Probability: [tex]\( \frac{111}{220} \)[/tex] or approximately 0.5045. Pita has about a 50.45% chance of drawing exactly £2.50.
To find the probability that Pita takes out exactly £2.50, we need to consider the different combinations of coins that would result in that amount.
First, let's calculate the total number of ways Pita can choose 3 coins out of 12:
[tex]\[ \text{Total number of ways} = \binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \][/tex]
Now, let's determine the combinations that would result in exactly £2.50. She can achieve this in two ways:
1. Selecting two 1 pound coins and one 50p coin.
2. Selecting five 50p coins.
For the first scenario, she can choose 2 out of the 3 1 pound coins and 1 out of the 9 50p coins. This can be calculated as:
[tex]\[ \binom{3}{2} \times \binom{9}{1} = \frac{3!}{2!(3-2)!} \times \frac{9!}{1!(9-1)!} = 3 \times 9 = 27 \][/tex]
For the second scenario, she can choose all 3 coins from the 9 50p coins:
[tex]\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \][/tex]
Now, let's add up these two possibilities to find the total number of favorable outcomes:
[tex]\[ \text{Total favorable outcomes} = 27 + 84 = 111 \][/tex]
Finally, we can calculate the probability:
[tex]\[ \text{Probability} = \frac{\text{Total favorable outcomes}}{\text{Total number of ways}} = \frac{111}{220} \][/tex]
[tex]\[ \text{Probability} \approx 0.5045 \][/tex]
So, the probability that Pita takes out exactly £2.50 is approximately 0.5045 or 50.45%.
The probability that Pita takes out exactly £2.50 when selecting 3 coins at random from her bag is approximately 42.27%.
To find the probability that Pita takes out exactly £2.50 when selecting 3 coins at random from her bag, we need to consider the different combinations of coins that would result in this amount.
Pita has 12 coins in total:
- 3 coins worth £1 each
- 9 coins worth 50p each
For a total of £2.50, she would need to select either:
1. Two £1 coins and one 50p coin, or
2. Five 50p coins
Let's calculate the probability for each case:
1. Probability of selecting two £1 coins and one 50p coin:
- Probability of selecting a £1 coin on the first draw: [tex]\( \frac{3}{12} = \frac{1}{4} \)[/tex]
- Probability of selecting another £1 coin on the second draw: [tex]\( \frac{2}{11} \)[/tex] (as there are 2 £1 coins left out of 11 total coins)
- Probability of selecting a 50p coin on the third draw: [tex]\( \frac{9}{10} \)[/tex] (as there are 9 50p coins left out of 10 total coins)
- Total probability for this case: [tex]\( \frac{1}{4} \times \frac{2}{11} \times \frac{9}{10} \)[/tex]
2. Probability of selecting five 50p coins:
- Probability of selecting a 50p coin on the first draw: [tex]\( \frac{9}{12} = \frac{3}{4} \)[/tex]
- Probability of selecting another 50p coin on the second draw: [tex]\( \frac{8}{11} \)[/tex] (as there are 8 50p coins left out of 11 total coins)
- Probability of selecting another 50p coin on the third draw: [tex]\( \frac{7}{10} \)[/tex] (as there are 7 50p coins left out of 10 total coins)
- Total probability for this case: [tex]\( \frac{3}{4} \times \frac{8}{11} \times \frac{7}{10} \)[/tex]
Now, we sum up the probabilities for both cases to find the total probability:
[tex]\[ \text{Total probability} = \left( \text{Probability of case 1} \right) + \left( \text{Probability of case 2} \right) \\\[ \text{Total probability} = \left( \frac{1}{4} \times \frac{2}{11} \times \frac{9}{10} \right) + \left( \frac{3}{4} \times \frac{8}{11} \times \frac{7}{10} \right) \][/tex]
Now, let's calculate these probabilities.
