Answer:
the answer to the question is - 2 over 7 and - 12 over 7
Peter Piper picked a pickled pepper out of a pepper jar. If the probability of drawing a pickled pepper was 2/5,
how many total peppers could be in the jar (psst. you can't have a half of a pepper)?
Answer:
5
Step-by-step explanation:
2/5 pickled peppers means that there are 2 pickled peppers out of 5 total peppers.
Which of the following is the square of a binomial?
Answer:
[tex]16 {x}^{2} + 24xy + 9 {y}^{2} [/tex]
Step-by-step explanation:
U can reduce this into
[tex](4x + 3y) {}^{2} [/tex]
Which is a square.
Now,
16x^2 + 24xy + 9y^2
16x^2 + 12xy + 12xy + 9y^2
4x ( 4x + 3y ) + 3y ( 4x + 3y )
( 4x + 3y ) ( 4x + 3y )
( 4x + 3y )^2
I hope you understand...
Mark me as brainliest...
please help me...........
Defined the total variation distance to be a distance TV(P,Q) between two probability measures P and Q. However, we will also refer to the total variation distance between two random variables or between two pdfs or two pmfs, as in the following.
Compute TV(X,X+a) for any a∈(0,1), where X∼Ber(0.5).
TV(X,X+a) = ?
Answer:
1
Step-by-step explanation:
Computing Tv(X, X + a ) for any a∈(0,1)
Given that : X∼Ber(0.5)
∴ The probability mass function
P(X = 1 ) = 0.5
P(X = 0) = ( 1 - 0.5 )
and expectation E[X] = 0.5
hence ; TV ( X, X + a ) = 1
Someone please help with the questions on this picture!! URGENT!!!
Answer:
A) Independent
B) Dependent
Step-by-step explanation:
A) If we take a marble out and put the marble back, it means we have restored the sample to what it was initially and thus it doesn't affect probability of making another selection.
Thus, this is an independent event.
B) A card is taken from a deck of cards without replacement and set aside. Then after that another card is taken from the first sample, this means that the first sample size has now reduced and thus the first card taken affects the probability of the second card to be picked. Thus, this is a dependent event.
Point B has coordinates (4,2). The x-coordinate of point A is - 1. The distance between point A and
point B is 13 units. What are the possible coordinates of point A?
Answer:
A (-1,-10) ; A (-1,14)
Step-by-step explanation:
[tex]\sqrt{(-1-4)^2 + (y-2)^2} = 13 \\ 25 + y^2 + 4 -4y = 169[/tex]
y^2 -4y - 140 = 0
Δ/4 = 4 + 140 = 144
y1 = 2 + 12 = 14
y2 = 2 -12= -10
A customer buys a different book that has an original selling price of $38. The book is discounted 25%. The customer must pay a 6% sales tax on the discounted price of the book.
What is the total amount, in dollars, the customer pays for the discounted book? Explain and SHOW how you arrived at your answer.
Answer:
$30.21
Step-by-step explanation:
100% -25%= 75%
Discounted price of the book
= 75% ×$38
= $28.50
Since the customer must pay an additional 6% of the discounted price,
percentage of discounted price paid
= 100% +6%
= 106%
Total amount paid
= 106% × $28.50
= $30.21
_________________________________
Alternative working:
Original selling price= $38
Since the book is discounted 25%,
100% ----- $38
1% ----- $0.38
75% ----- 75 ×$0.38= $28.50
Since the sales tax is based on the discounted price, we let the discounted price be 100%.
100% ----- $28.50
1% ----- $0.285
106% ----- 106 ×$0.285= $30.21
∴ The total amount the customer pays for the discounted book is $30.21.
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Khan Academy
Dependent probability
In a class of 7, there are 4 students who play soccer.
If the teacher chooses 3 students, what is the probability that none of the three of them play soccer?
Answer:
[tex]\frac{12}{49}[/tex]
Step-by-step explanation:
[tex]\frac{4}{7} *\frac{3}{7} = \frac{12}{49}[/tex]
Hope this helps.
Mô hình quy hồi tuyến tính và ứng dụng
Answer:
where are you from Korea or not
Hope it helps you!
-miraculousfanx-
. A population of rabbits oscillates 25 above and below an average of 129 during the year, hitting the lowest value in January (t = 0). a. Find an equation for the population, P, in terms of the months since January, t. b. What if the lowest value of the rabbit population occurred in April instead?
Answer:
Because we know that here we have an oscillation, we can model this with a sine or cosine function.
P = A*cos(k*t) + M
where:
k is the frequency
A is the amplitude
M is the midline
We know that at t = 0, we have the lowest population.
We know that the mean is 129, so this is the midline.
