By taking x as a common factor we get:
2x - 1 + 3x = 0
(2 + 3)*x - 1 = 0
5x - 1 = 0
How to get equation B from equation A?Let's start with equation A, it is:
2x - 1 + 3x = 0
If we group like terms, we will get:
(2x + 3x) - 1 = 0
Now we can take x as a common factor in the left term, so we get:
(2x + 3x) - 1 = 0
(2 + 3)*x - 1 = 0
Now we simplify the sum in the left term:
(2 + 3)*x - 1 = 0
5x - 1 = 0
This is equation B.
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a+b=0
What does b equal
Answer: D) -a
Step-by-step explanation:
a+b=0 , add -a for both side
- a + a + b = 0 - a
we cancel (- a + a) and we get
b = 0 - a => b = -a
Answer:
option d -a
if you take a to the opposite side of the equal mark it's sign is going to change since it's positive it's going to be negative on the other side 0 -a = -a
baam answer
Dilate this figure by a scale factor of 1.5. Is the image a stretch or compression?
Answer:
compression
Step-by-step explanation:
If b>1 then the graph will compress
The equation y =1/2x represents a proportional relationship. What is the constant of proportionality? A. 1/2B.XC. 0D. 2
You have the following equation:
y = 1/2 x
In order to determine what is the constant of proportionality, consider the general fom of proportional relatioship, given by:
y = kx
by comparing the previous equation with y=1/2x, you can notice that k=1/2. Then, the constant of proportionality is k = 1/2
A. 1/2
4x5y² how would I solve this
multiply the numbers
Answer is 20y2
What is 541,000 rounded to the nearest 10,000
Answer:
540,000
Step-by-step explanation:
Since you are rounding to the 10,000s place you must look at the 1,000s place to decide what to round to. Since in the 1,000s place is a 1, you will round the number down (1 < 5).
An item has a listed price of $90. If the sales tax rate is 3%, how much is the sales tax (in dollars)?
Please help soon thanks
Answer:
x = -7
Step-by-step explanation:
Hi!
Ok we know a few things here :
x + 37 + x + 67 + 90 = 180
2x + 104 = 90
2x = -14
x = -7
Hope this helps!
Have a great day! :)
In a class of 25 students, 15 of them have a cat, 16 of them have a dog and 3 of them have neither. Find the probabillity that a student chosen at random has both
9/25
Step-by-step explanation:
students who have either cat or dog =25-3=22
students who have both= 15+16-22= 9
probabillity that a student chosen at random has both= 9/25
Number 6. For questions 5-7, (a) use synthetic division to show that x is a zero.(b) find the remaining factors of f(x).(c) use your results to find the complete factorization of f(x).(d) list all zeros of f(x).(e) graph the function.
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given polynomials
[tex]f(x)=x^3+6x^2-15x-100[/tex]One of the zeroes is:
[tex]\begin{gathered} x=-5 \\ \text{this implies that:} \\ (x+5)=0 \end{gathered}[/tex]STEP 2: Use synthetic division to divide the polynomials
[tex]\frac{x^3+6x^2-15x-100}{x+5}[/tex]Write the coefficients of the numerator
[tex]1\:\:6\:\:-15\:\:-100[/tex][tex]\begin{gathered} \mathrm{Write\:the\:problem\:in\:synthetic\:division\:format} \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \\ Carry\:down\:the\:leading\:coefficient,\:unchanged,\:to\:below\:the\:division\: \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \\ \end{gathered}[/tex][tex]\begin{gathered} Multiply\:the\:carry-down\:value\:by\:the\:zero\:of\:the\:denominator,\:and\:carry\:the\:result\:up\:into\:the\:next\:column \\ 1\left(-5\right)=-5 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:\:\:\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} \mathrm{Add\:down\:the\:column:} \\ 6-5=1 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:\:\:\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} Multiply\:the\:carry-down\:value\:by\:the\:zero\:of\:the\:denominator,\:and\:carry\:the\:result\:up\:into\:the\:next\:column: \\ 1\left(-5\right)=-5 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} \mathrm{Add\:down\:the\:column:} \\ -15-5=-20 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:-20\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} Multiply\:the\:carry-down\:value\:by\:the\:zero\:of\:the\:denominator,\:and\:carry\:the\:result\:up\:into\:the\:next\:column: \\ \left(-20\right)\left(-5\right)=100 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:100}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:-20\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} \mathrm{Add\:down\:the\:column:} \\ -100+100=0 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:100}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:-20\:\:\:\:\:0}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} \mathrm{The\:last\:carry-down\:value\:is\:the\:remainder} \\ 0 \end{gathered}[/tex]The last carry-down value is the remainder and it is 0 (zero)