First, let's calculate the probability for case 1:
[tex]\[ \text{Probability of selecting two £1 coins and one 50p coin} \\\[ = \frac{1}{4} \times \frac{2}{11} \times \frac{9}{10} \\\[ = \frac{1}{4} \times \frac{18}{110} \\\[ = \frac{9}{220} \][/tex]
Now, let's calculate the probability for case 2:
[tex]\[ \text{Probability of selecting five 50p coins} \\\[ = \frac{3}{4} \times \frac{8}{11} \times \frac{7}{10} \\\[ = \frac{3}{4} \times \frac{56}{110} \\\[ = \frac{42}{110} \][/tex]
Now, let's sum up the probabilities for both cases to find the total probability:
[tex]\[ \text{Total probability} = \frac{9}{220} + \frac{42}{110} \][/tex]
Now, let's calculate the total probability.
[tex]\[ \text{Total probability} = \frac{9}{220} + \frac{42}{110} \][/tex]
First, let's find a common denominator for the fractions:
[tex]\[ \text{Total probability} = \frac{9}{220} + \frac{42 \times 2}{110 \times 2} \\\[ = \frac{9}{220} + \frac{84}{220} \][/tex]
Now, add the fractions:
[tex]\[ \text{Total probability} = \frac{9 + 84}{220} \\\[ = \frac{93}{220} \][/tex]
Now, let's simplify this fraction:
[tex]\[ \text{Total probability} = \frac{93}{220} \][/tex]
To express the probability as a decimal, we divide the numerator by the denominator:
[tex]\[ \text{Total probability} \approx \frac{93}{220} \approx 0.422727 \][/tex]
Now, let's express this probability as a percentage:
[tex]\[ \text{Total probability} \approx 0.422727 \times 100\% \approx 42.27\% \][/tex]
So, the probability that Pita takes out exactly £2.50 when selecting 3 coins at random from her bag is approximately 42.27%.
A rectangle has perimeter 78 cm and length 34 cm. What is its? width?
The perimeter of a rectangle is twice the sum of the dimensions:
[tex] 2p = 2(w+l) [/tex]
We know two of the three unkowns, so we can solve for w:
[tex] 78 = 2(w+34) \iff 78 = 2w+68 \iff 10=2w \iff w=5 [/tex]
The sum of four consecutive even integers is 300.
integer 1: x
integer 2: x + 2
integer 3: x + 4
integer 4: x + 6
integer 1 + integer 2 + integer 3 + integer 4 = sum
x + x + 2 + x + 4 + x + 6 = 300
4x + 12 = 300
4x = 288
x = 72
Answer: 72, 74, 76, 78
Write as a decimal . Tell whether the decimal terminates or repeats. 7/9
The fraction 7/9 converts to the decimal 0.77777778, which is a repeating decimal as the digit '7' repeats infinitely.
To convert the fraction 7/9 to a decimal, divide 7 by 9. So, when you divide 7 by 9, the answer you get is 0.77777778. This decimal does not terminate (come to an end after a specific number of decimal spaces), instead it repeats. Specifically, the digit '7' repeats indefinitely, so we can say its a repeating decimal.
As an example, if you consider 1/2, when you divide 1 by 2 it gives 0.5, which is a decimal that terminates. However, in our case - when we divided 7 by 9 it gave us 0.77777778 which repeats, hence it is a repeating decimal, not a terminating decimal. This is the key difference between repeating and terminating decimals.
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Aurora has a cell phone plan that charges
$65
for phone service plus
$12
for each gigabyte of data used each month. If her bill for September is
$113,
how many gigabytes of data did she use in September?
Aurora used 4 gigabytes of data in September.
Explanation:To determine how many gigabytes of data Aurora used in September, we need to subtract the fixed phone service charge from her total bill. The difference will represent the additional charges for the data usage. Let's say she used x gigabytes. The equation for her bill can be written as: 65 + 12x = 113. To solve for x, we can subtract 65 from both sides and then divide both sides by 12 to isolate x. Solving this equation, we find that x = 4. So, Aurora used 4 gigabytes of data in September.
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The tables below show running hours of three printers that produce greeting cards and the total number of greeting cards produced over three weeks.
Printer B uses $15 in ink every hour. What is the ink cost for each card coming from printer B?
A. $0.30
B. $0.50
C. $1.00
D. $1.50
ANSWER
The correct answer is A
EXPLANATION
To find the ink cost of each card coming from printer B, we need to find the total cost from printer B and the total number of cards printed by Printer B.
