Answer:
The answer is below
Step-by-step explanation:
The box contains 5 red and 4 white balls.
A) The probability that at least 1 ball was red = P(both are red) + P(first is red and second is white) + P(first is white second is red)
Given that the first ball was (Upper A )Replaced before the second draw:
P(both are red) = P(red) × P(red) = 5/9 × 5/9 = 25/81
P(first is red and second is white) = P(red) × P(white) = 5/9 × 4/9 = 20/81
P(first is white and second is red) = P(white) × P(red) = 4/9 × 5/9 = 20/81
The probability that at least 1 ball was red = 25/81 + 20/81 + 20/81 = 65/81
B) The probability that at least 1 ball was red = P(both are red) + P(first is red and second is white) + P(first is white second is red)
Given that the first ball was not Replaced before the second draw:
P(both are red) = P(red) × P(red) = 5/9 × 4/8 = 20/72 (since it was not replaced after the first draw the number of red ball remaining would be 4 and the total ball remaining would be 8)
P(first is red second is white) = P(red) × P(white) = 5/9 × 4/8 = 20/72
P(first is white and second is red) = P(white) × P(red) = 4/9 × 5/8 = 20/72
The probability that at least 1 ball was red = 20/72 + 20/72 + 20/72 = 60/72
How do I solve these equations:
sin(2θ) + sin θ = 0
sin(2θ) = sqrt 3 cos θ
for intervals 0=< θ < 2pi
Answer:
Step-by-step explanation:
Given that
sin(2θ)+sinθ=0
We know that
sin(2θ)=2 sinθ x cosθ
Therefore
2 sinθ x cosθ + sinθ=0
sinθ(2 cosθ+1)=0
sinθ= 0
θ=0
2 cosθ+1=0
cosθ= - 1/2
θ=120°
_______________________________________________________
[tex]sin 2\theta=\sqrt{3cos\theta}[/tex]
By squaring both sides
[tex]sin^2 2\theta={3cos\theta}[/tex]
4 sin²θ x cos²θ=3 cosθ
4 sin²θ x cos²θ - 3 cosθ=0
cos θ = 0
θ= 90°
4 sin²θ=3
θ=60°
find area of a square garden having a length 45m
Answer:
A = 2025 ft^2
Step-by-step explanation:
The area of a square is given by
A = s^2 where s is the side length
A = 45^2
A = 2025 ft^2
Answer:
Step-by-step explanation:
The area of any square is s^2
s in this case is 45
Area = s^2
Area = 45 * 45
Area = 2025 m^2
Which of the proportions below could be used to convert 2500 kilograms to grams?
A. 1000/1 = x/2500
B. 100/1 = x/2500
C. 1/100 = x/2500
D. 1/10 = x/2500
Answer:
A. 1000/1 = x/2500
Step-by-step explanation:
1000 g = 1 kg
Using ratios
1000 g x g
------------ = --------------
1 kg 25 00 kg
Answer:
A. 1000/1 = x/2500
Step-by-step explanation:
KHD - DCM
Kilo - Hecta - Deca - Base - Deci - Centi - Mili
*When moving from left to right, each spot you pass multiplis your number by 10.
2500 kilo → x grams (base)
Proportions:
grams/kilos = x/2500
*For every 1 kilo, there is 1000 grams.
1000/1 = x/2500
Need help!!!! Show work plz
Answer:
24 units²
Step-by-step explanation:
A rhombus is divided into 4 right triangles when it's two diagonals intersect at right angles. All the sides are of equal lengths.
Therefore, a simple method to use to find the area of the given rhombus is to calculate the area of one of the right triangles, and multiply by 4.
