the prime factorization of a certain number is 2 to the 3rd power times 3 times 5 to the second power what is the number and how did you get it
Answers
First, we can write out the value of each number with an indicie:
[tex] {2}^{3} = 8 \\ {5}^{2} = 25[/tex]
Next, we replace the numbers with indicies in the question with their values:
8×3×25
Now, just multiply these numbers together to get your answer:
8×3×25=600
So, the number is 600
The number that corresponds to the prime factorization '2 to the 3rd power times 3 times 5 to the 2nd power' is 600. Calculate this by resolving each component then multiply the results.
Explanation:The question revolves around finding the result of a number given its prime factorization, specifically when the prime factorization is '2 to the 3rd power times 3 times 5 to the 2nd power'. Let's start by resolving each of those components:
2 to the 3rd power, also known as cubing, would give us 8 because we multiply 2 three times (2*2*2).3 remains as it is because it is a prime number and is not raised to any power.5 to the 2nd power, also known as squaring, would give us 25 because we multiply 5 by itself (5*5).So, with these resolved components, we only have to multiply the results, which is 8 (from 2 cubed) times 3 times 25 (from 5 squared).
This would give us the answer: 8 * 3 * 25 = 600.
Therefore, the number that corresponds to the prime factorization '2 to the 3rd power times 3 times 5 to the 2nd power' is 600.
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Related Questions
A triangle on a coordinate plane is translated according to the rule T–3, 5(x, y). Which is another way to write this rule?
(x, y) → (x – 3, y + 5)
(x, y) → (x – 3, y – 5)
(x, y) → (x + 3, y – 5)
(x, y) → (x + 3, y + 5)
Answers
the original (x,y) coordinates are being moved to the left by 3 units and up by 5 units.
therefore
the answer is the option
(x, y) → (x – 3, y + 5)
Another way to write the given rule is (x, y) → (x – 3, y + 5).
Given that,
There are the original x and y coordinates that are being moved to the left-hand side by 3 units and go up by 5 units.Based on the information, we can conclude that another way to write the given rule is (x, y) → (x – 3, y + 5).
Hence, the other options are incorrect.
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Solve the equation by completing the square. Round to the nearest hundredth if necessary. x2 – 4x = 5
Answers
x^2 - 4 x = 5
Subtract 5 from both sides:
x^2 - 4 x - 5 = 0
x = (4 ± sqrt((-4)^2 - 4 (-5)))/2 = (4 ± sqrt(16 + 20))/2 = (4 ± sqrt(36))/2:
x = (4 + sqrt(36))/2 or x = (4 - sqrt(36))/2
sqrt(36) = sqrt(4×9) = sqrt(2^2×3^2) = 2×3 = 6:
x = (4 + 6)/2 or x = (4 - 6)/2
(4 + 6)/2 = 10/2 = 5:
x = 5 or x = (4 - 6)/2
(4 - 6)/2 = -2/2 = -1:
Answer: x = 5 or x = -1
Answer:
x = 5 or x = -1
Step-by-step explanation:
x^2 – 4x = 5
Solve it by completing the square method
In completing the square method, we take the coefficient of x that is -4, divide it by 2 and then square it
-4/2 = -2
square it (-2)^2 = 4
Now add 4 on both sides
x^2 – 4x +4 = 5+4
x^2 – 4x +4 = 9
Now factor the left hand side
(x-2)(x-2)=9
(x-2)^2 = 9
Take square root on both sides
x-2 = +-3
x-2 = 3 , so x= 5
x-2 = -3, so x= -1
Find geometric mean of the pair of number 6 and 10
Answers
6 x 10 = 60
2nd step find the square root
sqrt(60) = 7.745
round the answer as needed.
The Geometric mean of the pair of number 6 and 10 is [tex]7.74[/tex]
Geometric mean :The geometric mean of two number m and n is given as,
[tex]G.M=\sqrt{m*n}[/tex]
Geometric mean of the pair of number 6 and 10 is,
[tex]G.M=\sqrt{6*10} =\sqrt{60} \\\\G.M=7.74[/tex]
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HELP PLEASE
Translate the answer, T2 = A3, into words:
The BLANK of the orbital period, T, of a
planet is equal to the BLANK of the average distance, A, of the planet from the Sun.