Since the remainder is a zero, hence, x=-5 is a zero
Step 3: Answer question b
To get the factors, the remainder of the division in step 2 is given as:
The remaining factors of f(x) is:
[tex]x^2+x-20[/tex]STEP 4: Answer Question c
[tex]\begin{gathered} roots=(x+5)(x^2+x-20) \\ Factorize\text{ the other root to have:} \\ Using\text{ factorization methods:} \\ (x^2+x-20)=(x^2+5x-4x-20) \\ x(x+5)-4(x+5)=0 \\ (x-4)(x+5)=0 \end{gathered}[/tex]The complete factorization will give:
[tex](x+5)(x-4)(x+5)[/tex]STEP 5: Answer question d
The zeroes of f(x) will be:
[tex]\begin{gathered} zeroes\text{ of f\lparen x\rparen=?, we equate the roots to 0} \\ zeroes\Rightarrow x=-5,4,-5 \end{gathered}[/tex]zeroes are: -5,4,-5
STEP 6: Plot the graph
Is the ordered pair a solution of the equation?
y = x + 16; (-1,-17)
a. yes
b. no
Answer:
b.
Step-by-step explanation:
No, because -1 is x and -17 is y. if You plug it in, it will look like this:
-17=-1+16
-1+16=15
15 is not equal to -17, so the answer is no, b.
Hope this helps!
Answer:
a.)yes. у=х+16
Answer:
(-1)
discuss the advantages and disadvantages of using sampling to reduce the number of data objects. would simple random sampling (without replacement) be a good approach to sampling? why or why not? what kind of sampling method that you would like to use? g
The advantage and disadvantages are sampling.
We have to find the advantages and disadvantages of sampling.
The advantage is:
It provides an opportunity to do data analysis with a lower likelihood of carrying an error.
Researchers can analyze the data that is collected with a smaller margin of error thanks to random sampling. This is permitted since the sampling procedure is governed by predetermined boundaries. The fact that the entire procedure is random ensures that the random sample accurately represents the complete population, which enables the data to offer precise insights into particular topic matters.
The disadvantage is:
No extra information is taken into account.
Although unconscious bias is eliminated by random sampling, deliberate bias remains in the process. Researchers can select areas for random sampling in which they think particular outcomes can be attained to confirm their own bias. The random sampling does not take into account any other information, however sometimes the data collector's supplementary information is retained.
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suppose that a tall child with arm span 120 cm and height 118 cm was added to the sample used in this study. what effect will this addition have on the correlation and the slope of the least-squares regression line?
A tall child with an arm span of 120 cm and height of 118 cm was added to the sample used in this study then the correlation will increase whereas the slope stays the same.
Correlation is a relation between two variables, by shifting one independent variable, how the dependent variable will shift, that's the degree of correlation since Correlation can be fitted in any shape or condition. It can be fitted nonlinear, straight or curvy straight, etc whereas slope is related to a straight variety of two variables, it is characterized as the rate of altering of the dependent variable in order with the independent variable. By and large you'll discover equation incline =( y2-y1)/(x2-x1).
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Use the figure to find measures of the numbered angles.
The measure of the numbered angles formed by the common transversal to the two parallel lines are;
[tex] \angle 1 = 113^{ \circ} [/tex]
[tex] \angle 2 = 67^{ \circ} [/tex]
[tex] \angle 3 = 67^{ \circ}[/tex]
[tex]\angle 4 = 113^{ \circ} [/tex]
What are the relationships between the angles formed by the common transversal to two parallel lines?The relationships between the angles are;
Corresponding angles are congruentVertical angles are congruentSame side exterior angles are supplementarySame side interior angles are supplementaryAlternate interior angles are congruentAlternate exterior angles are congruentThe given angle is 113°
According to corresponding angles theorem, we have;
[tex] \angle 1 = 113^{ \circ} [/tex]
[tex]\angle 1 \: and \: \angle 4 [/tex] are vertical angles
According to vertical angles theorem, we have;
[tex]\angle 1 = \angle 4 = 113^{ \circ} [/tex]
[tex] \angle 3 \: and \: 113^{ \circ} [/tex] are same side exterior angles.