Area of right triangle = ½*base*height
Height = 3
Base = [tex]\sqrt{5^2 - 3^2} = \sqrt{16} = 4[/tex] (Pythagorean theorem)
Area of right triangle = ½*4*3 = 2*3 = 6 units²
Area of rhombus = 4(6 units²) = 24 units²
a can complete a piece of work in 10 days B can do it in 12 days C can do it in 15 days in how many days will they finish the work if they work together
hope it helps you
The width of a rectangle is three units less than the length if the area is 28 square units then find the dimensions of the rectangle
Let
width be xLength=x+3ATQ
[tex]\\ \sf \longmapsto Area=Length\times Width[/tex]
[tex]\\ \sf \longmapsto x(x+3)=28[/tex]
[tex]\\ \sf \longmapsto x^2+3x=28[/tex]
[tex]\\ \sf \longmapsto x^2+3x-28=0[/tex]
[tex]\\ \sf \longmapsto x^2+7x-4x-28=0[/tex]
[tex]\\ \sf \longmapsto x(x+7)-4(x+7)=0[/tex]
[tex]\\ \sf \longmapsto (x-4)(x+7)=0[/tex]
[tex]\\ \sf \longmapsto x=4\:or\:x=-7[/tex]
Ignore negative value[tex]\\ \sf \longmapsto Width=4units[/tex]
[tex]\\ \sf \longmapsto Length=4+3=7units[/tex]
1
Drag and drop the
labels to the correct
sides using Angle A
as a reference.
A
boy
3
4
hypotenuse
adjacent
opposite
5
6
hypotenuse goes to the line across from the right angle. adjacent is the bottom one. lastly opposite is the left one.
Step-by-step explanation:
Hi, there!!
According to the question, we should find the hypotenuse, adjacent, and opposite to the refrence angle A ,right.
so, let's simply work with it,
hypotenuse (h)= AC {side opposite to the 90° is always a hypotenuse}.
opposite (p)= BC { as the side opposite to the refrence angle is always perpendicular or opposite}
adjacent (b)= AB { as remaining side is always base or adjacent}
Hope it helps....
Twenty-seven minus 3/2 of a number (x) os not mpre than 36. What is the number?
Answer:
-6
Step-by-step explanation:
27-3/2x=36
53-3x=72
3x=-18
x=-6
In a lottery game, a player picks shox numbers from 1 to 20. If the player matches all six numbers, they win 20,000 dollars. Otherwise, they lose $1. What is the expected value of this game? $
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ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ
ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ
What are the solutions of the equation x4 + 6x2 + 5 = 0? Use u substitution to solve.
x = i and x = i5
x=+ i and x
x= +115
O x=V-1 and x = = -5
x=+ -1 and x = = -5
Answer:
A; The first choice.
Step-by-step explanation:
We have the equation [tex]x^4+6x^2+5=0[/tex] and we want to solve using u-substitution.
When solving by u-substitution, we essentially want to turn our equation into quadratic form.
So, let [tex]u=x^2[/tex]. We can rewrite our equation as:
[tex](x^2)^2+6(x^2)+5=0[/tex]
Substitute:
[tex]u^2+6u+5=0[/tex]
Solve. We can factor:
[tex](u+5)(u+1)=0[/tex]
Zero Product Property:
[tex]u+5=0\text{ and } u+1=0[/tex]
Solve for each case:
[tex]u=-5\text{ and } u=-1[/tex]
Substitute back u:
[tex]x^2=-5\text{ and } x^2=-1[/tex]
Take the square root of both sides for each case. Since we are taking an even root, we need plus-minus. Thus:
[tex]x=\pm\sqrt{-5}\text{ and } x=\pm\sqrt{-1}[/tex]
Simplify:
[tex]x=\pm i\sqrt{5}\text{ and } x=\pm i[/tex]
Our answer is A.
Prove that A.M, G.M. and H.M between any two unequal positive numbers satisfy the following relations.
i. (G.M)²= (A.M)×(H.M)
ii.A.M>G.M>H.M
Answer:
See below
Step-by-step explanation:
we want to prove that A.M, G.M. and H.M between any two unequal positive numbers satisfy the following relations.