Square
Square Root
Cube
Answers
Answer: The square of the orbital period, T, of a planet is equal to the cube of the average distance, A, of the planet from the Sun.
Step-by-step explanation:
In mathematics, A number or variable 'a' to its second power is said to be square of 'a' i.e. [tex]\ a^2[/tex] is the square of a.
A number or variable 'b' to its third power is said to be cube of 'b' i.e. [tex]\ b^3[/tex] is the cube of b.
In the given equation: [tex]T^2 = A^3[/tex]
Here, [tex]T^2 [/tex] is square of T.
And [tex]A^3[/tex] is cube of A.
Therefore, The square of the orbital period, T, of a planet is equal to the cube of the average distance, A, of the planet from the Sun.
What is the mean of the values in the stem-and-leaf plot? Enter your answer in the box.Key: 2|5 means 25
Enter your answer in the box.
Key: 2|5 means 25
A stem-and-leaf plot with a stem value of 1 with leaf values of 5 and 8, a stem value of 2, a stem value of 3, a stem value of 4 with a leaf value of 6, a stem value of 5 with leaf values of 0, 0, 0, 0, 7, a stem value of 6, a stem value of 7, a stem value of 8, and a stem value of 9.
Answers
The mean of the values in the stem-and-leaf plot is 42.
To find the mean of the values in the stem-and-leaf plot, we first need to interpret the plot correctly. Here's the stem-and-leaf plot description:
Stem | Leaves
1 | 5, 8
2 |
3 |
4 | 6
5 | 0, 0, 0, 0, 7
6 |
7 |
8 |
9 |
Let's list the individual values from the plot:
- From stem 1: Leaves are 5 and 8
- From stem 4: Leaf is 6
- From stem 5: Leaves are 0, 0, 0, 0, 7
Now, we calculate the mean (average) of these values.
Step-by-step calculation:
1. List of values:
- Values from stem 1: 15, 18
- Values from stem 4: 46
- Values from stem 5: 50, 50, 50, 57
2. Count the number of values:
- There are 2 values from stem 1, 1 value from stem 4, and 5 values from stem 5.
- Total count = 2 + 1 + 5 = 8
3. Calculate the sum of all values:
- Sum = 15 + 18 + 46 + 50 + 50 + 50 + 50 + 57
- Sum = 336
4. Calculate the mean:
- Mean = Sum / Count
- Mean = 336 / 8
- Mean = 42
Therefore, the mean of the values in the stem-and-leaf plot is [tex]{42} \).[/tex]
Suppose that the probabilities of a customer purchasing 0, 1, or 2 books at a book store are 0.20.2, 0.30.3, and 0.50.5, respectively. what is the standard deviation of this customer's book purchases?
Answers
μ = average
σ = standard deviation
V (X) = Variance
σ = (V(X)) ^ 0.5
E [X] = X * P (x)
Assuming that the number of books purchased is a discrete random variable with mean μ = E [X]
Then the variance of X can be written as V (X) = E [X-μ]^2
We started finding the average μ
μ = 0 * 0.20 + 1 * 0.30 + 2 * 0.50
μ = 1.3
Once the average is found, we can calculate the value of the variance
V (X) = 0.20 * (0-1.3) ^ 2 + 0.30 * (1-1.3) ^ 2 + 0.50 * (2-1.3) ^ 2
V (X) = 0.61
Now we know that from the variance the standard deviation can be obtained by doing:
σ = (V (X)) ^ 0.5
Finally
σ = 0.781
The standard deviation of the customer's book purchases is 0.72.