According to same side exterior angles theorem, we have;
[tex] \angle 3 \: and \: 113^{ \circ} [/tex] are supplementary angles. Which gives;
[tex] \angle 3 + 113^{ \circ} = 180^{ \circ} [/tex]
[tex] \angle 3 = 180^{ \circ} - 113^{ \circ} = 67^{ \circ}[/tex]
[tex] \angle 2 \: and \: \angle 3 [/tex] are vertical angles, which gives;
[tex] \angle 2 = \angle 3 [/tex] = 67°
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Use the following function rule to find g(r + 2). Simplify your answer.
g(k)= k-4
g(r + 2) =
Using the function rule, the value of g(r +2) is r-2
How can we evaluate and simply g(r + 2) using the function rule?Given the function rule g(k) = k-4.
Then g(r + 2) can be found by replacing k with r+2 in the function g(k):
g(r+2) = r + 2 - 4
= r - 2
Therefore, g(r+2) gives r - 2 when evaluated using the function rule.
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Analyze each situation. Identify a reasonable
domain and range. Is the domain continuous of discrete? A car has a 15-gallon gas tank and gets a
maximum average mileage of 24 miles per gallon
The reasonable domain and range are 0 ≤ x ≤ 15 and 0 ≤ y ≤ 360 and the domain is continuous
How to determine the domain?From the question, we have
Gallon gas tank = 15 gallons
This means that the domain is from 0 to 15
This can be represented as 0 ≤ x ≤ 15
For the range, we have
0 * 24 ≤ y ≤ 15 * 24
Evaluate
0 ≤ y ≤ 360
This means that
Range: 0 ≤ y ≤ 360
Lastly, the domain of the situation is continuous
This is so because the gas in the tank can take decimal values
Take for instance, the gas tank can be 13.79 gallons full
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Select whether the pair of lines is parallel, perpendicular, or neither.
y = 5, y = −3
The given pair of lines are parallel lines.
What are parallel lines?Parallel lines in geometry are coplanar, straight lines that don't cross at any point. In the same three-dimensional space, parallel planes are any planes that never cross. Curves with a fixed minimum distance between them and no contact or intersection are said to be parallel. A line with the same slope is said to be parallel. To put it another way, these lines won't ever cross. At the interception, perpendicular lines always intersect and form right angles.So, graph the lines on the given equation:
y = 5y = -3(Refer to the graph attached below)
We can easily look at the graph and tell that the given lines of the equations are parallel to each other.
Therefore, the given pair of lines are parallel lines.
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State if the three side lengths form an acute obtuse or a right triangle
Given three side lengths form an acute obtuse or a right triangle 17, 21, & 28
Check Right triangle
[tex]\begin{gathered} Hyp^2=opp^2+adj^2 \\ 28^2=17^2+21^2 \\ 28^2\text{ = 289 +441} \\ 784\text{ }\ne\text{ 730} \end{gathered}[/tex]Not a Right triangle
An obtuse triangle is a triangle with one obtuse angle and two acute angles. Since a triangle's angles must sum to 180° in Euclidean geometry.
Not an Obtuse triangle
[tex]\begin{gathered} \sin \text{ A= }\frac{opp}{hyp} \\ \sin \text{ A = }\frac{17}{28} \\ A=sin^{-1}\text{ }\frac{17}{28} \\ A=37.4^0\text{ (less than 90)} \end{gathered}[/tex][tex]\begin{gathered} \cos \text{ B = }\frac{adj}{hyp} \\ \cos \text{ B = }\frac{21}{28} \\ B=cos^{-1}\frac{21}{28} \\ B=41.4^0\text{ (less than 90)} \end{gathered}[/tex][tex]\begin{gathered} \tan \text{ C = }\frac{opp}{adj} \\ \tan \text{ C = }\frac{17}{21} \\ C=tan^{-1\text{ }}\frac{17}{21}\text{ } \\ C=38.9^0\text{ (less than 90) } \end{gathered}[/tex]Hence it is acute angle because all angles are less than 90°
Find the difference between the product of 2.5 and 7.5and the sum of 2.7 and 9.55
I'm not good with graphs so I really need help solving and understanding this
4x + 3y = -24 (option D)
Explanation:To determine the orrect option, we would find the equation of the line.