(G.M)²= (A.M)×(H.M) A.M>G.M>H.Mwell, to do so let the two unequal positive numbers be [tex]\text{$x_1$ and $x_2$}[/tex] where:
[tex] x_{1} > x_{2}[/tex]the AM,GM and HM of [tex]x_1[/tex] and[tex] x_2[/tex] is given by the following table:
[tex]\begin{array}{ |c |c|c | } \hline AM& GM& HM\\ \hline \dfrac{x_{1} + x_{2}}{2} & \sqrt{x_{1} x_{2}} & \dfrac{2}{ \frac{1}{x_{1} } + \frac{1}{x_{2}} } \\ \hline\end{array}[/tex]
Proof of I:[tex] \displaystyle \rm AM \times HM = \frac{x_{1} + x_{2}}{2} \times \frac{2}{ \frac{1}{x_{1} } + \frac{1}{x_{2}} } [/tex]
simplify addition:
[tex] \displaystyle \frac{x_{1} + x_{2}}{2} \times \frac{2}{ \dfrac{x_{1} + x_{2}}{x_{1} x_{2}} } [/tex]
reduce fraction:
[tex] \displaystyle x_{1} + x_{2} \times \frac{1}{ \dfrac{x_{1} + x_{2}}{x_{1} x_{2}} } [/tex]
simplify complex fraction:
[tex] \displaystyle x_{1} + x_{2} \times \frac{x_{1} x_{2}}{x_{1} + x_{2}} [/tex]
reduce fraction:
[tex] \displaystyle x_{1} x_{2}[/tex]
rewrite:
[tex] \displaystyle (\sqrt{x_{1} x_{2}} {)}^{2} [/tex]
[tex] \displaystyle AM \times HM = (GM{)}^{2} [/tex]
hence, PROVEN
Proof of II:[tex] \displaystyle x_{1} > x_{2}[/tex]
square root both sides:
[tex] \displaystyle \sqrt{x_{1} }> \sqrt{ x_{2}}[/tex]
isolate right hand side expression to left hand side and change its sign:
[tex]\displaystyle\sqrt{x_{1} } - \sqrt{ x_{2}} > 0[/tex]
square both sides:
[tex]\displaystyle(\sqrt{x_{1} } - \sqrt{ x_{2}} {)}^{2} > 0[/tex]
expand using (a-b)²=a²-2ab+b²:
[tex]\displaystyle x_{1} -2\sqrt{x_{1} }\sqrt{ x_{2}} + x_{2} > 0[/tex]
move -2√x_1√x_2 to right hand side and change its sign:
[tex]\displaystyle x_{1} + x_{2} > 2 \sqrt{x_{1} } \sqrt{ x_{2}}[/tex]
divide both sides by 2:
[tex]\displaystyle \frac{x_{1} + x_{2}}{2} > \sqrt{x_{1} x_{2}}[/tex]
[tex]\displaystyle \boxed{ AM>GM}[/tex]
again,
[tex]\displaystyle \bigg( \frac{1}{\sqrt{x_{1} }} - \frac{1}{\sqrt{ x_{2}}} { \bigg)}^{2} > 0[/tex]
expand:
[tex]\displaystyle \frac{1}{x_{1}} - \frac{2}{\sqrt{x_{1} x_{2}} } + \frac{1}{x_{2} }> 0[/tex]
move the middle expression to right hand side and change its sign:
[tex]\displaystyle \frac{1}{x_{1}} + \frac{1}{x_{2} }> \frac{2}{\sqrt{x_{1} x_{2}} }[/tex]
[tex]\displaystyle \frac{\frac{1}{x_{1}} + \frac{1}{x_{2} }}{2}> \frac{1}{\sqrt{x_{1} x_{2}} }[/tex]
[tex]\displaystyle \rm \frac{1}{ HM} > \frac{1}{GM} [/tex]
cross multiplication:
[tex]\displaystyle \rm \boxed{ GM >HM}[/tex]
hence,
[tex]\displaystyle \rm A.M>G.M>H.M[/tex]
PROVEN
22 tons is equivalent to ______ kilograms.