The standard deviation of a discrete probability distribution can be calculated using the following formula:
σ = √(p[tex](x - \mu)^2[/tex])
where:
σ is the standard deviation
p(x) is the probability of the event x
μ is the mean of the distribution
In this case, the probabilities of the customer purchasing 0, 1, or 2 books are 0.2, 0.3, and 0.5, respectively. So, the mean of the distribution is:
μ = (0 * 0.2) + (1 * 0.3) + (2 * 0.5) = 1.2
The standard deviation is then:
σ = √([tex](0 - 1.2)^2 (0.3)^2[/tex] + [tex](1 - 1.2)^2 (0.3)^2[/tex] + [tex](2 - 1.2)^2 (0.5)^2)[/tex] = 0.72
So, the standard deviation of the customer's book purchases is 0.72.
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How many bits are required to represent the decimal numbers in the range from 0 to 999 in straight binary code?
Answers
[tex]2^0=1_2[/tex]
[tex]2^1=10_2[/tex]
[tex]2^2=100_2[/tex]
and so on. Observe that [tex]n+1[/tex] digits are required to represent the [tex]n[/tex]-th power of 2 in binary.
Also observe that
[tex]\log_2(2^n)=n\log_22=n[/tex]
so we need only add 1 to the logarithm to find the number of binary digits needed to represent powers of 2. For any other number (non-power-of-2), we would need to round down the logarithm to the nearest integer, since for example,
[tex]2_{10}=10_2\iff\log_2(2^1)=\log_22=1[/tex]
[tex]3_{10}=11_2\iff\log_23=1+(\text{some number between 0 and 1})[/tex]
[tex]4_{10}=100_2\iff\log_24=2[/tex]
That is, both 2 and 3 require only two binary digits, so we don't care about the decimal part of [tex]\log_23[/tex]. We only need the integer part, [tex]\lfloor\log_23\rfloor[/tex], then we add 1.
Now, [tex]2^9=512<1024=2^{10}[/tex], and 999 falls between these consecutive powers of 2. That means
[tex]\log_2999=9+\text{(some number between 0 and 1})[/tex]
which means 999 requires [tex]\lfloor\log_2999\rfloor+1=9+1=10[/tex] binary digits.
Your question seems to ask how many binary digits in total you need to represent all of the numbers 0-999. That would depend on how you encode numbers that requires less than 10 digits, like 1. Do you simply write [tex]1_2[/tex]? Or do you pad this number with 0s to get 10 digits, i.e. [tex]0000000001_2[/tex]? In the latter case, the answer is obvious; [tex]1000\times10=10^4[/tex] total binary digits are needed.
In the latter case, there's a bit more work involved, but really it's just a matter of finding how many number lie between successive powers of 2. For instance, 0 and 1 both require one digit, 2 and 3 require two, while 4-7 require three, while 8-15 require four, and so on.
Final answer:
To represent the decimal numbers from 0 to 999 in binary code, we require 10 bits. This is determined by finding the binary equivalent of the largest number, 999, which necessitates at least 10 powers of 2 (from [tex]2^0 \ to \ 2^9[/tex]).
Explanation:
To determine how many bits are required to represent the decimal numbers in the range from 0 to 999 in straight binary code, we need to find the binary equivalent of the largest number in that range, which is 999. In binary, this would require the highest power of 2 that is less than or equal to 999. The largest power of 2 less than 999 is 512 (29), and we continue to add powers of 2 to represent the number:
512 (29)
256 (28)
128 (27)
64 (26)
32 (25)
16 (24)
8 (23)
4 (22)
2 (21)
1 (20)
Adding these up, we see that we need at least 10 bits to represent the number 999 in binary, because adding powers of 2 from 20 to 29 will give us the required range. Therefore, we need 10 bits to represent any number from 0 to 999 in binary.