Equation of line: y = mx + c
m = slope, c = y-intercept
Using the slope formula:
[tex]m\text{ = }\frac{y_2-y_1}{x_2-x_1}[/tex]We would use any two points on the graph.
Using points: (-6, 0) and (0, -8)
[tex]\begin{gathered} x_1=-6,y_1=0,x_2=0,y_2\text{ = -}8 \\ m\text{ = }\frac{-8-0}{0-(-6)} \\ m\text{ =}\frac{-8}{0+6}=\frac{-8}{6}=-\frac{4}{3} \end{gathered}[/tex]The y intercept is the point where the line crosses the y axis and the value of x is zero. It crosses the line at y = -8
The equation of line becomes:
y = -4/3 x + (-8)
y = -4/3x - 8
Multiply both sides by 3:
3(y) = 3(-4/3 x) - 3(8)
3y = -4x - 24
3y + 4x = -24
4x + 3y = -24 (option D)
robert john and paul start to run from the same point in the same direction around a circular track.
if they take 126 seconds 154 seconds and 198 seconds respectively to complete one round along the track, when will they next meet again at the starting point?
They will next meet again after 1386 seconds or 23.1 minutes from the starting point.
Given that Robert, John, and Paul begin running in the same direction from the same starting point around a circular track If they take 126 seconds, 154 seconds, and 198 seconds respectively to complete one round of the track.
What is HCF?The highest common factor is defined as the set of numbers with the highest common multiple. The highest positive integer with more than one factor in the set is HCF.
To determine whether they will next meet again at the starting point.
We have to calculate the LCM of the numbers 126, 154, and 198
⇒ Prime factors of 126 = 2 × 2 × 3 ×7
⇒ Prime factors of 154 = 2 × 11 ×7
⇒ Prime factors of 198 = 2 × 3 × 3 ×11
So the LCM of the numbers 126, 154, and 198 will be as:
⇒ 2 × 3 × 3 × 7 × 11 = 1386
Therefore, they will meet after 1386 seconds or 23.1 minutes.
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You are designing a poster with an area of 625 cm2 to contain a printing area in the middle and have the margins of 4cm at the top and bottom and 7cm on each side. Find the largest possible printing area. Round your answer to the nearest four decimal places.
Explanation
Step 1
diagram
so
Step 2
let
a)total area= side1*side 2
[tex]side1*side2=625[/tex]and
[tex]printing\text{ area=\lparen side1-14\rparen\lparen side2-8\rparen}[/tex]Step 3
solve
let
[tex]\begin{gathered} side1\text{ = x} \\ side\text{ 2 =}y \\ xy=625 \\ y=\frac{625}{x}\Rightarrow equation(1) \end{gathered}[/tex]and in teh second equation we have
[tex]\begin{gathered} pr\imaginaryI nt\imaginaryI ng\text{area=}\operatorname{\lparen}\text{s}\imaginaryI\text{de\times1-14}\operatorname{\rparen}\operatorname{\lparen}\text{s}\imaginaryI\text{de\times2-8}\operatorname{\rparen} \\ A=(x-14)(y-8) \\ replace\text{ the y value} \\ A=(x-14)(\frac{625}{x}-8) \\ A=\frac{625x}{x}-\frac{8750}{x}-8x+112) \\ A=-\frac{8750}{x}+625-8x+112 \\ dA=\frac{8750}{x^2}-8 \\ \frac{8750}{x^2}=8 \\ x^2=\frac{8750}{8} \\ x=\sqrt[\placeholder{⬚}]{\frac{8750}{8}} \\ x=33 \end{gathered}[/tex]replace to find y
[tex]\begin{gathered} y=\frac{625}{x}\operatorname{\Rightarrow}equat\imaginaryI on(1) \\ y=\frac{625}{33} \\ y=18.9 \end{gathered}[/tex]so
the maximum area is
[tex]\begin{gathered} x-14=33-14=19 \\ y-8=18.9=10.9 \end{gathered}[/tex]so, the greates area is
[tex]18.9*19=359.1[/tex]therefore, the answer is
359.1 square cm
I hope this helps you
The table shows the number of cars and trucks that used a certain toll road on a particular day. The number of cars and trucks that used, and did not use, an electronic toll pass on that same day was also recorded.Toll PassCars Trucks TotalUsed537330867Did not use9046491553Total14419792420a) If one of these vehicles is selected at random, determine the probability that the vehicle is a car.b) If one of these vehicles is selected at random, determine the probability that the vehicle is a car, given that it used the toll passa) The probability that the vehicle was a car is(Round to four decimal places as needed.)
