Answer:
x = 11
Step-by-step explanation:
3/x+1=2/x-3
Solve by using cross products
2 (x+1) = 3 (x-3)
Distribute
2x+2 = 3x-9
Subtract 2x
2x+2-2x = 3x-2x-9
2 = x-9
Add 9 to each side
2+9 =x-9+9
11 =c
All human blood can be "ABO-typed" as O, A, B, or AB, but the distribution of the types varies a bit among groups of people. Here are the distributions of blood types for a randomly chosen person in China and in the United States:The probability O A B ABChinese 0.35 0.27 0.26 0.12American 0.45 0.4 0.11 0.04Suppose we randomly select an American and a Chinese, independently of each other, apply multiplication and addition probability rules, compute:a. Pr(They both have type O)b. Pr( they both have the same blood type)c. Pr( at least one person has type O)
Answer:
a. Pr(They both have type O)
= Pr(They both have type O)
= 0.35 x 0.45
= 0.1575 = 15.75%
b. Pr( they both have the same blood type)
= Pr( they both have the same blood type)
= 2/8
= 0.25 = 25%
c. Pr( at least one person has type O)
= Pr (at least one person has type O)
= 1 - 0.3575
= 0.6425 = 64.25%
Step-by-step explanation:
a) Data:
O A B AB
Chinese 0.35 0.27 0.26 0.12
American 0.45 0.4 0.11 0.04
b) Calculations:
i. Pr(They both have type O)
= Probability of Chinese with O multiplied by Probability of American with O
= 0.35 * 0.45
= 0.1575 = 15.75%
ii. Pr( they both have the same blood type)
= Probability of two out of 8 outcomes
= 2/8
= 0.25 = 25%
iii. Pr( at least one person has type O)
= Probability of (1 – p(none) )
The probability of none = p(none O blood type)
= p(none)
for Chinese = (0.27 + 0.26 + 0.12) * for American ( 0.4 + 0.11 + 0.04)
= 0.65 * 0.55 = 0.3575
Pr (at least one person has type O) = 1 - 0.3575
= 0.6425
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Answer:
[tex] {x}^{2} + 5x + 10[/tex]
Answer:
[tex]\large \boxed{x^2 +5x+10}[/tex]
Step-by-step explanation:
A polynomial is an expression that has variables, coefficients, and constants.
An example of a polynomial can be x² - 6x + 2.
An experimental probability is ______ likely to approach the theoretical probability if the number of trials simulated is larger. A. as B. more C. less D. not
Answer:
B. More
Step-by-step explanation:
This is according to the law of large numbers
An experimental probability is more likely to approach the theoretical probability if the number of trials simulated is larger.
What is an experimental probability and theoretical probability?Experimental probability is an experimental outcome whereas theoretical probability is a possible or expected outcome.
An experimental probability is more likely to approach the theoretical probability if the number of trials increased because of the law of large numbers which states that the average of the results obtained from a large number of trials should be close to the expected value and tends to become closer to the expected value as more trials are performed
Thus using the concept of the law of large numbers we can say that an experimental probability is more likely to approach the theoretical probability.
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Which expression is equivalent to (jk)l? A. (j + k) + l B. j(kl) C. (2jk)l D. (j + k)l
Answer:
B. j(kl)
Step-by-step explanation:
(jk)l
We can change the order we multiply and still get the same result
j(kl)
Answer:
Step-by-step explanation:
its B i did it
Reduce the following fraction to lowest terms: 8/14
Answer:
4/7
Step-by-step explanation:
divide both by two for its simplest form
Answer:4/7
Step-by-step explanation
Divide both the numerator and denominator by 2
The result for the numerator is 8/2=4
that of the denominator is 14/2=7
Therefore the resultant answer is 4/7
What is the solution to the following system of equations? 3x-2y=12 6x - 4y = 24
Answer:
D question,somewhat confusing, itsit's like simultaneous equation,but values are different
Answer:
x = 4 + 2y/3
Step-by-step explanation:
Brian needs to paint a logo using two right triangles. The dimensions of the logo are shown below. What is the difference between the area of the large triangle and the area of the small triangle?
