Arranging the terms with p² and p, we get
[tex]p^2(n+q^2_{\alpha /2)-p(2n \bar R _n+q^2_{\alpha / 2})+n \bar R ^2 _n <0[/tex]
Hence, the inequality is of the form
Ap² + Bp + c < 0
2. A quadratic equation of the form
Ap² + Bp + c < 0 with A > 0 looks like
Check the attached image
The region where the values are negative lies between p₁ and p₂ ...
The p₁ < p < p₂
Can anyone help???????
Answer:
80
Step-by-step explanation:
For every additional 10 hrs, you get 200 more dollars.
Merely needs to add enough water to 11 gallons of an 18% detergent solution to make 12% detergent solution which equation can she used to find g the number of gallon of water she should add?
1 × 18/100 = 12/100(g+11), is the equation. The answer is 12/100 gallons
Analyze the diagram below and answer the question that follows.
Ruby has a bird feeder which is visited by an average of 13 birds every 2 hours during daylight hours. What is the probability that the bird feeder will be visited by more than 3 birds in a 40 minute period during daylight hours? Round your answer to three decimal places.
Answer:
62.93%
Step-by-step explanation:
We have to solve it by a Poisson distribution, where:
p (x = n) = e ^ (- l) * l ^ (x) / x!
Where he would come being the number of birds that there would be in 40 minutes, we know that in 2 hours, that is 120 minutes there are 13, therefore in 40 there would be:
l = 13 * 40/120
l = 4,333
Now, we have p (x> 3) and that is equal to:
p (x> 3) = 1 - p (x <= 3)
So, we calculate the probability from 0 to 3:
p (x = 0) = 2.72 ^ (- 4.33) * 4.33 ^ (0) / 0! = 0.01313
p (x = 1) = 2.72 ^ (- 4.33) * 4.33 ^ (1) / 1! = 0.0568
p (x = 2) = 2.72 ^ (- 4.33) * 4.33 ^ (2) / 2! = 0.12310
p (x = 3) = 2.72 ^ (- 4.33) * 4.33 ^ (3) / 3! = 0.17767
If we add each one:
0.01313 + 0.0568 + 0.12310 + 0.17767 = 0.3707
replacing:
p (x> 3) = 1 - 0.3707
p (x> 3) = 0.6293
Which means that the probability is 62.93%
A high school student took two college entrance exams, scoring 1070 on the SAT and 25 on the ACT. Suppose that SAT scores have a mean of 950 and a standard deviation of 155 while the ACT scores have a mean of 22 and a standard deviation of 4. Assuming the performance on both tests follows a normal distribution, determine which test the student did better on.
Answer:
Due to the higher z-score, he did better on the SAT.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
Determine which test the student did better on.
He did better on whichever test he had the higher z-score.
SAT:
Scored 1070, so [tex]X = 1070[/tex]
SAT scores have a mean of 950 and a standard deviation of 155. This means that [tex]\mu = 950, \sigma = 155[/tex].
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1070 - 950}{155}[/tex]
[tex]Z = 0.77[/tex]
ACT:
Scored 25, so [tex]X = 25[/tex]
ACT scores have a mean of 22 and a standard deviation of 4. This means that [tex]\mu = 22, \sigma = 4[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{25 - 22}{4}[/tex]
[tex]Z = 0.75[/tex]
Due to the higher z-score, he did better on the SAT.
Shiva brought $39.00 to the state fair. She bought a burger, a souvenir, and a pass. The burger was 1/3 as much as the souvenir, and the souvenir cost 1/2 the cost of the pass. Shiva had $4.00 left over after buying these items.
Pass = $21 , Souvenir = 2x/3 = $14 , Burger = x/6 =$3.5 .
Alex is planning to surround his pool ABCD with a single line of tiles. How many units of tile will he need to surround his pool? Round your answer to the nearest hundredth.
Answer:
19.82 units
Step-by-step explanation:
The number of units of tile simply refers to the perimeter.
So, we need to find all the sides of the rectangle.
Now, we have AB = 4.24 units and BD = 7.07 units.
So, we can find AD using pythagoras theorem.
So,
(AD)² + 4.24² = 7.07²
(AD)² + 17.978 = 49.985
(AD)² = 49.985 - 17.978
AD = √32.007
AD = 5.66 units
AD = BC = 5.66 units
Likewise, AB = DC = 4.24 units
Thus,
perimeter = 2(5.66) + 2(4.24) = 19.8 units
Closest answer among the options is approximately 19.82
Answer:
19.82 units
Step-by-step explanation:
just took test and got it right
A multiple-choice standard test contains total of 25 questions, each with four answers. Assume that a student just guesses on each question and all questions are answered independently. (a) What is the probability that the student answers more than 20 questions correctly
Answer:
[tex]P(x>20)=9.67*10^{-10}[/tex]
Step-by-step explanation:
If we call x the number of correct answers, we can said that P(x) follows a Binomial distribution, because we have 25 questions that are identical and independent events with a probability of 1/4 to success and a probability of 3/4 to fail.