We are given a two-way probability table
Part a)
If one of the vehicles is selected at random, determine the probability that the vehicle is a car.
From the table, we see that the total number of cars are 1441
Also, the total number of vehicles is 2420
Then the probability of selecting a car is
[tex]P(car)=\frac{1441}{2420}=0.5955[/tex]Therefore, the probability that the vehicle was a car is found to be 0.5955
Part b)
If one of the vehicles is selected at random, determine the probability that it used the electronic toll pass, given that it was a car.
This is a conditional probability problem.
The conditional probability is given by
[tex]P(used\: |\: car)=\frac{n(used\: and\: car)}{n(car)}[/tex]From the table, we see that,
[tex]\begin{gathered} n(car\: and\: used)=537 \\ n(car)=1441 \end{gathered}[/tex]So, the probability is
[tex]\begin{gathered} P(used\: |\: car)=\frac{n(used\: and\: car)}{n(car)} \\ P(used\: |\: car)=\frac{537}{1441} \\ P(used\: |\: car)=0.3727 \end{gathered}[/tex]Therefore, the probability that it used the electronic toll pass, given that it was a car is found to be 0.3727
The probability that the randomly chosen vehicle is a car is 59.54 %.
The probability that the randomly chosen vehicle is a car given that it used the toll pass is 61.94%.
What is conditional probability?Conditional probability is a term used in probability theory to describe the likelihood that one event will follow another given the occurrence of another event.
The probability that the randomly chosen vehicle is a car is the total no of cars divided by the total no. of vehicles which is,
= 1441/2420.
= (1441/2420)×100%.
= 59.54 %.
The probability that the randomly chosen vehicle is a car given that it used the toll pass is the total no. of cars that used the toll pass divided by the no. of vehicles that used the toll pass which is,
= (537/867)×100%.
= 61.94%.
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suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 1.5 and a mean diameter of 205 inches. if 79 shafts are sampled at random from the batch, what is the probability that the mean diameter of the sample shafts would differ from the population mean by less than 0.3 inches? round your answer to four decimal places.
The probability that the mean diameter of the sample shafts would vary from the population mean by less than 0.3 inches exists 0.0757.
How to estimate population mean?Let X the random variable that represent the diameters of interest for this case, and for this case we know the following info
Where [tex]$\mu=205$[/tex] and [tex]$\sigma=1.5$[/tex]
Let the probability be
[tex]$P(205-0.3=204.7 < \bar{X} < 205+0.3=205.3)$$[/tex]
For this case they select a sample of n = 79 > 30, so then we have enough evidence to utilize the central limit theorem and the distribution for the sample mean can be approximated with:
[tex]$\bar{X} \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right)$$[/tex]
To simplify this problem is utilizing the normal standard distribution and the z score given by:
[tex]$z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$$[/tex]
To find the z scores for each limit and we got:
[tex]$z=\frac{204.7-205}{\frac{1.5}{\sqrt{79}}}=-1.778$$[/tex]
[tex]$z=\frac{205.3-205}{\frac{1.5}{\sqrt{79}}}=1.778$$[/tex]
To find this probability:
P(-1.778 < Z < 1.778) = P(Z < 1.778) - P(Z < -1.778)
simplifying the above equation, we get
= 0.962 - 0.0377
= 0.9243
The interest exists the probability that the mean diameter of the sample shafts would vary from the population mean by more than 0.3 inches utilizing the complement rule we got:
P = 1 - 0.9243 = 0.0757
The probability that the mean diameter of the sample shafts would vary from the population mean by less than 0.3 inches exists 0.0757.