Jorge is standing at a horizontal distance of 25 feet away from a building. his eye level is 5.5 feet above the ground and looking up he notices a window washer on the side of the building at an angle of elevation of 65, how high is the window washer above the ground
Answers
tan 65°=y/25------------> y=tan 65°*25 ( distance of the washer at the eye level)
[the window washer high above the ground]=[(tan 65°)*25+5.5]
[the window washer high above the ground]=[53.61+5.5]=59.11 ft
the answer is 59.11 ft
How to start the problem off. (substitution)
Answers
8. The U.S. population was 309 million in 2010. If the U.S. had had the same firearm death rate as Australia in 2010, how many firearm deaths would the U.S. have expected to have that year? Australia’s population was 22.3 million people in 2010. Australia had 236 firearm deaths in 2010. (round to the nearest person)
Answers
As per linear equation, in 2010 the expected firearm death U.S. had is 3270.
What is a linear equation?A linear equation represents an equation that has one or multiple variables with the highest power of the variable is 1.
Australia’s population was 22.3 million people in 2010.
Australia had 236 firearm deaths in 2010.
Therefore, portion of Australian population had firearm deaths is
[tex]= \frac{236}{22.3(10)^{6} } \\= 0.000011[/tex]
The U.S. population was 309 million in 2010.
Therefore, the expected firearm deaths in U.S. in 2010 was
[tex]= (309) (10^{6}) (0.000011)\\= 3270.13\\= 3270[/tex]
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What is the measure of angle z in this figure?
Enter your answer in the box.
z = ___°
Answers
I've attached a photo for reference.
Lines A and B form straight angles, which measure 180 degrees. That means that -
m∠x + m∠y = 180°
m∠y + m∠z = 180°
m∠z + 43° = 180°
43° + m∠x = 180°
Since you're trying to find z, use the solvable equation with z in it:
m∠z + 43° = 180°
180 = z + 43
137 = z
Answer:
m∠z = 137°
With a short time remaining in the day a delivery driver has to make deliveries at 5 locations among the 6 locations remaining. How many different routes are possible
Answers
So, you need to use permutation for this problem. The calculation would be:
6P5= 6!/ (6-5)!= 720 different routes
To find the number of different routes the delivery driver can take to make the deliveries at 5 out of 6 locations, we can use the concept of permutations.
Explanation:To find the number of different routes the delivery driver can take, we can use the concept of permutations. Since the delivery driver has to make deliveries at 5 out of the remaining 6 locations, we need to calculate the number of ways to arrange these locations.
We can use the formula for permutations to find the number of different routes:
nPr = n! / (n - r)!
Using this formula, we can calculate the number of different routes as:
6P5 = 6! / (6 - 5)! = 6! / 1! = 6 × 5 × 4 × 3 × 2 × 1 / 1 = 720
Therefore, there are 720 different routes the delivery driver can take to make the deliveries.
a paragraph of; compare savings and investments
Answers
A gardener plants a bed of flowers such that he plants twenty day lilies in the first row, twenty-six day lilies in the second row, and thirty-two day lilies in the third row. He continues to plant lilies in the bed with this pattern for a total of twelve rows. How many day lilies did he plant?
Answers
There are 86 lilies he did plant in 12th row of the garden.
What is Arithmetic Sequence?Arithmetic Progression (AP) is a sequence of numbers in order, in which the difference between any two consecutive numbers is a constant value. It is also called Arithmetic Sequence.
Here, The number of lilies plants in first, second third,..., and last row are respectively.
20,26,32,........
the number of rows of lilies plants is 12.
The sequence 20,26,32,........ is an A.P. with first term a =20, common difference d = 6 and n =12
formula for nth term.
∴aₙ = a+(n−1)d
aₙ=20+ (12-1).6
aₙ = 20 + 11 X 6
aₙ = 86
Thus, there are 86 lilies he did plant in 12th row of the garden.
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The perimeter of a quilt is 34 ft. If the quilt is 8 ft long, what is the area
Answers
2 x 8 = 16 (Finding out the length)
16 + 2w = 34 (Problem with the length)
-16 -16 (Using the subtraction property of equality to isolate the variable)
2w = 18 (New problem)
/2 /2 (Using division property of equality)
w = 9. (Width is found)
9 x 8 = 72. The area is 72 squared feet.