So, the probability can be calculated as:
[tex]P(x)=nCx*p^{x}*q^{n-x}=25Cx*0.25^{x}*0.75^{25-x}[/tex]
Where n is 25 questions, p is the probability to success or 0.25 and q is the probability to fail or 0.75.
Additionally, [tex]25Cx=\frac{25!}{x!(25-x)!}[/tex]
So, the probability that the student answers more than 20 questions correctly is equal to:
[tex]P(x>20)=P(21)+P(22)+P(23)+P(24)+P(25)[/tex]
Where, for example, P(21) is equal to:
[tex]P(21)=25C21*0.25^{21}*0.75^{25-21}=9.1*10^{-10}[/tex]
Finally, P(x>20) is equal to:
[tex]P(x>20)=9.67*10^{-10}[/tex]
Among fatal plane crashes that occurred during the past 65 years, 627 were due to pilot error, 64 were due to other human error, 113 were due to weather, 382 were due to mechanical problems, and 481 were due to sabotage. Construct the relative frequency distribution.
What is the most serious threat to aviation safety, and can anything be done about it?
A. Sabotage is the most serious threat to aviation safety. Airport security could be increased.
B. Mechanical problems are the most serious threat to aviation safety. New planes could be better engineered.
C. Weather is the most serious threat to aviation safety. Weather monitoring systems could be improved.
D. Pilot error isPilot error is the most serious threat to aviation safety. Pilots could be better trained.
Answer:
D. Pilot error isPilot error is the most serious threat to aviation safety. Pilots could be better trained.
Step-by-step explanation:
We can construct the relative frequency distribution dividing the amount of incidents for each category by the total amount of incidents.
This amount is:
[tex]\sum x_i=627+64+113+382+481=1667[/tex]
Then, the relative frequency for each category is:
[tex]\text{Pilot error}=627/1667=0.38\\\\\text{Human error}=64/1667=0.04\\\\\text{Weather}=113/1667=0.07\\\\\text{Mechanical problems}=382/1667=0.23\\\\\text{Sabotage}=481/1667=0.29\\\\[/tex]
As the pilot error has the largest relative frequency, we can conclude that pilot error is the most serious threat to aviation safety.
SOMEONE PLEASE HELP ME ASAP PLEASE!!!
Answer:
105.12 ft^2
Step-by-step explanation:
Area of a rectangle: bh
In this case 8*10.... so area of the rectangle is 80
Area of a circle: pir^2
Half it for a semicircle.
so 1/2 pi r^2
radius is 4 cuz its half of 8.
so 1/2(3.14)(4^2)=(0.5)(3.14)(16)=25.12
Now add up 80+25.12
Total is 105.12
Hope I helped :)
Evaluate: 5-^2 =
pls help
Answer:
1/25
Step-by-step explanation:
5^-2
We know that a^ -b = 1/ a^b
5^-2 = 1/ 5^2
= 1/25
We are interested in finding an estimator for Var (Xi), and propose to use V=-n (1-Xn). 0/2 puntos (calificable) Now, we are interested in the bias of V. Compute: E [V]-Var (Xi)-[n Using this, find an unbiased estimator V for p (1 - p) if n22. rite barX_n for Л n . 72 1--X 7t
Here is the full question .
We are interested in finding an estimator for [tex]Var (X_i )[/tex] and propose to use :
[tex]\hat {V} = \bar {X}_n (1- \bar {X} )_n[/tex]
Now; we are interested in the basis of [tex]\hat V[/tex]
Compute :
[tex]E \ \ [ \bar V] - Var (X_i) =[/tex]
Using this; find an unbiased estimator [tex][ \bar V][/tex] for [tex]p(1-p) \ if \ n \geq 2[/tex]
Write [tex]bar \ x{_n} \ for \ X_n[/tex]
Answer:
Step-by-step explanation:
[tex]\bar X_n = \dfrac{1}{n} {\sum ^n _ {i=1} } \\ \\ E(X_i) = - \dfrac{1}{n=1} \sum p \dfrac{1}{n}*np = \mathbf{p}[/tex]
[tex]V(\bar X_n) = V ( \dfrac{1}{n_{i=1} } \sum ^n \ X_i )} = \dfrac{1}{n^2} \sum ^n_{i=1} Var (X_i) \\ \\ = \dfrac{1}{n^2} \ \sum ^n _{i=1} p(1-p) \\ \\ = \dfrac{1}{n^2}*np(1-p) \\ \\ = \dfrac{p(1-p)}{n}[/tex]