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Find the distance between the points (-16, 20) and (-16, -14).2007(-16, 20)161284-20-16-12-8-4048121620-4-812(-16, -14)-16-20units
Point A
(-16, 20)
Point B
(-16, -14)
The Distance Formula itself is actually derived from the Pythagorean Theorem which is a
[tex]\begin{gathered} d=\sqrt[]{(y2-y1)^2+(x2-x1)^2} \\ d=\sqrt[]{(-14-20)^2+(-16-(-16}))^2 \\ d=\sqrt[]{(-34)^2} \\ d=34 \end{gathered}[/tex]The distance would be 34 units
IMAGE ATTACHED,please help me with this one please.
Answer:
Step-by-step explanation:
$17.64 for 147 text messages
Therefore 1 text message will cost $17.64 ÷ 147 Text messages = $0.12 per text message
Use inverse trigonometric functions to solve the following equations. If there is more than one solution, enter all solutions as a comma-separated list (like "1, 3"). If an equation has no solutions, enter "DNE".Solve tan(θ)=1 for θ (where 0≤θ<2π).θ=Solve 7tan(θ)=−15 for θ (where 0≤θ<2π).θ=
Starting with the equation:
[tex]\tan (\theta)=1[/tex]take the inverse tangent function to both sides of the equation:
[tex]\begin{gathered} \arctan (\tan (\theta))=\arctan (1) \\ \Rightarrow\theta=\arctan (1) \\ \therefore\theta=\frac{\pi}{4} \end{gathered}[/tex]Yet another value can be found for this equation to be true since the period of the tangent function is π:
[tex]\begin{gathered} \theta_1=\frac{\pi}{4} \\ \theta_2=\frac{\pi}{4}+\pi=\frac{5}{4}\pi \end{gathered}[/tex]Starting with the equation:
[tex]7\tan (\theta)=-15[/tex]Divide both sides by 7:
[tex]\Rightarrow\tan (\theta)=-\frac{15}{7}[/tex]Take the inverse tangent to both sides of the equation:
[tex]\begin{gathered} \Rightarrow\arctan (\tan (\theta))=\arctan (-\frac{15}{7}) \\ \Rightarrow\theta=\arctan (-\frac{15}{7}) \\ \therefore\theta=-1.13416917\ldots \end{gathered}[/tex]The tangent function has a period of π. Since the value that we found for theta is not between 0 and 2π, then we can add π to the value:
[tex]\begin{gathered} \theta_1=-1.13416917\ldots+\pi \\ =2.007423487\ldots \end{gathered}[/tex]We can find another value for theta such that its tangent is equal to -15/7 by adding π again, provided that the result is less than 2π:
[tex]\begin{gathered} \theta_2=\theta_1+\pi \\ =5.14901614\ldots \end{gathered}[/tex]Therefore, for each equation we know that:
[tex]\begin{gathered} \tan (\theta)=1 \\ \Rightarrow\theta=\frac{\pi}{4},\frac{5\pi}{4} \end{gathered}[/tex][tex]\begin{gathered} 7\tan (\theta)=-15 \\ \Rightarrow\theta=2.007423487\ldots\text{ , }5.14901614\ldots \end{gathered}[/tex]Starting with the equation:
[tex]\tan (\theta)=1[/tex]take the inverse tangent function to both sides of the equation:
[tex]\begin{gathered} \arctan (\tan (\theta))=\arctan (1) \\ \Rightarrow\theta=\arctan (1) \\ \therefore\theta=\frac{\pi}{4} \end{gathered}[/tex]Yet another value can be found for this equation to be true since the period of the tangent function is π:
[tex]\begin{gathered} \theta_1=\frac{\pi}{4} \\ \theta_2=\frac{\pi}{4}+\pi=\frac{5}{4}\pi \end{gathered}[/tex]Starting with the equation:
[tex]7\tan (\theta)=-15[/tex]Divide both sides by 7:
[tex]\Rightarrow\tan (\theta)=-\frac{15}{7}[/tex]Take the inverse tangent to both sides of the equation:
[tex]\begin{gathered} \Rightarrow\arctan (\tan (\theta))=\arctan (-\frac{15}{7}) \\ \Rightarrow\theta=\arctan (-\frac{15}{7}) \\ \therefore\theta=-1.13416917\ldots \end{gathered}[/tex]The tangent function has a period of π. Since the value that we found for theta is not between 0 and 2π, then we can add π to the value:
[tex]\begin{gathered} \theta_1=-1.13416917\ldots+\pi \\ =2.007423487\ldots \end{gathered}[/tex]We can find another value for theta such that its tangent is equal to -15/7 by adding π again, provided that the result is less than 2π:
[tex]\begin{gathered} \theta_2=\theta_1+\pi \\ =5.14901614\ldots \end{gathered}[/tex]Therefore, for each equation we know that:
[tex]\begin{gathered} \tan (\theta)=1 \\ \Rightarrow\theta=\frac{\pi}{4},\frac{5\pi}{4} \end{gathered}[/tex][tex]\begin{gathered} 7\tan (\theta)=-15 \\ \Rightarrow\theta=2.007423487\ldots\text{ , }5.14901614\ldots \end{gathered}[/tex](6-6)x 6=0
(6 + 6):6=2
6? 6?6 = 4
Answer:
I don't know if you can use permutations but
(6P2-6)/6=4 ????
H=4(x+3y)+2 make x the subject
When the formula is changed to x, the equation is x = 1/4(H - 2) - 3y
How to change the subject of formula to x?From the question, the equation is given as
H=4(x+3y)+2
Subtract 2 from both sides of the equation
So, we have the following equation
H - 2 =4(x+3y)+2 -2
Evaluate
So, we have the following equation
H - 2 =4(x+3y)
Divide by 4
So, we have the following equation
1/4(H - 2) = x +3y
Subtract 3y from both sides
x = 1/4(H - 2) - 3y
Hence, the solution for x is x = 1/4(H - 2) - 3y
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The simplified solution for the x is given as x = [H - 2]/4 - 3y.
As mentioned in the question, for an equation H=4(x+3y)+2, we have to transform the equation in the subject of x.
The operational processes in mathematics for the functions to make the function or expression.
Here,
Simplification of the given function in a manner to the transposition of the variables from left to right,
H=4(x+3y)+2
H - 2 = 4(x+3y)
{H - 2}/4 = (x + 3y)
x = [H - 2]/4 - 3y
Thus, the simplified solution for the x is given as x = [H - 2]/4 - 3y.
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Which is another way to represent
five hundred six and ninety-two
thousandths?
Select all that apply.
(5 x 100) + (6 × 1)
+ X
(9 × 10) + (2x 100)
(5 x 100) + (6 × 1)
+(9 × 100) + (2.× 1,000)
506.092
506.902
506.92
In a state's lottery, you can bet $3 by selecting three digits, each between 0 and 9 inclusive. If the same three numbers are drawn in the same order, you win and collect $500. Complete parts (a) through (e).
Using the Fundamental Counting Theorem, it is found that:
a) 1000 combinations are possible.
b) The probability of winning is of 0.001.
c) If you win, the net profit is of $497.
d) The expected value of a $3 bet is of -$2.5.
What is the Fundamental Counting Theorem?It is a theorem that states that if there are n trials, each with [tex]n_1, n_2, \cdots, n_n[/tex] possible results, each thing independent of the other, the number of results is given as follows:
[tex]N = n_1 \times n_2 \times \cdots \times n_n[/tex]
In the context of this problem, three digits that can be repeated are chosen, hence the parameters are:
[tex]n_1 = n_2 = n_3 = 10[/tex]
Hence the number of combinations is:
N = 10³ = 10 x 10 x 10 = 1000.
The order also has to be correct, hence the there is only one winning outcome and the probability is:
p = 1/1000 = 0.001.
You bet $3, and if you win you collect $500, hence the net profit is of:
500 - 3 = $497.
Then the distribution of earnings are as follows:
P(X = -3) = 0.999 -> losing.P(X = 497) = 0.001. -> winning.Hence the expected value is:
E(X) = -3 x 0.999 + 497 x 0.001 = -$2.5.
Missing informationThe problem is given by the image at the end of the answer.
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