Answer:
7.5 cm²
Step-by-step explanation:
Dimensions of the large ∆:
[tex] base (b) = 3cm, height (h) = 9cm [/tex]
[tex] Area = 0.5*b*h = 0.5*3*9 = 13.5 cm^2 [/tex]
Dimensions of the small ∆:
[tex] base (b) = 2cm, height (h) = 6cm [/tex]
[tex] Area = 0.5*b*h = 0.5*2*6 = 6 cm^2 [/tex]
Difference between the area of the large and the small ∆ = 13.5 - 6 = 7.5 cm²
A maker of microwave ovens advertises that no more than 10% of its microwaves need repair during the first 5 years of use. In a random sample of 50 microwaves that are 5 years old, 12% needed repairs at a=.04 can you reject the makers claim that no more than 10% of its microwaves need repair during the first five years of use?
Answer:
We conclude that no more than 10% of its microwaves need repair during the first five years of use.
Step-by-step explanation:
We are given that a maker of microwave ovens advertises that no more than 10% of its microwaves need repair during the first 5 years of use.
In a random sample of 50 microwaves that are 5 years old, 12% needed repairs.
Let p = population proportion of microwaves who need repair during the first five years of use.
So, Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 10% {means that no more than 10% of its microwaves need repair during the first five years of use}
Alternate Hypothesis, [tex]H_A[/tex] : p > 10% {means that more than 10% of its microwaves need repair during the first five years of use}
The test statistics that will be used here is One-sample z-test for proportions;
T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of microwaves who need repair during the first 5 years of use = 12%
n = sample of microwaves = 50
So, the test statistics = [tex]\frac{0.12-0.10}{\sqrt{\frac{0.10(1-0.10)}{50} } }[/tex]
= 0.471
The value of z-test statistics is 0.471.
Now, at a 0.04 level of significance, the z table gives a critical value of 1.751 for the right-tailed test.
Since the value of our test statistics is less than the critical value of z as 0.471 < 1.751, so we have insufficient evidence to reject our null hypothesis as the test statistics will not fall in the rejection region.
Therefore, we conclude that no more than 10% of its microwaves need repair during the first five years of use.
A rectangle has an area of 81 square centimeters. Which of the following would be the rectangle's length and width? (Area = equals length×times width)
Answer:
length: 9cm
width: 9cm
Step-by-step explanation:
9×9=81
Write "six and thirty-four thousandths" as a decimal
Answer:
6.034
Step-by-step explanation:
6 is a whole number.
.034 because it is 34 thousandths, not 34 hundredths.
The area of the circle x² + y2 - 6x-4y +9 = 0 is
Answer:
Your answer is here.Enjoy dude
Answer:
12.56 unit²
Step-by-step explanation:
Given:x² + y² - 6x - 4y + 9 = 0To find:The area of circleSolution:The form of the circle is:
(x- h)² + (y-k)² = r²Let's bring the given to the form of a circle as above:
x² + y² - 6x - 4y + 9 = 0x² - 6x + y²- 4y + 9 = 0 ⇒ combining like terms and completing squarex² - 6x + 9 + y²- 4y + 4 = 4 ⇒ adding 4 to both sides(x-3)² + (y - 2)² = 2² ⇒ got the form of this circleAs per the form, we got r² = 2², so the radius of circle is 2 units.
The area of circle:
A= πr² = 3.14×2² = 12.56 unit²Find the fourth roots of 16(cos 200° + i sin 200°).
Answer:
See below.
Step-by-step explanation:
To find roots of an equation, we use this formula:
[tex]z^{\frac{1}{n}}=r^{\frac{1}{n}}(cos(\frac{\theta}{n}+\frac{2k\pi}{n} )+\mathfrak{i}(sin(\frac{\theta}{n}+\frac{2k\pi}{n})),[/tex] where k = 0, 1, 2, 3... (n = root; equal to n - 1; dependent on the amount of roots needed - 0 is included).
In this case, n = 4.
Therefore, we adjust the polar equation we are given and modify it to be solved for the roots.
Part 2: Solving for root #1
To solve for root #1, make k = 0 and substitute all values into the equation. On the second step, convert the measure in degrees to the measure in radians by multiplying the degrees measurement by [tex]\frac{\pi}{180}[/tex] and simplify.