Answer:
20000 kg
Step-by-step explanation:
Recall that 1 kg = 2.2 lb approximately. Then:
22 tons 1 kg 2000 lb
------------ * ------------ * -------------- = 20000 kg
1 2.2 lb 1 ton
For a given confidence level, t ? df is larger than z ? . Explain how t ∗ df being slightly larger than z ∗ affects the width of the confidence interval.
Answer:
Answer is below
Step-by-step explanation:
The width of the CI is directly proportional to critical value. When t* is greater than z value, the t value would then cause the margin of error to be larger and this will in turn cause the width of the confidence interval to be larger.
Greater t*df than z* gives us a bigger margin of error. This would in turn give bigger width of confidence interval. t distribution has greater width confidence interval compared to z distribution.
The width of confidence interval is a function of the margin of error. If the critical value of t(t*) is slightly larger than the critical value of z(z*), then the width of the confidence interval will be larger.
The margin of error is the product of the critical value and the standard error. Therefore, given the same standard error value, the value of the margin of error will increases based on the value of the critical value.
Since, t* is slightly larger than z*, then the confidence interval, t will be wider.
Learn more : https://brainly.com/question/18405415
Ae
A Man travels distance
84km in Ihr 20mins. Find
His speed average find-
Answer:
1.05km/min
Step-by-step explanation:
[tex]Distance = 84 km\\\\Time = 1hr 20 min= 80minuttes\\\\Average \: Speed = \frac{Distance}{Time} \\\\A.S =\frac{84}{80} \\\\A.S = 1.05km/min[/tex]
we, know that
➺ Average Speed = Total Distance/Total Time taken
where,
Total Distance = 84km = 84000mTotal time = 1hr20mins = 4800sec♛ Substituting these values ♛
◗Average Speed = Distance/Time
◗Average Speed = 84000/4800
◗Average Speed = 840/48
◗Average Speed = 70/4
◗Average Speed = 17.5m/s
Hence, Average Speed of the man is 17.5m/sDescribe how the graph of y= x2 can be transformed to the graph of the given equation. y = (x - 6)2 Shift the graph of y = x2 down 6 units. Shift the graph of y = x2 right 6 units. Shift the graph of y = x2 up 6 units. Shift the graph of y = x2 left 6 units.
Answer:
Shift the graph of y = x2 right 6 units.
What is the area of polygon EFGH?
Tony rounded each of the numbers 1, 143 and 1, 149 to the nearest hundred which word correctly compares the rounding numbers
Full Question:
Tony rounded each of the numbers 1,143 and 1,149 to the nearest hundred. Which choice correctly compares the rounded numbers?
[tex]1,000 = 1,000[/tex]
[tex]1,140< 1,150[/tex]
[tex]1,100 = 1.100[/tex]
[tex]1,140>1,150[/tex]
Answer:
[tex]1,100 = 1,100[/tex]
Step-by-step explanation:
Given
1,143 and 1,149
Required
Which of the option is correct
We start by approximating both numbers to nearest digit
1,143; when approximated to nearest hundred is 1,100
1,149; when approximated to nearest hundred is also 1,100
Hence;
1,143 ≅ 1,100
1,149 ≅ 1,100
Comparing both results, we have that
[tex]1,100 = 1,100[/tex]
From the list of given options, option C is correct;
Factor completely 6x - 18.
6(x + 3)
6(x-3)
6X (-18)
Prime
Answer:
6(x-3)
Step-by-step explanation:
the common number for 6 and 18 is 6 so if you extract that from the expression then it turns to 6(x-3) which cannot be factored further
Answer:
Option B: 6(x - 3)
Step-by-step explanation:
At Jefferson Middle School, eighty-two students were asked which sports they plan to participate in for the coming year. Twenty students plan to participate in track and cross country; six students in cross country and basketball; and eight students in track and basketball. Twelve students plan to participate in all three sports. A total of thirty students plan to participate in basketball, and a total of forty students plan to participate in cross country. Ten students don't plan to participate in any of the three sports. How many students plan to just participate in cross country? 2 4 40 30
Answer:
40
Step-by-step explanation:
In the question only lies the answer:
"and a total of forty students plan to participate in cross country."