The total hours of all the three printers over the three weeks
[tex]=40+45+55+50+50+30+45+40+60=415[/tex]
The total hours of printer B only
[tex]=50+50+30=130[/tex]
Total number of cards produced over the three weeks
[tex]=7950+7800+9600=25350[/tex]
We can use ratio and proportion to determine the total cards produce by printer B only.
[tex]415:25350[/tex]
[tex]130:x[/tex]
If less, more divides
[tex]x=\frac{130\times 25350}{415} \approx7941[/tex]
Total cost of Printer B in 130 hours
$[tex]=130 \times 15=1950[/tex]
The ink cost
[tex]=\frac{1950}{7941}[/tex]
[tex]=0.2456[/tex]
$[tex]\approx 0.30[/tex]
to 2 decimal places
i need to remember how to solve this problem
1 + 4x = -5 + 7x
1st step: you need to move 4x to the right and -5 to the left so that all the x's are in one place. 4x will be negative and -5 will be positive when you move them. So it will be:
1+5=7x-4x
2nd step:
3x=6
3rd step: you need to divide both sides into 3, because it is the common divisor.
x=2
To sum up:1 + 4x = -5+7x
7x - 4x = 1 + 5
3x = 6
x = 2You can try to memorize the summary. However, it is better to understand the solution as a whole.
Yolanda spends 8 1/2 hour per month playing soccer how many hrs does she play in a year
0.00000000005 in scientific notation
show your work
The simple interest I on an investment of P dollars at an interest rate r for t years is given by I=Prt. Find the time it would take to earn $1400 in interest on an investment of $27,000 at a rate of 4.2%.
we know that
[tex]I=Prt[/tex]
where
I is the amount in interest
P is the investment
r is the interest rate
t is the time in years
Solve for t
[tex]t=I/(Pr)[/tex]
we have
[tex]I=\$1,400\\ P=\$27,000\\ r=4.2\%=0.042[/tex]
substitute
[tex]t=1,400/(27,000*0.042)=1.23\ years[/tex]
therefore
the answer is
[tex]1.23\ years[/tex]
What is 673.983 rounded to the nearest hundredth
I believe it would be 673.98
I hope this helps :)
673.983 rounded to the nearest hundredth is 673.98, as the third decimal digit (3) is less than 5.
To round the number 673.983 to the nearest hundredth, we need to look at the third decimal place, which is in the thousandths place. If this digit is 5 or greater, we round up. If it is less than 5, we keep the second decimal place as it is. In the case of 673.983, the third decimal digit is a 3, which is less than 5. Therefore, the number rounded to the nearest hundredth is 673.98.
Please help 20 points. Explain.
We can write this line in y-intercept form, and then solve to convert it to standard form.
Y = mx + b
m = slope
b = y-intercept
Y = -1/4x - 2 is the slope-intercept form of the equation.
Now, we can convert this to standard form.
Add 2 to both sides.
y + 2 = -1/4x
Subtract y from both sides.
-1/4x - y = 2
Multiply all terms by -4.
x + 4y = -8 (This is your answer.)
Y=1/2x + 3/4
Select all of the points that are contained in the graph of the equation. There may be more than one.
(0, 1/2)
(0, 3/4)
(3/4, 0)
(3/4, 1/2)
(1/2, 1)
to test the points, substitute the x-coordinate into the equation and if the value of y equals the y-coordinate then the point lies on the graph
(0, [tex]\frac{1}{2}[/tex]) → y = 0 + [tex]\frac{3}{4}[/tex] = [tex]\frac{3}{4}[/tex] ≠ [tex]\frac{1}{2}[/tex]
(0 , [tex]\frac{3}{4}[/tex]) → y = [tex]\frac{3}{4}[/tex] ← on graph
([tex]\frac{3}{4}[/tex], 0) → y = [tex]\frac{3}{8}[/tex] + [tex]\frac{3}{4}[/tex] ≠ 0
([tex]\frac{3}{4}[/tex],[tex]\frac{1}{2}[/tex] → y = [tex]\frac{3}{8}[/tex] + [tex]\frac{3}{4}[/tex] ≠ [tex]\frac{1}{2}[/tex]
([tex]\frac{1}{2}[/tex], 1) → y = [tex]\frac{1}{4}[/tex] + [tex]\frac{3}{4}[/tex] = 1 ← on graph
the points (0, [tex]\frac{3}{4}[/tex]) and ([tex]\frac{1}{2}[/tex], 1) are on the graph
The points that are contained in the graph of the straight line y = (1/2)x + 3/4 are (0,3/4) and (1/2,1).