We know that the population oscillates 25 above and below this midline,
And we know that for t = 0 we have the lowest population, so:
P = A*cos(k*0) + 129 = 129 - 25
P = A + 129 = 129 - 25
A = -25
So, for now, our equation is
P = -25*cos(k*t) + 129
Because this is a yearly period, we should expect to see the same thing for t = 12 (because there are 12 months in one year).
And remember that the period of a cosine function is 2*pi
Then:
k*12 = 2*pi
k = (2*pi)/12 = pi/6
Finally, the equation is:
P = -25*cos(t*pi/6) + 129
Now we want to find the lowest population was in April instead:
if January is t = 0, then:
February is t = 2
March is t = 3
April is t = 4
Then we would have that the minimum is at t = 4
If we want to still use a cosine equation, we need to use a phase p, such that now our equation is:
P = -25*cos(k*t + p) + 129
Such that:
cos(k*4 + p) = 1
Then:
k*4 + p = 0
p = -k*4
So our equation now is:
P = -25*cos(k*t - 4*k) + 129
And for the periodicity, after 12 months, in t = 4 + 12 = 16, we should have the same population.
Then, also remembering that the period of the cosine function is 2*pi:
k*12 - 4*k = 2*pi
k*8 = 2*pi
k = 2*pi/8 = pi/4
And remember that we got:
p = -4*k = -4*(pi/4) = -pi
Then the equation for the population in this case is:
P = -25*cos( t*pi/4 - pi) + 129
Help. Urgent. Skenekeks
Answer:
D
Step-by-step explanation:
Plug in the numbers for the x-value and solve the equation.
| 3(80/3)/4 + 1 | = 16
| 3(-40/3)/4 + 1 | = 16
If an average-sized man with a parachute jumps from an airplane, he will fall
12.5(0.2t − 1) + 21t feet
in t seconds. How long will it take him to fall 150 feet? (Round your answer to two decimal places.)
Answer:
It will take him 5.85 seconds.
Step-by-step explanation:
12.5 (0.2t - 1) + 21t = 150
Use Distributive Property:
2.5t - 12.5 + 21t = 150
Combine like terms:
23.5t - 12.5 = 150
Subtract 12.5 from both sides:
23.5t = 137.5
Divide both sides by 23.5 to isolate variable t:
5.851063.....
Round to two decimal places (hundredths place):
5.85
a system of regular payments for when something bad happens
a. Directly b. Reasonable c. Insurance
d. Tuition
the answer is b. insurance
Neglecting air resistance and the weight of the propellant, determine the work done in propelling a five-ton satellite to a height of (a) 100 miles above Earth and (b) 300 miles above Earth.
Answer:
a) the work done in propelling a five-ton satellite to a height of 100 miles above Earth is 487.8 mile-tons
b) the work done in propelling a five-ton satellite to a height of 300 miles above Earth is 1395.3 mile-tons
Step-by-step explanation:
Given the data in the question;
We know that the weight of a body varies inversely as the square of its distance from the center of the earth.
⇒F(x) = c / x²
given that; F(x) = five-ton = 5 tons
we know that the radius of earth is approximately 4000 miles
so we substitute
5 = c / (4000)²
c = 5 × ( 4000 )²
c = 8 × 10⁷
∴ Increment of work is;
Δw = [ ( 8 × 10⁷ ) / x² ] Δx
a) For 100 miles above Earth;
W = ₄₀₀₀∫⁴¹⁰⁰ [ ( 8 × 10⁷ ) / x² ] Δx
= (8 × 10⁷) [tex][[/tex] [tex]-\frac{1}{x}[/tex] [tex]]^{4100}_{4000[/tex]
= (8 × 10⁷) [tex][[/tex] [tex]-\frac{1}{4100}[/tex] [tex]+\frac{1}{4000}[/tex] [tex]][/tex]
= (8 × 10⁷ ) [ 6.09756 × 10⁻⁶ ]
= 487.8 mile-tons
Therefore, the work done in propelling a five-ton satellite to a height of 100 miles above Earth is 487.8 mile-tons
b) For 300 miles above Earth.
W = ₄₀₀₀∫⁴³⁰⁰ [ ( 8 × 10⁷ ) / x² ] Δx
= (8 × 10⁷) [tex][[/tex] [tex]-\frac{1}{x}[/tex] [tex]]^{4300}_{4000[/tex]
= (8 × 10⁷) [tex][[/tex] [tex]-\frac{1}{4300}[/tex] [tex]+\frac{1}{4000}[/tex] [tex]][/tex]
= (8 × 10⁷ ) [ 1.744186 × 10⁻⁵ ]
= 1395.3 mile-tons
Therefore, the work done in propelling a five-ton satellite to a height of 300 miles above Earth is 1395.3 mile-tons
NO LINKS OR ELSE YOU'LL BE REPORTED! Only answer if you're very good at Math.No guessing please.