Answer:
2
Step-by-step explanation:
2
7. Over the past 50 years, the number of hurricanes that have been reported are as follows: 9 times there were 6 hurricanes, 13 times there were 8 hurricanes, 16 times there were 12 hurricanes, and in the remaining years there were 14 hurricanes. What is the mean number of hurricanes is a year
Answer:
Step-by-step explanation:
Let us first generate the frequency table from the information given:
Hurricane number(X) Frequency(f) f(X)
6 9 54
8 13 104
12 16 192
14 12 168
Total ∑(f) = 50 ∑f(x) =518
In order to determine the last frequency (the remaining years), we will add the other frequencies and subtract the answer from 50, which is the total frequency (50 years). This is done as follows:
Let the last frequency be f
9 + 3 + 16 + f = 50
38 + f = 50
f = 50 - 38 = 12
Now, calculating mean:
[tex]\bar {X} = \frac{\sum f(x)}{\sum(f)} \\\\\bar {X} = \frac{518}{50} \\\\\bar {X} = 10.36[/tex]
Therefore mean number of hurricanes = 10.4 (to one decimal place)
Select the line segment.
O A.
P
OB.
R
S
Ос.
M
N
OD.
G
H
Answer: Choice A
Explanation: A segment (or line segment) is whenever we have two endpoints connected with a straight line. It does not go on forever in either direction.
Answer:
Step-by-step explanation:
B is a ray. One of its ends is stationary.
C is an arc which means it bends
D is a line.
That means that A is the answer. Both its ends are stationary.
If a= 3 and b= 5; find
(a+b)
Answer: 3/8
Step-by-step explanation:
A=3
B=5
3/3+5=3/8
Find the length of side
x to the nearest tenth.
look at the image to find the question
Answer:
yes, the volume = 16 ft^3
hi hi hi please help ♀️
Answer:
Cooper read the least because
Step-by-step explanation:
1/5 is less than
Answer: Cooper read more
Step-by-step explanation:
help with math ASAP!
Answer:
1.) [tex]\frac{1}{9^4}*9^3[/tex]
2.) [tex]\frac{1}{w^7}[/tex]
3.)
Step-by-step explanation:
When you have a negative exponent, rewrite:
[tex]x^{-a}=\frac{1}{x^a}[/tex]
Rewrite using this to change all negative exponents.
Answer:
Multiple Answers
Step-by-step explanation:
Note: When multiplying numbers with exponents, you add the exponents. When dividing numbers with exponents, you subtract exponents.When you have a negative exponent, flip the fraction and write it as a positive exponent.
1) -4 + 3= -1
So we have (9^-4) + (9^3)= (1/(9^1)
2) (1/w)^7
3) cannot read problem, but just apply the rules I wrote under "Note"
4) 14/y
5) cannot read problem,but just apply the rules I wrote under "Note"
6) 20d^4 n^? --Cannot read n exponents--.
7) cannot read problem
8) Cannot read problem
9) 90/z^4---only if exponents are 5,-3,and-6
10) 1/(9^5)
11) 54b^4
12) Cannot read problem
13) 16d^8c^8 ---if exponents are 5,3,6,2--
14) s^8
Hope this helps! Plz give brainly, I kinda need it.
PLS HELP !!
Define two terms, each containing the variables x and y, with exponents on each. (For Example : 10x³y–⁵)Find the quotient of the two terms. Explain step-by-step how you found the quotient
Answer:
Step-by-step explanation:
Two such terms are 7x^3*y^9 and -3x*y^5
Their quotient is
7x^3*y^9
--------------
-3x*y^5
This can be simplified as follows:
The numerical coefficients become -7/3.
x^3/x = x^3*x^1 = x^(3 - 1) = x^2 (we subtract the exponent of x in the denominator from the exponent of x in the numerator).