18÷2=9
8x9=72
describe the graph of y=3/4x-12 as compared to the graph of y=1/x
Answers
Answer:
Parent function is compressed by a factor of 3/4 and shifted to right by 3 units.
Step-by-step explanation:
We are asked to describe the transformation of function [tex]y=\frac{3}{4x-12}[/tex] as compared to the graph of [tex]y=\frac{1}{x}[/tex].
We can write our transformed function as:
[tex]y=\frac{3}{4(x-3)}[/tex]
[tex]y=\frac{3}{4}*\frac{1}{(x-3)}[/tex]
Now let us compare our transformed function with parent function.
Let us see rules of transformation.
[tex]f(x-a)\rightarrow\text{Graph shifted to the right by a units}[/tex],
[tex]f(x+a)\rightarrow\text{Graph shifted to the left by a units}[/tex],
Scaling of a function: [tex]a*f(x)[/tex]
If a>1 , so function is stretched vertically.
If 0<a<1 , so function is compressed vertically.
As our parent function is multiplied by a scale factor of 3/4 and 3/4 is less than 1, so our parent function is compressed vertically by a factor of 3/4.
As 3 is being subtracted from x, so our parent function is shifted to right by 3 units or a horizontal shift to right by 3 units.
Therefore, our parent graph is compressed by a factor of 3/4 and shifted to right by 3 units to get our new graph.
Final answer:
The equation y=3/4x-12 represents a linear function with a straight line graph, while y=1/x depicts a rectangular hyperbola with a curved graph showcasing an inverse relationship. These graphs exhibit fundamentally different behaviors, one being constant and the other varying.
Explanation:
The student's question involves comparing the graph of y=3/4x-12 to the graph of y=1/x. The first equation, y=3/4x-12, describes a linear function with a slope of 3/4 and a y-intercept of -12. This means it is a straight line that increases as x increases, crossing the y-axis at (0, -12).
On the other hand, the graph of y=1/x is a rectangular hyperbola. This curve approaches the x-axis and y-axis but never touches them, known as asymptotes. The hyperbola is divided into two parts, one in the first quadrant where both x and y are positive, and another in the third quadrant where both x and y are negative. This graph does not resemble a straight line at any part.
Comparatively, the key difference lies in the nature of their graphs - one is a straight line and the other a curved hyperbola, showcasing completely different behaviors. The linear equation reflects constant change, whereas the hyperbola’s rate of change varies with x, indicating an inverse relationship.
factor, x^4y+8x^3y-6x^2y^2-48xy^2
Answers
[tex]\boxed{\boxed{((((x^4)*y)+((8*(x^3))*y))-((6*(x^2))*(y^2)))-(24*3xy^2)}} \\ \\ (let's \ simplify \ some \ number's \ down) \\ \\ \boxed{\boxed{ ((((x4)*y)+((8*(x3))*y))-((2*3x2) *y2))-(24*3xy2)}} \\ \\ \\ Pull \ out \ like \ factors : \\ \\ \boxed{\boxed{ x4y + 8x^3y - 6x2y^2 - 48xy^2 = xy * (x3 + 8x^2 - 6xy - 48^y) }} \\ \\ \\ (TAKE \ NOTICE) \swarrow\swarrow\swarrow\swarrow \swarrow\swarrow\swarrow\swarrow\swarrow\swarrow \\ \\ \\ [/tex]
[tex]\boxed{\boxed{\boxed{ (((((x^4)*y)+(23x^3*y))-(2*3x2y2))}}} \\ \\ \\ \boxed{\boxed{\boxed{-(24*3xy2))-xy*(x3+8x2-6xy-48y) = \boxed{0}}}}}[/tex]
I hope this helps you!
see attached picture for steps:
If you take out a loan that costs $130.50 over nine years at an interest rate of 10%, how much was the loan for?
Answers
The original loan amount for which the final amount payable is $130.50 over nine years at an interest rate of 10% was approximately $55.39.