[tex]z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(0)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(0)\pi}{4}))[/tex]
[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{4}))[/tex]
[tex]z^{\frac{1}{4}} = 2(sin(\frac{5\pi}{18}+\frac{\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{4}))[/tex]
Root #1:
[tex]\large\boxed{z^\frac{1}{4}=2(cos(\frac{19\pi}{36}))+\mathfrack{i}(sin(\frac{19\pi}{38}))}[/tex]
Part 3: Solving for root #2
To solve for root #2, follow the same simplifying steps above but change k to k = 1.
[tex]z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(1)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(1)\pi}{4}))[/tex]
[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{2\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{2\pi}{4}))\\[/tex]
[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{\pi}{2}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{2}))\\[/tex]
Root #2:
[tex]\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{7\pi}{9}))+\mathfrak{i}(sin(\frac{7\pi}{9}))}[/tex]
Part 4: Solving for root #3
To solve for root #3, follow the same simplifying steps above but change k to k = 2.
[tex]z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(2)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(2)\pi}{4}))[/tex]
[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{4\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{4\pi}{4}))\\[/tex]
[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\pi))+\mathfrak{i}(sin(\frac{5\pi}{18}+\pi))\\[/tex]
Root #3:
[tex]\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{23\pi}{18}))+\mathfrak{i}(sin(\frac{23\pi}{18}))}[/tex]
Part 4: Solving for root #4
To solve for root #4, follow the same simplifying steps above but change k to k = 3.
[tex]z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(3)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(3)\pi}{4}))[/tex]
[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{6\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{6\pi}{4}))\\[/tex]
[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{3\pi}{2}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{3\pi}{2}))\\[/tex]
Root #4:
[tex]\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{16\pi}{9}))+\mathfrak{i}(sin(\frac{16\pi}{19}))}[/tex]
The fourth roots of 16(cos 200° + i(sin 200°) are listed above.
A box of chocolates contains five milk chocolates, three dark chocolates, and four white chocolates. You randomly select and eat three chocolates. The first piece is milk
chocolate, the second is white chocolate, and the third is milk chocolate. Find the probability of this occuring.
Answer:
60/220
Step-by-step explanation:
we use combination,
[tex] (\frac{5}{1} ) \times ( \frac{4}{1} ) \times ( \frac{3}{1} )[/tex]
[tex]5 \times 4 \times 3 = 60[/tex]
then, all divided by,
[tex] (\frac{12}{3}) = 220 [/tex]
[tex]60 \div 220[/tex]
The probability of the first piece being milk chocolate, the second being white chocolate, and the third being milk chocolate is 0.06.
What is Probability?The probability helps us to know the chances of an event occurring.
[tex]\rm Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]
The sample contains five milk chocolates, three dark chocolates, and four white chocolates. Therefore, the probability that the first piece is milk chocolate is
[tex]\rm Probability=\dfrac{\text{Number of Milk choclates}}{\text{Total number of choclates}}[/tex]
[tex]\rm Probability=\dfrac{5}{12}[/tex]
Now, since the chocolate is been eaten the sample size will reduce from 12 chocolates in total to 11 chocolates in total (four milk chocolates, three dark chocolates, and four white chocolates). Therefore, the probability of the second piece being white chocolate is
[tex]\rm Probability=\dfrac{\text{Number of White choclates}}{\text{Total number of choclates}}[/tex]
[tex]\rm Probability=\dfrac{4}{11}[/tex]
Now, as the chocolate is been eaten the sample size will reduce from 11 chocolates in total to 10 chocolates in total (four milk chocolates, three dark chocolates, and three white chocolates). Therefore, the probability of the third piece being milk chocolate is
[tex]\rm Probability=\dfrac{\text{Number of Milk choclates}}{\text{Total number of choclates}}[/tex]
[tex]\rm Probability=\dfrac{4}{10}[/tex]
Thus, the probability of the first piece being milk chocolate, the second being white chocolate, and the third being milk chocolate is
[tex]\rm Probability=\dfrac{5}{12}\times \dfrac{4}{11} \times \dfrac{4}{10} = \dfrac{80}{1320} = 0.06[/tex]
Hence, the probability of the first piece being milk chocolate, the second being white chocolate, and the third being milk chocolate is 0.06.