Next, y^9*y^5 = y^4.
The quotient in final reduced form is then (-7/3)x^2*y^4
A boutique wants to make at least $127 profit from purses this week. The boutique earns $7 profit from each purse. How many purses must be sold?
Answer:
19 purses
Step-by-step explanation:
Set up an inequality where x represents the number of purses:
127 [tex]\geq[/tex] 7x
Solve for x by dividing each side by 7:
18.14 [tex]\geq[/tex] x
Round up to 19 because purses have to be whole
So, 19 purses have to be sold.
A bus company has contracted with a local high school to carry 450 students on a field trip. The company has 18 large buses which can carry up to 30 students and 19 small buses which can carry up to 15 students. There are only 20 drivers available on the day of the field trip.
The total cost of operating one large bus is $225 a day, and the total cost of operating one small bus is $100 per day.
Determine the minimum cost of transporting all 450 students.
Answer:
Step-by-step explanation:
From the given information;
Let consider p to be the number of large buses and q to be the number of small buses.
The company has 18 large buses which can carry up to 30 students and 19 small buses which can carry up to 15 students.
The linear inequality represent the students with respect to the total students s:
30p + 15q ≥ 450 -----(1)
They are only 20 drivers available on the day of the field trip, therefore only 20 buses can be used. So,
p + q ≤ 20 ----- (2)
Also , There are only 19 small buses and 18 large buses:
∴
0 ≤ p ≤ 18
0 ≤ q ≤ 19
The total cost of operating one large bus is $225 a day, and the total cost of operating one small bus is $100 per day
In order to minimize the cost of transporting all 450 students, let z be the minimal cost. i.e
z = 225p + 100q
From equation 1 and 2 ; if we plot them graphically on the graph, the following points of intersection were obtained as shown in the sketch below, in which the shaded region lies the answer.
The point of intersection between equation (1) and (2) is (p,q) = (10,10)
From the critical point in the sketch of our graph attached below, the following values of z can be determined.
Point(s) Value for Z
(15,0) 15 × 225 = 3375
(18,0) 18 × 225 = 4050
(18,2) (18 × 225 )+ (2 × 100 ) = 4250
(10,10) (10 × 225) + (10 × 100) = 3250
Thus ; the minimum cost of transporting all 450 students = $3250
a. Assume that the selections are made with replacement. Are the events independent? The probability of getting two orders from Restaurant D is . The events (1) independent because choosing the first order (2) the choice of the second order. (Round to four decimal places as needed.) b. Assume that the selections are made without replacement. Are the events independent? The probability of getting two orders from Restaurant D is . The events (3) independent because choosing the first order (4) the choice of the second order.
Answer: = a = 0.0206
b = 0.0205.
Step-by-step explanation:
From the question, given that;
Order Accurate = 328 273 242 142
Order Not Accurate = 32 54 37 20
Let us make the Total orders given be
T.O = 328+273+242+142+32+54+37+20 = 1128.
a) Let the Prob. that the first order is from restaurant D be
= Number of order from restaurant D / Total number of orders
= 162 / 1128 = 0.1436
Probability of the second order is 0.1436.
This is because, from the question we can tell that the selections are made with replacement, that means the order is the same.
So, the probability of getting 2 orders =
= 0.1436 * 0.1436 = 0.0206
NB: The probability of getting two orders from restaurant B is 0.0206.
This is because choosing the first order does not affect the second order
(independent events).
b) Assuming that the selections are made without replacement , the probability of getting both the orders from restaurant D =
Probability of getting 1st order from restaurant D = 162/1128 = 0.1436.Probability of getting 2nd order from restaurant D = 161 / 1127 = 0.1428This gives the Total Probability of getting both the orders from restaurant D, without replacement to be = 0.1436*0.1428
= 0.0205.
That is to say choosing the first order affects the second order because of the events are not independent as compared to the first question.
cheers i hope tis helps