The sample space,S,of a coin being tossed three times is shown below, where H and T denote the coin landing on heads and tails respectively.
Answer: Bottom left corner
=======================================================
Explanation:
There are only four possible outcomes here
A) we get all tails, ie getting 0 headsB) we get exactly one head (the rest tails)C) we get exactly 2 headsD) we get all three headsBased on this so far, the answer is either the table in the bottom left corner or in the top right corner. It's not possible for X = 4 since we only flipped 3 coins.
The probability of case A happening is 1/8 since we have 1 scenario that's all tails (TTT) out of 8 items in the sample space. Similarly, the probability for case D is the same probability. We only have one HHH out of 8 total items.
The probabilities of cases B and C are the same. Both are 3/8. Note that for case B, we have HTT, THT, TTH which is three occurrences in which we get exactly 1 head. So that explains the 3/8.
x( 3x - 2y + 4z)x = -2, y = 4, and z = -3
x^3y+2x^2y^2+xy^3 and 2x^3+4x^2y+2xy^2 Find the HCF.
Answer:
[tex]x(x+y)^2[/tex]
Step-by-step explanation:
We are given that
[tex]x^3y+2x^2y^2+xy^3[/tex] and [tex]2x^3+4x^2y+2xy^2[/tex]
We have to find HCF.
[tex]x^3y+2x^2y^2+xy^3=xy(x^2+2xy+y^2)[/tex]
=[tex]xy(x+y)^2[/tex]
By using the formula
[tex](x+y)^2=x^2+2xy+y^2[/tex]
[tex]xy(x+y)^2=x\times y\times (x+y)^2[/tex]
[tex]2x^3+4x^2y+2xy^2=2x(x^2+2xy+y^2)[/tex]
[tex]=2x(x+y)^2[/tex]
[tex]2x(x+y)^2=2\times x\times (x+y)^2[/tex]
HCF of ([tex]x^3y+2x^2y^2+xy^3,2x^3+4x^2y+2xy^2[/tex])
[tex]=x(x+y)^2[/tex]
19. Which of the following would best be solved using factoring the difference of squares?
O x^3 + 5x^2 - 9x - 45 = 0
O 3x² + 12x = 8
O x^2 - 25 = 0
O x^2 + 3x – 10 = 0
Please hurry!
Answer:
x² + 3x - 10 = 0
x² - 25 = 0
Use the limit definition of the derivative to find the instantaneous rate of change of
f(x)=5x^2+3x+3 at x=4
[tex]f'(4) = 43[/tex]
Explanation:
Given: [tex]f(x)=5x^2 + 3x + 3[/tex]
[tex]\displaystyle f'(x)= \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}[/tex]
Note that
[tex]f(x+h) = 5(x+h)^2 + 3(x+h) + 3[/tex]
[tex]\:\:\;\:\:\:\:= 5(x^2 + 2hx + h^2) + 3x + 3h +3[/tex]
[tex]\:\:\;\:\:\:\:= 5x^2 + 10hx + 5h^2 + 3x + 3h +3[/tex]
Substituting the above equation into the expression for f'(x), we can then write f'(x) as
[tex]\displaystyle f'(x) = \lim_{h \to 0} \dfrac{10hx + 3h + 5h^2}{h}[/tex]
[tex]\displaystyle\:\:\;\:\:\:\:= \lim_{h \to 0} (10x +3 +5h)[/tex]
[tex]\:\:\;\:\:\:\:= 10x + 3[/tex]
Therefore,
[tex]f'(4) = 10(4) + 3 = 43[/tex]
What is the cube root of -1,000p12q3?
O-1004
O - 10pta
O 1004
O 10pta
Answer:
Your options are not clear
Step-by-step explanation:
[tex]\sqrt[3]{-1000 \times p^{12} \times q^3} \\\\(-1 \times 10^3 \times p^{12} \times q^3)^{\frac{1}{3} }\\\\(-1^3)^{\frac{1}{3} }\times 10^{3 \times \frac{1}{3} } \times p^{12 \times \frac{1}{3}} \times q^{3 \times \frac{1}{3}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ (-1)^3 = - 1 \ ] \\\\- 1 \times 10 \times p^4 \times q\\\\-10p^4q[/tex]
Find the value of k, if (x - 2) is a factor of the polynomial p(x) = 2x2 + 3x - k
Answer:
The value of k is 14.
Step-by-step explanation:
(x - 2) is a factor of the polynomial
[tex]x - 2 = 0 \rightarrow x = 2[/tex]
This means that [tex]p(2) = 0[/tex]
p(x) = 2x² + 3x - k
[tex]p(2) = 2(2)^2 + 3(2) - k[/tex]
[tex]0 = 8 + 6 - k[/tex]
[tex]14 - k = 0[/tex]
[tex]k = 14[/tex]
The value of k is 14.