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An online polling site posed this question: "How much stock do you put in long-range weather forecasts?" Among its Web site users, 38, 528 chose to respond Complete parts (a) through (c) below.
a. Among the responses received, 3% answered with "a lot". What is the actual number of responses consisting of "a lot"?
b. Among the responses received, 18, 566 consisted of "very little or none". What percentage of responses consisted of "very little or none"?
c. Because the sample size of 38, 528 is so large, can we conclude that about 3% of the general population puts "a lot" of stock in long-range weather forecasts? Why or why not?
A. No, because the sample is a voluntary response sample, so the sample is not likely to be representative of the population.
B. Yes, because the sample is so large, the margin of error is negligible.
C. No, because even though the sample size is so large, there is still a margin of error.
D. Yes, because the sample size is large enough so that the sample is representative of the population.
Answer:
(a) 1155.84
(b) 48.2%
(c) D
Step-by-step explanation:
The number of total responses is, N = 38,528.
(a)
It is provided that 3% answered with "a lot".
Compute the actual number of responses consisting of "a lot" as follows:
n (a lot) = N × P (a lot)
= 38528 × 0.03
= 1155.84
Thus, the actual number of responses consisting of "a lot" is 1155.84.
(b)
The number of responses consisting of "very little or none" is,
n (very little or none) = 18,566
Compute the percentage of responses consisted of "very little or none" as follows:
[tex]P(\text{very little or none})=\frac{n(\text{very little or none})}{N}[/tex]
[tex]=\frac{18566}{38528}\\\\=0.481883\\\\\approx 0.482[/tex]
The percentage is: 0.482 × 100% = 48.2%.
Thus, the percentage of responses consisted of "very little or none" is 48.2%.
(c)
As the sample size increases the sample statistic value gets closer and closer to the actual population parameter value.
Thus, making the sample statistic an unbiased estimator of the population parameter.
And proving that the sample is a true representative of the population.
Thus, the correct option is (D).
what is the distance between the first and third quartiles of a data set called?
Answer:
Interquartile range is the distance between the first and third of a data.
Step-by-step explanation:
Hope it will help you :)
PLEASE HELP!!!
Evaluate the expression when b=4 and y= -3
-b+2y
Answer: -10
Step-by-step explanation: All you have to do is plug the values into the equation. -4+2(-3). Then you solve the equation using PEDMAS.
1. -4+2(-3)
2. -4+(-6)
3.-4-6
4.-10
Answer:
8
Step-by-step explanation:
-b + 2y
if
b = 4
and
y = 3
then:
-b + 2y = -4 + 2*6 = -4 + 12
= 8
Stock prices used to be quoted using eighths of a dollar. Find the total price of the transaction. 400 shares of national semi at 135 1/2
Answer:
The value is [tex]T = \$54200[/tex]
Step-by-step explanation:
From the question we are told that
The number of shares is n = 400
The rate of each share is [tex]k = 135\frac{1}{2} = 135.5[/tex]
Generally the total price is mathematically represented as
[tex]T = 400 * 135.5[/tex]
[tex]T = \$54200[/tex]
Give the domain and range of each relation using set notation
Answer:
See below.
Step-by-step explanation:
First, recall the meanings of the domain and range.
The domain is the span of x-values covered by the graph.
And the range is the span of y-values covered by the graph.
1)
So, we have here an absolute value function.
As we can see, the domain of the function is all real numbers because the graph stretches left and right infinitely. Therefore, the domain of the function is:
[tex]\{x|x\in\textbb{R}\}[/tex]
(You are correct!)
For the range, notice how the function stops at y=7. The highest point of the function is (-2,7). There graph doesn't and won't ever reach above y=7. Therefore, the range of the graph is all values less than or equal to 7. In set notation, this is:
[tex]\{y|y\leq 7\}[/tex]
2)
We have here an ellipse.
First, for the domain. We can see the the span of x-values covered by the ellipse is from x=-4 to x=6. In other words, the domain is all values in between these two numbers and including them. Therefore, we can write it as such:
[tex]-4\leq x\leq 6[/tex]
So x is all numbers greater than or equal to -4 but less than or equal to 6. This describes the span of x-values. In set notation, this is:
[tex]\{x|-4\leq x\leq 6\}[/tex]
For the range, we can see that the span of x values covered by the ellipse is from y=-5 to y=1. Just like the domain, we can write it like this:
[tex]-5\leq y\leq 1[/tex]
This represents all the y-values between -5 and 1, including -5 and 1.