[tex]E( \bar X^2 _ n) = Var (\bar X_n) + [E(\bar X_n)]^2 \\ \\ = \dfrac{p(1-p)}{n}+ p \\ \\ = p^2 + \dfrac{p(1-p)}{n} \\ \\ \\ \hat V = \bar X_n (1- \bar X_n ) = \bar X_n - \bar X_n ^2 \\ \\ E [ \hat V] = E [ \bar X_n - \bar X_n^2] \\ \\ = E[\bar X_n ] - E [\bar X^2_n] \\ \\ = p-(p^2 + \dfrac{p(1-p)}{n}) \\ \\ = p-p^2 -\dfrac{p(1-p)}{n}[/tex]
[tex]=p(1-p)[1-\dfrac{1}{n}] = p(1-p)\dfrac{n-1}{n}[/tex]
[tex]Bias \ (\bar V ) = E ( \hat V) - Var (X_i) \\ \\ = p(1-p) [1-\dfrac{1}{n}] - p(1-p) \\ \\ - \dfrac{p(1-p)}{n}[/tex]
Thus; we have:
[tex]E [\hat V] = p(1-p ) \dfrac{n-1}{n}[/tex]
[tex]E [\dfrac{n}{n-1} \ \ \bar V] = p(1 -p)[/tex]
[tex]E [\dfrac{n}{n-1} \ \ \bar X_n (1- \bar X_n )] = p (1-p)[/tex]
Therefore;
[tex]\hat V ' = \dfrac{n}{n-1} \bar X_n (1- \bar X_n)[/tex]
[tex]\mathbf{ \hat V ' = \dfrac{n \bar X_n (1- \bar X_n)} {n-1}}[/tex]
ASAP! GIVING BRAINLIEST! Please read the question THEN answer CORRECTLY! NO guessing. I say no guessing because people usually guess on my questions.
Answer:
D. G(x) = |x| + 7
Step-by-step explanation:
→For the function G(x) to shift upwards, there needs to be a number being added to the whole function.
→The answer isn't "A," because the 1 is being subtracted, making it shift downwards 1 unit, not upwards.
→The answer isn't "B," because adding the 2 there would cause the function to shift to the left for 2 units, not upwards.
→The answer isn't "C," because 10 is being multiplied, which would cause the function to narrow, and not shift upwards.
This means the correct answer is "D," because the 7 is being added, making the function shift upwards 7 units.
5. Si P(x)=2x+4a , Q(x)=4x-2 y P[Q(4)]=60 , Calcular el valor de a
Answer:
a = 8
Step-by-step explanation:
Explanation:-
Given P(x) = 2 x+4 a
Q(x)=4 x - 2
P( Q(4)) = 60
P(4 (4) - 2) = 60
P( 14 ) = 60
2 (14) + 4 a = 60
4 a + 28 = 60
Subtracting '28' on both sides , we get
4 a +28 - 28 = 60 - 28
4 a = 32
Dividing '4' on both sides , we get
a = 8
Hey can anyone help me with this 3 3/5 x (-8 1/3)?
Answer:
[tex]-30[/tex]
Step-by-step explanation:
[tex]3\frac{3}{5} \times (-8 \frac{1}{3} )[/tex]
[tex]\frac{18}{5}\times \left(-\frac{25}{3}\right)[/tex]
[tex]\frac{18}{5}\times -\frac{25}{3}[/tex]
[tex]-\frac{18\times \:25}{5\times \:3}[/tex]
[tex]-\frac{450}{15}[/tex]
[tex]=-30[/tex]
Answer:-30
Step-by-step explanation:
3 3/5 x -8 1/3
18/5 x -25/3
-30
Question 7 (5 points)
Which of the following is the simplified fraction that's equivalent to 0.3
OA) 35/999
OB) 31/99
C) 105
7333
OD) 35
D) 35/111
Answer: B. although none are exactly 0.3 B is closest
Step-by-step explanation:
a. 35/999 = .0350
b. 31/99 = .3153
c. 105/7333 = .0143
d. 35/111 = .3135
PLEASE HELLLLP!!!! If x+y−z=8 and x−y+z=12, then x=
Answer:
(C) 10
Step-by-step explanation:
x+y−z=8
Subtract y from both sides
x-z= -y+8
Add z to both sides
x= -y+z+8
Subtract 8 from both sides
x-8= -y+z
x−y+z=12
Add y to both sides
x+z=12+y
Subtract z from both sides
x=y-z+12
Subtract 12 from both sides
x-12=y-z
Multiply both sides by -1
-x+12= -y+z
Combine equations:
x-8= -x+12
Add x to both sides
2x-8+12
Add 8 to both sides
2x=20
Divide both sides by 2
x=10
The answer is (C) 10.
Evaluate the expression 2x-7 for x = -4
Answer:
-15
Step-by-step explanation:
The solution of expression for x = - 4 is,
⇒ - 15
We have to given that,
An expression is,
⇒ 2x - 7
Now, We can simplify the expression for x = - 4 as,
An expression is,
⇒ 2x - 7
Plug x = - 4;
⇒ 2 × - 4 - 7
⇒ - 8 - 7
⇒ - 15
Thus, The solution of expression for x = - 4 is,
⇒ - 15
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