Based on the Pythagorean theorem , find the missing length for each of the given right triangles
Answer:
See Explanation
Step-by-step explanation:
The question is incomplete, as the right trianglea are not given. The general explanation is as follows.
Using Pythagoras Theorem, we have:
a² = b² + c²
Where:
a => hypotenuse
Assume that the opposite and the adjacent sides are given as 3 and 4, respectively.
The hypotension becomes
a² = 3² + 4²
a² = 9 + 16.
a² = 25
Take square roots.
a = 5
If any of the other side lengths is missing; you make that side the subject and then solve.
A bacteria culture is growing at a rate of
r(t) = 7e^0.6t
thousand bacteria per hour after t hours. How much did the bacteria population increase during the first two hours? (Round your answer to three decimal places.)
Answer:
[tex]{ \bf{r(t) = 7e {}^{0.6t} }} \\ { \tt{r(2) = 7 {e}^{0.6 \times 2} }} \\ = { \tt{7 {e}^{1.2} }} \\ = 23.241 \: thiusand bacteria \: per \: hour[/tex]
Use a net to find the surface area of the cone
to the nearest square centimeter. Use 3.14 for
20 cm
TT.
Answer:
4444
Step-by-step explanation:
Answer:
819
Step-by-step explanation:
addinh jndenf,r fm,fd vm,fd jngtjgntftb n
bm bm tm mt m
tmknmenmgv
etab
etbbbbbbehgeb
tbbbbbbbbbbb
If 12 girls can sweep a room in 20hours, how many hours will it take 8 girls to perform the same task, assuming they are sweeping at the same rate?
Answer:
30 hour
Step-by-step explanation:
girls time
12 20 hour
8 x(let)
now,
12/8=x/20
12×20=8×x
240=8x
x=240/8
x=30,,
What’s the solution
Answer:
x ≥ 12
Step-by-step explanation:
-3/4x +2 ≤ -7
Subtract 2 from each side
-3/4x +2-2 ≤ -7-2
-3/4x ≤ -9
Multiply each side by -4/3, remembering to flip the inequality
-3/4x * -4/3 ≥ - 9 *(-4/3)
x ≥ 12
Answer:
x>=12
Step-by-step explanation:
-3/4x + 2<=-7
-3/4x <= -7 -2
-3/4x<=-9
cross multiply
-3x<=-36
dividing throughout by -3
x>=12
The equation of the line passing through (2, 3) with a slope of 5 is y = [] x - []
what are the answers to []
Answer:
y = 5x - 7
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Here m = 5, then
y = 5x + c ← is the partial equation
To find c substitute (2, 3) into the partial equation
3 = 10 + c ⇒ c = 3 - 10 = - 7
y = 5x - 7 ← equation of line
find the equation of the line passing through points A(3,4) and B(1,10)
Answer:
y = -3x + 13
Step-by-step explanation:
First, find the slope:
[tex]m=\frac{y_1-y_2}{x_1-x_2}\\\\m=\frac{4-10}{3-1}\\\\m=\frac{-6}{2}\\\\m=-3[/tex]
Finally, find the equation:
[tex]y-y_1=m(x-x_1)\\\\y-4=-3(x-3)\\\\y-4=-3x+9\\\\y=-3x+13[/tex]
HELP ME PLEASE!!!
GIVEN sin0= √23/12
tan0= √23/11
Find cos0
Answer:
[tex]cos \theta = \frac{11}{12}[/tex]
Step-by-step explanation:
[tex]sin \theta = \frac{\sqrt{23}}{12} \ , \ tan \theta = \frac{\sqrt{23}}{11}\\\\tan \theta = \frac{sin \theta }{cos \theta }\\\\ \frac{\sqrt{23}}{11} = \frac{\frac{\sqrt{23}}{12} }{cos \theta}\\\\cos \theta = \frac{\frac{\sqrt{23}}{12} }{\frac{\sqrt{23}}{11} }\\\\cos \theta = \frac{\sqrt{23}}{12 } \times \frac{11}{\sqrt{23}}\\\\cos \theta = \frac{11}{12}[/tex]
Finding the Area of a Circle Given the Radius Th It The area in terms of pi isi mi? The approximated value for the area is A circle has a radius of 3 miles. Use the work shown below to identify the area in terms of pi and the approximate area of the circle. Use 3.14 for a and round the answer to the nearest tenth. A = 2 A= T(3 mi) A = 3.14(9 mi)
Answer:
I'd use A = πr^2
The area is 28.3 if we're using 3.14 as pi (rounded to the nearest tenth)