In set notation, thi is:
[tex]\{y|-5\leq y\leq 1\}[/tex]
A cabinet door has a perimeter of 76 inches. Its area is 357 square inches. What are the dimensions of the door?
Answer:
17 by 21 inches
Step-by-step explanation:
The perimeter is twice the sum of the dimensions, and the area is their product, so you have ...
L + W = 38
LW = 357
__
Solution:
W(38 -W) = 357 . . . . . substitute for L
-(W^2 -76W) = 357 . . expand on the left
-(W^2 -38 +19^2) = 357 -19^2 . . . . complete the square
(W -19)^2 = 4 . . . . . . . write as a square
W -19 = ±√4 = ±2 . . . take the square root; next, add 19
W = 19 ±2 = {17, 21} . . . . if width is one of these, length is the other
The dimensions are 17 by 21 inches.
Help me and I will for real give u brainleist
should be 2 3 andd 5
think of the - (- as a plus sign (this is what i was always taught) to add them so it would in turn be (-5) + 12 which equals 7 and choice 3 and 5 also equal this
The table shows the height, in meters, of an object that is dropped as time passes until the object hits the ground. A 2-row table with 10 columns. The first row is labeled time (seconds), x with entries 0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.6. The second row is labeled height (meters), h with entries 100, 98.8, 95.1, 89.0, 80.4, 69.4, 55.9, 40.0, 21.6, 0. A line of best fit for the data is represented by h = –21.962x + 114.655. Which statement compares the line of best fit with the actual data given by the table? According to the line of best fit, the object would have hit the ground 0.6 seconds later than the actual time the object hit the ground. According to the line of best fit, the object was dropped from a lower height. The line of best fit correctly predicts that the object reaches a height of 40 meters after 3.5 seconds. The line of best fit predicts a height of 4 meters greater than the actual height for any time given in the table.
Answer: A. According to the line of best fit, the object would have hit the ground 0.6 seconds later than the actual time the object hit the ground.
The statement first "According to the line of best fit, the object would have hit the ground 0.6 seconds later than the actual time the object hit the ground" is correct.
What is the line of best fit?A mathematical notion called the line of the best fit connects points spread throughout a graph. It's a type of linear regression that uses scatter data to figure out the best way to define the dots' relationship.
We have a line of best fit:
h = –21.962x + 114.655
As per the data given and line of best fit, we can say the object would have impacted the ground 0.6 seconds later than it did according to the line of best fit.
Thus, the statement first "According to the line of best fit, the object would have hit the ground 0.6 seconds later than the actual time the object hit the ground" is correct.
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On a class trip with 40 students, 14 are male. What percentage of the class is female?
66%
60%
65%
58%
Answer:
65%
Step-by-step explanation:
If 14 are male, then 26 are female.
To find the percent female, divide the number of females by the total.
26/40 = 0.65
So, the percentage of the class that is female is 65%
Answer:
C. 65%
Step-by-step explanation:
We know that of the 40 total students, 14 are male, which means the remaining students are female.
To find how many are female, we subtract 14 from 40:
40 - 14 = 26 females
Percentage is simply a part divided by a whole, multiplied by 100. Here, the "part" is the number of females, which is 26. The "whole" is the total number of students, which is 40. So, we have:
(26 / 40) * 100 = 65
The answer is thus C, 65%.
~ an aesthetics lover
if f(x)=3x-3 and g(x)=-x2+4,then f(2)-g(-2)=
Answer:
3
Step-by-step explanation:
f(x)=3x-3
g(x)=-x^2+4,
f(2) = 3(2) -3 = 6-3 =3
g(-2) = -(-2)^2+4 = -4+4 = 0
f(2)-g(-2)= = 3-0 = 3
What does "C" represent and how do you evaluate this?
[tex]_9C_7=\dfrac{9!}{7!2!}=\dfrac{8\cdot9}{2}=36[/tex]
Find the area of the shaded regions:
area of Arc subtending [tex]360^{\circ}[/tex] (i.e. the whole circle) is $\pi r^2$
so area of Arc subtending $\theta^{\circ}$ is, $\frac{ \pi r^2}{360^{\circ}}\times \theta^{\circ}$
$\theta =72^{\circ}$ so the area enclosed by one such arc is $\frac{\pi (10)^272}{360}$
abd there are 2 such arcs, so double the area.