Explanation:To find out the original amount of the loan for which the final amount payable is $130.50 over nine years at an interest rate of 10%, we will use the formula for the present value of a loan.
The formula given is P = A / (1 + r/n)^nt, where P is the principal (initial loan amount), A is the final amount, r is the annual interest rate (expressed as a decimal), n is the number of times that interest is compounded per year, and t is the time the money is borrowed for in years.
As the question doesn't specify how many times the interest is compounded per year, we will assume it is compounded once a year (n=1).
We have:
A = $130.50,r = 10% or 0.10,n = 1,t = 9 years.Plugging these values into the formula:
P = $130.50 / (1 + 0.10/1)^(1*9)
P = $130.50 / (1 + 0.10)^9
P = $130.50 / (1.10)^9
P = $130.50 / 2.35794769
P = $55.39 approximately
Therefore, the original loan amount was approximately $55.39.
If the test scores of a class of 34 students have a mean of 73.1 and the test scores of another class of 26 students have a mean of 66.4, then the mean of the combined group is
Answers
= (73.1 * 34 + 66.4 * 26) / (34 + 26)
= (2485.4 + 1726.4) / 60
= 70.2 (rounded)
The Mean of the combined class is 70.19.
MeanMean is the average of a set of two or more numbers. The arithmetic and geometric mean are the types of mean that can be calculated.
Mean formula[tex]\rm Mean = \dfrac{sum\ of\ the\ term}{total\ number\ of\ term}[/tex]
Given
The test scores of a class of 34 students have a mean of 73.1 and
The test scores of another class of 26 students have a mean of 66.4.
To find
The mean of the combined group.
How to get the mean of the combined group?The sum of the total of the test score = Total number of students x Mean
Class 1 sum will be
The sum of the total test score = 34 x 73.1
The sum of the total test score = 2485.4
Class 2 sum will be
The sum of the total test score = 26 x 66.4
The sum of the total test score = 1726.4
Than Mean for the combine will be
[tex]\rm Mean = \dfrac{sum\ of\ the\ term}{total\ number\ of\ term}\\\\Mean = \dfrac{2485.4+1726.4}{34+26} \\\\Mean = \dfrac{4211.8}{60} \\\\Mean = 70.19[/tex]
Thus the Mean of the combined class is 70.19.
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Please help me out with #8 surface area and please explain
Answers
angle symbol is meant to connect to the vertical line closest to us.
What's all the writing to the left and right. What does it say?
Finally what do you know? Can you find area using Heron's Formula for area. I'm going to hold off saying much. You can put a remark somewhere. The magic editor will pick it up and tell me you've done it.
You track fuel accounts for q shipping line. One vessel consumes 2 1/2 tons of fuel per day during transport. About how many fuel in tons should the vessel consume during 15-day transport?
Answers
2 1/2
Rewriting we have:
2.5 ton / day
The total amount after 15 days is:
(2.5) * (15) = 37.5 Ton
Rewriting we have:
37 1/2 Ton
Answer:
the vessel should consume 37 1/2 Ton during 15-day transport
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let f(x)=4x+3 and g(x)=-2+5. find (fog) (5)
Answers
let f(x)=4x+3 and g(x)=-2x+5. find (fog) (5)
we have that
f(x)=4x+3
g(x)=-2x+5
(fog) (x)---------> f(g(x))------> 4*[-2x+5]+3---------> f(g(x))=-8x+20+3
f(g(x))=-8x+23
then
(fog) (5)=-8*5+23---------> (fog) (5)=-40+23
(fog) (5)=-17
the answer is (fog) (5)=-17
Hdc produces microcomputer hard drives at four different production facilities (f1, f2, f3, and f4) hard drive production at f1, f2, f3, and f4 is 20%, 25%, 15%, and 40%, respectively. quality control records indicate that 1.5%, 2%, 1%, and 3% of the hard drives are defective at f1, f2, f3, and f4, respectively.
a. if a defective hdc hard drive is picked at random, what is the probability that it was produced at f2?
b. if a defective hdc hard drive is picked at random, what is the probability that it was produced at f4?
c. if an hdc hard drive is picked at random, what is the probability that it is non-defective? g
Answers
a. Probability defective hard drive from F2 ≈ 0.3226.
b. Probability defective hard drive from F4 ≈ 0.7742.
c. Probability non-defective hard drive ≈ 0.9785.