What is an equation?An equation describes that both sides of the variables and constants are in different forms but they are equal.
To determine if each given point is contained in the graph of the equation Y = (1/2)x + 3/4 we will put the values of x in the equation and determine the corresponding value of y.
When x = 0, y = (3/4).
When x = (3/4), y = 9/8.
When x = 1/2, y = 1.
So, the points that are contained in the graph Y = (1/2)x + 3/4 are (0,3/4), (1/2,1).
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a^2-2ab-6by+ay
Factor the expression
Factor a^2−2ab−6by+3ay
a^2−2ab+3ay−6by
=(a−2b)(a+3y)
Answer:
(a−2b)(a+3y)
the factored expression is:
a(a + y) - 2b(a + 3y)
We can factor the expression [tex]a^2-2ab-6by+ay[/tex] using a combination of grouping and factoring by difference of squares:
We can group the terms based on common factors:
[tex](a^2 + ay) - (2ab + 6by)[/tex]
In the first group, a is the common factor: a(a + y)
In the second group, 2b is the common factor: 2b(a + 3y)
Then we combine the terms
a(a + y) - 2b(a + 3y)
Therefore, the factored expression is:
a(a + y) - 2b(a + 3y)
solve for x in the equation is x^2-4x-9=29
Step 1. Move all terms to one side
[tex]x^2-4x-9-29=0[/tex]
Step 2. Simplify [tex]x^2-4x-9-29[/tex] to [tex]x^2-4x-38[/tex]
[tex]x^2-4x-38=0[/tex]
Step 3. Use the Quadratic Formula
[tex]x=\frac{4+2\sqrt{42} }{2} ,\frac{4-2\sqrt{42} }{2 }[/tex]
Step 4. Simplify solutions
[tex]x = 2 \sqrt{42,} 2-\sqrt{42}[/tex]
Let's solve your equation step-by-step.
[tex]x^{2}[/tex]−4x−9=29
Step 1: Subtract 29 from both sides.
[tex]x^{2}[/tex]−4x−9−29=29−29
[tex]x^{2}[/tex]−4x−38=0
Step 2: Use quadratic formula with a=1, b=-4, c=-38.
[tex]x=\frac{-b+-\sqrt{b^{2} -4ac} }{2a}[/tex]
[tex]x=\frac{-(-4)+-\sqrt({-4})^{2} -4(1)(-38)}{2(1)}[/tex]
[tex]x=\frac{4-+\sqrt{168}}{2}[/tex]
Answer:
[tex]x=2+\sqrt{42}[/tex] or x = -2 [tex]\sqrt{42}[/tex]
A 13.5-gasoline tank is 4 / 5 full how many gallons will it take to fill the tank
A farmer has 180 bushels of wheat to sell at her roadside stand. She sells an average of 16 and three fourths bushels each day. Represent the total change in the number of bushels she has for sale after 5 days.
To find the total change in the number of bushels the farmer has for sale after 5 days, multiply the average number of bushels sold per day by the number of days.
Explanation:To find the total change in the number of bushels the farmer has for sale after 5 days, we need to multiply the average number of bushels sold per day by the number of days.
The average number of bushels sold per day is 16 and three fourths, which can be written as 16.75.
Multiplying 16.75 by 5 gives us the total change in bushels after 5 days, which is 83.75 bushels.
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the area of a quadrilateral is 8.4 square feet. find two decimals that give a product close to 8.4
Answer:
2.81 and 0.3 are two decimals that give a product of 8.43 which is close to 8.4
Step-by-step explanation:
let consider a decimal number 0.3
now consider another decimal number 2.81
now if we multiply these two numbers then we get 8.43 which is also a decimal number and it is close to 8.4.