[tex] \LARGE{ \underline{ \boxed{ \rm{ \purple{Solution}}}}}[/tex]
Given:-Radius of the circle = 10 inchesAngle of each sector = 72°Number of sectors = 2To FinD:-Find the area of the shaded regions....?How to solve?For solving this question, Let's know how to find the area of a sector in a circle?
[tex] \large{ \boxed{ \rm{area \: of \: sector = \frac{\theta}{360} \times \pi {r}^{2} }}}[/tex]
Here, Θ is the angle of sector and r is the radius of the circle. So, let's solve this question.
Solution:-We have,
No. of sectors = 2Angle of sector = 72°By using formula,
⇛ Area of shaded region = 2 × Area of each sector
⇛ Area of shaded region = 2 × Θ/360° × πr²
⇛ Area of shaded region = 2 × 72°/360° × 22/7 × 10²
⇛ Area of shaded region = 2/5 × 100 × 22/7
⇛ Area of shaded region = 40 × 22/7
⇛ Area of shaded region = 880/7 inch. sq.
⇛ Area of shaded region = 125.71 inch. sq.
☄ Your Required answer is 125.71 inch. sq(approx.)
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The probability density function for random variable W is given as follows: Let x be the 100pth percentile of W and y be the 100(1 – p)th percentile of W, where 0
Answer:
Step-by-step explanation:
A probability density function (pdf) is used for continuous random variables. That is why p is between 0 and 1 (the two extremes - 0 and 1 - exclusive).
X = 100pth percentile of W
Y = 100(1-p)th percentile of W
Expressing Y as a function of X;
Y = 100(1-p)th = 100th - 100pth
Recall that 100pth is same as X, so substitute;
Y = 100th - X
where 100th = hundredth percentile of W and X = 100pth percentile of W
Find the rectangular coordinates of the point with the given polar coordinates.
Answer:
[tex]( - \sqrt{3} \: an d \: 1)[/tex]
Which of the following is an arithmetic sequence? A.-2, 4, -6, 8, ... B.2, 4, 8, 16, ... C.-8, -6, -4, -2, ...
Answer:
C. -8, -6, -4, -2, ...
Step-by-step explanation:
An arithmetic sequence increases by the same amount every time through addition or subtraction. There is a common difference.
A: -2, 4, -6, 8, ... If there were a common difference, the numbers would not switch between being positive and back to negative. The numbers would either keep going positive or keep going negative.
B: 2, 4, 8, 16, ... The common difference between 16 and 8 is 16 - 8 = 8. The difference between 8 and 4 is 8 - 4 = 4. Since the difference changes between the numbers, this is not an arithmetic sequence.
C. -8, -6, -4, -2, ... The common difference between -2 and -4 is -2 - (-4) = -2 + 4 = 2. The difference between -4 and -6 is -4 - (-6) = -4 + 6 = 2. The difference between -6 and -8 is -6 - (-8) = -6 + 8 = 2. Since the common difference is always two, this is an arithmetic sequence.
Hope this helps!
How do i do this equation
-3(-2y-4)-5y-2=
Answer:
combined like terms and then follow the order of operations.
Step-by-step explanation:
Shawna finds a study of American men that has an equation to predict weight (in pounds) from
height (in inches): y = -210 + 5.6x. Shawna's dad's height is 72 inches and he weighs 182 pounds.
What is the residual of weight and height for Shawna's dad?
a. 11.2 pounds
b. -11.2 pounds
c. 193.2 pounds
d. 809.2 pounds
Answer:
-11.2 pounds
Step-by-step explanation:
It is given that,
Shawna finds a study of American men that has an equation to predict weight (in pounds) from height (in inches):
y = -210 + 5.6x
Height of Shawna's dad is 72 inches
Weight is 182 pounds
We need to find the residual of weight and height for Shawna's dad.
Predicted weight of 72 inches men,
y' = -210 + 5.6(72)
y' = 193.2 pounds
So, residual is :
Y = 182 - 193.2
Y = -11.2 pounds
So, the residual of weight and height for Shawna's dad is -11.2 pounds.
Answer:
-11.2 pounds
Step-by-step explanation:
Got it right on the test.