To solve this problem, we can use Bayes' theorem, which states:
[tex]\[ P(A|B) = \frac{P(B|A) \times P(A)}{P(B)} \][/tex]
Where:
- [tex]\( P(A|B) \)[/tex] is the probability of event A happening given that event B has occurred.
- [tex]\( P(B|A) \)[/tex] is the probability of event B happening given that event A has occurred.
- [tex]\( P(A) \)[/tex] and [tex]\( P(B) \)[/tex] are the probabilities of events A and B, respectively.
Let's solve each part of the problem:
a. If a defective HDC hard drive is picked at random, what is the probability that it was produced at F2?
Let:
- A be the event that the hard drive is defective.
- B be the event that the hard drive was produced at F2.
We need to find [tex]\( P(B|A) \)[/tex], the probability that the hard drive was produced at F2 given that it is defective.
[tex]$\begin{aligned} & P(B \mid A)=\frac{P(A \cap B)}{P(A)} \\ & P(A \cap B)=P(A \mid B) \times P(B)=0.02 \times 0.25=0.005 \\ & P(A)=P(A \cap F 1)+P(A \cap F 2)+P(A \cap F 3)+P(A \cap F 4) \\ & P(A)=0.015 \times 0.20+0.02 \times 0.25+0.01 \times 0.15+0.03 \times 0.40=0.0155 \\ & P(B \mid A)=\frac{0.005}{0.0155} \approx 0.3226\end{aligned}$[/tex]
b. If a defective HDC hard drive is picked at random, what is the probability that it was produced at F4?
We need to find [tex]\( P(F4|A) \)[/tex], the probability that the hard drive was produced at F4 given that it is defective.
[tex]\[ P(F4|A) = \frac{P(A \cap F4)}{P(A)} \][/tex]
[tex]\[ P(A \cap F4) = P(A|F4) \times P(F4) = 0.03 \times 0.40 = 0.012 \][/tex]
[tex]\[ P(F4|A) = \frac{0.012}{0.0155} \approx 0.7742 \][/tex]
c. If an HDC hard drive is picked at random, what is the probability that it is non-defective?
Let P(non-defective) = P(non-defective at F1) + P(non-defective at F2) + P(non-defective at F3) + P(non-defective at F4)
P(non-defective) = [tex](1 - 0.015) \times 0.20 + (1 - 0.02) \times 0.25 + (1 - 0.01) \times 0.15 + (1 - 0.03) \times 0.40[/tex]
P(non-defective) = [tex]0.985 \times 0.20 + 0.98 \times 0.25 + 0.99 \times 0.15 + 0.97 \times 0.40[/tex]
P(non-defective) = 0.197 + 0.245 + 0.1485 + 0.388 = 0.9785
So, the probability that an HDC hard drive picked at random is non-defective is approximately [tex]\( 0.9785 \)[/tex].
Can you answer this one for my plz I need before Monday ty
Answers
.. -x^2 -4x +1
.. = -(x^2 +4x) +1
.. = -(x^2 +4x +(4/2)^2) +1 -(-(4/2)^2) . . . . complete the square
.. = -(x +2)^2 +5
The vertex is (-2, 5).
The axis of symmetry is x = -2.
A triangle has one 60 degree angle and one 20 degree angle. The third angle in the triangle must be
a.acute
b.obtuse
c.commplementary
d.straight
Answers
Which shows how the distributive property can be used to evaluate 7×84/5?
Answers
84/5 gives 16 4/5. I don't know how much saving in work there is in doing this but you could do this.
7 ( 16 + 4/5)
You still have to do something with 28/5 when you get done, but it is one way to do it.