Answer:
Hence the approximate values are 2.91 and 2.89
Step-by-step explanation:
The area of a quadrilateral is expressed as
A=L*B
Where L=length
B=breadth or width
I. e assuming it has two of its sides equal
Say given area is
A=L*B=8.4 square feet
Let's take the square root of the area A=2.898
=2.89
Approximately
L= 2.9 feet
Let's us divide the area by L to get B
8.4/2.9= 2.91
Cara is grocery shopping and needs chicken and steak for the barbeque she is having on Saturday. Chicken is $2.99 per pound and steak is $5.79 per pound. She knows she wants to spend exactly $50 on meat. If c represents pounds of chicken and s represents pounds of steak, which equation represents this situation?
2.99s + 5.79c = 50
8.78m = 50
c + s = 50
2.99c + 5.79s = 50
chicken: ($2.99) c
steak: ($5.79) s
chicken + steak = total
2.99c + 5.79s = 50
Answer: D
The probability of something can be expressed as a decimal or as a percent. We know that the probability of something happening can not be below 0% or above 100%. Background information: To turn a percent into a decimal (or an integer depending on the number you are working with), you move the decimal two places to the left. For example: 56% would be 0.56 If probability has to be between 0% - 100%, between what two integers does probability have to be after you convert from a percent to an integer? *
Answer: The probability lies between 0 and 1
Step-by-step explanation:
As from the given information probability of an event lies between 0% to 100% .
To convert percent into integer or fraction we have to divide it by 100
[tex] 0\%= \frac{0}{100} = 0 [/tex]
And
[tex] 100\% = \frac{100}{100} = 1 [/tex]
From the above the probability of an event lie between integers 0 to 1.
Two investments totaling $58,500 produce an annual income of $3795. One investment yields 5% per year, while the other yields 7% per year. How much is invested at each rate?
Total investment is =$58500
Annual income produced = $3795
Rate of one investment = 5%
Rate of second investment = 7%
Let the amount invested at 5% be = x
Then, amount invested at 7% rate will be = [tex]58500-x[/tex]
The equation becomes:
5%x + 7%(58500-x) = 3795
= 0.05x + 0.07(58500-x) = 3795
0.05x + 4095 - 0.07x = 3795
0.02x = 300
x = 15000
Hence, the amount invested at 5% is = $15000
Amount invested at 7% = 58500-15000 = $43500
What is the slope of the line shown below?
y = 4x – 10 y = 2 What is the solution to the system of equations?
ANSWER
The solution is
[tex](3,2)[/tex]
EXPLANATION
We were given the equations
[tex]y=4x-10--(1)[/tex]
and
[tex]y=2----equation (2)[/tex]
This is a simultaneous equation in [tex]x[/tex] and [tex]x[/tex].
But the coefficient of [tex]x[/tex] in the second equation is zero. that is why it seems to be absent.
We solve these two equations simultaneously
by putting [tex]y=2[/tex] in to equation (1).
Tis gives;
[tex]2=4x-10[/tex]
We add 10 to both sides to get;
[tex]12=4x[/tex]
we divide through by 4 to get
[tex]3=x[/tex]
Hence the solution for the system is [tex](3,2)[/tex]
Answer:
(3, 2)
Step-by-step explanation:
A conjecture and the paragraph proof used to prove the conjecture are shown.
Given: E is the midpoint of segment D F. Prove: 2 D E equals D F. A segment DF is horizontal with endpoints D and F. E is the midpoint of the segment.
Drag an expression or statement to each box to complete the proof.
Answer:
1. DF (option 3)
2. segment addition postulate
3. segment congruence postulate
4. DF (option 5)
Answer with explanation:
Given: E is the mid point of DF.
Prove: 2 DE=D F
Proof with explanation:
It is given that [tex]\Bar{DE}[/tex] is mid point of [tex]\Bar{DF}[/tex].