Evaluate g(n-5) if g(x)=x^2-6/2x
Answers
g( n-5) = ( [n-5]^2 -6) / 2(n-5)
= ( n^2 -10n +25 -6 ) / 2(n-5) or ( n^2 -10n +25 -6 ) / (2n-10)
Your Answer
I did foil first and got n^2-10n+25
then did 25-6 to get 19.
so n^2-10n+25 is on top.////// should have been n^2 -10n +19
and the bottom is 2(n-5) so I distributed the 2 and got 2n-10
so n^2-10n+25
2n-10
= (n^2 -10n +19) / 2(n-5)
Answer: We want to evaluate g(x) in x = n -5, where g(x) = (x^2-6)/2x
then, doing it step by step: [tex]g( n- 5)= \frac{(n-5)^{2} - 6 }{2*(n-5)}= \frac{(n^{2} - 2*5*n + 25) - 6 }{2n - 10} = \frac{n^{2} - 10n + 19 }{2n - 10}[/tex]
where i used that (a - b)^2 = a^2 -2ab + b^2
So the correct option is C.
A bank runs a contest to encourage new customers to open accounts. In the contest, each contestant draws a slip representing a different reward—$5, $3, or $x—from a jar. At the beginning of the contest the jar contains 60 slips for $5, 40 slips for $3, and 50 slips for $x.
If the expected value of the first draw from the jar is $5.8, the value of x is__
At one point in the contest, the jar contains 3 slips for $5, 7 slips for $3, and y slips for $x. If the expected value on the next draw is $6, the value of y is __
Answers
We'll use x = 9 from part 1.
Again let,
A = event that the $5 reward is drawn
B = event that the $3 reward is drawn
C = event that the $x reward is drawn (x is some positive number)
We can update event C to say
C = event that the $9 reward is drawn
The probabilities change to
P(A) = 3/(10+y)
P(B) = 7/(10+y)
P(C) = y/(10+y)
where y is some positive whole number. It represents the number of slips in jar C
The net values are
V(A) = 5
V(B) = 3
V(C) = x = 9
Like before, multiply the probabilities and net values to get
P(A)*V(A) = (3/(10+y))*5 = 15/(10+y)
P(B)*V(B) = (7/(10+y))*3 = 21/(10+y)
P(C)*V(C) = (y/(10+y))*9 = (9y)/(10+y)
The results add up to
[ 15/(10+y) ] + [ 21/(10+y) ] + [ (9y)/(10+y) ]
(15+21+9y)/(10+y)
(36+9y)/(10+y)
That last expression is the expected value. The expected value is also given to be 6, so set the two expressions equal to each other and solve for y.
(36+9y)/(10+y) = 6
36+9y = 6(10+y)
36+9y = 6(10)+6(y)
36+9y = 60+6y
9y-6y = 60-36
3y = 24
3y/3 = 24/3
y = 8
The statement "y slips of $x" turns into "8 slips of $9" since x = 9 and y = 8.
what is the probabilaty getting a sum of 7 if you rolled a pair of dice?
Answers
Out of these, six {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} give a sum of 7.
Therefore the probability of rolling a sum of 7 is 6/36=1/6
Which number line represents the solution set for the inequality-4(x+3)<-2-2x
Answers
The inequality -4(x+3)<-2-2x simplifies to x > -5. Therefore, the number line representing this solution should shade the values greater than -5, thus, highlighting that x is any value greater than -5.
Explanation:Let's solve the inequality -4(x+3)<-2-2x step by step:
Distribute -4 through (x+3) to get -4x-12<-2-2x. Add 4x to both sides to isolate x, which leads to -12 < 2x - 2. Finally, add 2 to both sides to isolate x to get -10 < 2x or equivalently, x > -5.
As a result, the number line that represents this solution is one where the values greater than -5 are shaded. This means the value of x is anything greater than -5. Any number line visualization should start at -5 and include everything to the right of that point.
Learn more about Inequalities here:https://brainly.com/question/30231190
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