1.→
So, [tex]\Bar{DE}\cong\Bar{DF}[/tex]
-------------------------By Definition of mid point
2.→ DE=E F----------------Segment Congruence Postulate
3.→DE+E F=D F ----------------Segment Addition Postulate
4.→DE+DE=D F-------------By Substitution
5.→Simplifying gives
2 DE=D F
Please help will give brainliest
A conjecture and the flowchart proof used to prove the conjecture are shown.
Given: Measure of angle D E G equals 51 degrees. Measure of angle G E F equals 39 degrees. Prove: Triangle D E F is a right triangle. Art: A right angle triangle D E F. Angle D E F is a right angle. Point G is on segment D F, and segment E G is drawn.
Drag an expression or phrase to each box to complete the proof.
box number one: given
box number two: angle addition postulate
box number three: 51 degrees + 39 degrees = m DEF
box number four: 90 degrees = m DEF
box number five: DEF is a right angle
box number six: definition of a right triangle
just took the test and these are the correct answers, hope i helped :)
Solution:
Given : ∠ D E G= 51°, ∠GEF=39°
To prove: Δ D E F is a right triangle.
Proof:
A right angle triangle D E F.
∠ D E F =90°.
Point G is on segment D F, and segment E G is drawn.
Now, taking ∠DEG and ∠GEF into consideration,
∠DEG=51° and ∠GEF=39°→→(Given)
∠DEG + ∠GEF =
∠DEF=51° + 39°
∠DEF=90°,showing Triangle D E F is a right triangle.
4-5 of the money collected from a fundraiser was dived equally among 8 grades.What fraction of money did each grade receive
If you are just dividing percent's, then each grade would have 12.5 percent.
Please anyone help me
How many students practiced the piano more then 3 hours a week
The number of students that practice the piano more than 3 hours a week is 6
From the dot plot (see attachment), we have the following highlights
3 students practice the piano for 4 hours2 students practice the piano for 5 hours1 student practices the piano for 6 hoursNext, we add the total number of students
[tex]Total = 3 + 2 +1[/tex]
[tex]Total = 6[/tex]
Hence, the number of students that practice the piano more than 3 hours a week is 6
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6. Which type of angle pair are LSM and OSN?
A. Adjacent angles
B. Linear pair
C. Complementary angles
D. Vertical angles
7. Which of the following statements are false?
A. LSM and MSN are adjacent angles
B. LSM and MSN form a linear pair of angles
C. LSM and MSN are vertical angles
D. M
8. Which angle is supplementary to LSM?
A. OSN
B. SLM
C. LMS
D. MSN
6. These angles are vertical angles, so D. is the correct answer!
7. These angles are not vertical angles, although I don't know what D. is....
8. D) MSN is the correct answer
help me please!! hurry up timed
The quickest way to do this is to plug the first b-value from each table into the equation to find p-value. Then compare your answer with the p-value from the table to find the one that is greater than your answer.
p = 3.25b
Equation Table Table compared to Equation
Table A: p = 3.25(1) = 3.25 p = 2.75 <
Table B: p = 3.25(1.5) = 4.875 p = 3.75 <
Table C: p = 3.25(2) = 6.50 p = 7.00 >
Table D: p = 3.25(2.5) = 8.125 p = 10.00 >
Table E: p = 3.25(3) = 9.75 p = 11.25 >
Table F: p = 3.25(3.5) = 11.375 p = 10.50 <
Answer: C, D, and E
The product of x and the sum of 6 and 8 times the square of x
"·" - product
"+" - sum
x² - square of x
Answer: x · (6 + 8 · x²) = x(6 + 8x²)
The expression "the product of x and the sum of 6 and 8 times the square of x" can be written as [tex]\( x \cdot (6 + 8x^2) \)[/tex].
The mathematical expression "the product of x and the sum of 6 and 8 times the square of x" is represented as [tex]\( x \cdot (6 + 8x^2) \)[/tex]. To simplify, distribute x into the parentheses, yielding 6x + 8x^3. This expression captures the specified operation, where x is multiplied by the sum of 6 and 8 times the square of x.
It is essential to follow the order of operations, ensuring that the square of x is calculated first, then multiplied by 8, and finally added to 6 before multiplying the result by x. This expression forms the algebraic representation of the described mathematical operation.