Answer:
Thousandths: 7
Ten-thousandths: 4
Tenths: 3
Step-by-step explanation:
The given number is 59.3274659
The place values of a numner is given in the followwing format:
Tens Ones.Tenths Hundredths Thousandths Tenthousandths and so on
We can fit our number
Tens Ones.Tenths Hundredths Thousandths Tenthousandths HT M
5 9 . 3 2 7 4
Thousandths: 7
Ten-thousandths: 4
Tenths: 3
Find an equation of a line with the x- and y-intercepts below. Use exact fractions when necessary.
x-intercept 7; y-intercept -5
Answer:
The line with the x- and y-intercepts below has the following equation:
[tex]f(x) = \frac{5x}{7} - 5[/tex]
Step-by-step explanation:
The equation of the line has the following format:
[tex]f(x) = ax + b[/tex]
We are given two points, we are going to substitute them into the above equation, and find the equation of the line given the conditions.
Solution
Starting from the y-intercept makes the solution easier, since the term a is multiplied by 0
y-intercept -5
This means that when [tex]x = 0, y = f(x) = -5[/tex], so:
[tex]f(x) = ax + b[/tex]
[tex]-5 = a(0) + b[/tex]
[tex]b = -5[/tex]
For now, the line has the following equation:
[tex]f(x) = ax - 5[/tex]
x-intercept 7
This means that when [tex]y = f(x) = 0,x = 7[/tex], so:
[tex]f(x) = ax - 5[/tex]
[tex]0 = 7(a) - 5[/tex]
[tex]7a = 5[/tex]
[tex]a = \frac{5}{7}[/tex]
So, the line with the x- and y-intercepts below has the following equation:
[tex]f(x) = \frac{5x}{7} - 5[/tex]
John sells hot dogs from a cart outside an office building 5 days a week. If the price of a hot dog is $3, what is John’s weekly revenue from hot dogs if he sells 47 each day.
Answer:
John’s weekly revenue from hot dogs = [tex]\$705[/tex]
Step-by-step explanation:
John sells hot dogs from a cart outside an office building 5 days a week .
Price of a hot dog = [tex]\$3[/tex]
Number of hot dogs sold each day = 47
So, total number of hot dogs sold on 5 days = 5 × Number of hot dogs sold each day = 5 × 47 = [tex]235[/tex]
We need to find John’s weekly revenue from hot dogs if he sells 47 each day.
John’s weekly revenue from hot dogs = Price of a hot dog × total number of hot dogs sold on 5 days = 3 × 235 = [tex]\$705[/tex]
Suppose that Jane borrows $8400 from a bank on April 9 at an annual rate of 9.2% simple interest. How much does she owe on August 20 of the same year?
Answer:
274.576 + 8400 = $8674.576
Step-by-step explanation:
Here, Number of days = 21 + 31 +30 +31 +20 = 123
We know that,
[tex]Simple Interest = \frac{P\timesT\timesR}{100}[/tex]
where, P = Principle = 8400
T = time = 123 ÷ 365
R = Rate = 9.2
⇒ [tex]Simple Interest = \frac{8400\times123\times9.2}{365\times100}[/tex]
⇒ Simple Interest = 274.576
Thus, total amount Jane has on 20 August = 274.576 + 8400 = $8674.576
Use a proof by contradiction to prove that the sum of two odd integers is even CM
Answer:
The sum of two odd integers is even
Step-by-step explanation:
Proof by contradiction:
We are going to assume that the sum of two odd integers is odd.
An odd integer is written as 2p+1 where p is an integer and an even integer is written as 2p where p is an integer
So, if the sum of two odd integers is odd we would have
[tex](2k+1) + (2p+1) = 2r+1\\2k+1+2p+1=2r+1\\2k+2p+2=2r+1\\2(k+p+1)=2r+1[/tex]
The left side of the equation is clearly an even number while the right side of the equation is odd. Therefore, our hypothesis is wrong and we can conclude that the sum of two odd integers is even.
How many grams are 2.7x10-^4 grams
Answer:
0,00027 grams!
Step-by-step explanation:
This is a scientific notation problem.
When a number is followed by a [tex]10^{-1}[/tex], it means that said number has one zero at the beginning.
In this case, the number is followed by [tex]10^{-4}[/tex], so that means that 2.7 has four zeros at the beginning. So, 2.7x[tex]10^{-4}[/tex] grams is equal to 0,00027 grams! (always the comma goes after the first zero).
The expression "2.7x10^-4 grams" is already given in grams, using scientific notation, and does not require a unit conversion since it is already in the unit we are interested in (grams). To understand what this quantity represents in standard decimal form, let's break down the scientific notation:
Scientific notation is a way of expressing very large or very small numbers in a compact form. The notation "2.7x10^-4" means that you take the number 2.7 and multiply it by 10 raised to the power of -4.
The term "10^-4" means "1 divided by 10 to the 4th power," which is the same as 0.0001 (1 followed by 4 zeros in the denominator).
To convert "2.7x10^-4" to its decimal form, you would perform the multiplication:
2.7 × 0.0001 = 0.00027
So, "2.7x10^-4 grams" is equal to 0.00027 grams in standard decimal notation.
In 1912, the RMS Titanic, a British passenger ship, sank in the North Atlantic Ocean after colliding with an iceberg. Historians do not know the exact passenger list, so the death toll is estimated. Here is data from the 2201 passengers on board, by cabin class. First Class Second Class Third Class Crew Row Totals Died 122 167 528 673 1490 Survived 203 118 178 212 711 Col Totals 325 285 706 885 2201 Source: Wikipedia, RMS Titanic (2015) If we randomly select a passenger who survived the Titanic, what is the probability that this passenger is in a second class cabin?
The probability that a randomly selected survivor was in second class is calculated by dividing the number of second class passengers who survived (118) by the total number of survivors (711), resulting in a probability of 16.6%.
Explanation:The subject of this question is probability. To calculate the probability that a randomly selected survival was in the second class cabin, we need to consider the number of second class passengers who survived compared to the total number of survivors. From the data provided, we can see that 118 passengers in second class survived the incident. The total number of survivors is 711.
Therefore, the probability (P) is calculated as follows:
P = Number of successful outcomes / Total number of outcomes.
Hence, P = 118 / 711 = 0.166.
This means that there is a 16.6% chance that a passenger who survived was from the second-class cabin.
Learn more about Probability here:
https://brainly.com/question/22962752
#SPJ12
The probability that a randomly selected survivor is from the second-class cabin is approximately 0.166, or 17% when rounded to the nearest percent.
Explanation:Any calculation of probability involves dividing the number of favorable outcomes by the total number of outcomes. In this case, you want to find the probability that a randomly selected passenger who survived the Titanic was in a second-class cabin.
From the chart, we can see that 118 second-class passengers survived. The total number of survivors is 711. Hence, the probability of a survivor being from the second-class cabin is given by the formula:
Probability = favorable outcomes / total outcomes = number of second class survivors / total number of survivors.
Substituting these values into the formula we get: Probability = 118 / 711 = 0.166, or around 17% when rounded to the nearest percent.
So, if we randomly select a passenger who survived the Titanic, there is a 17% probability that this passenger is in a second-class cabin.
Learn more about Probability here:
https://brainly.com/question/32117953
#SPJ2
A patient is ordered 60 mg/kg/day of an antibiotic to be divided in 3 doses per day. The patient weighs 60 kg. How many mg of drug should they receive per dose and how many mg of drug should they receive per day Round answers to the nearest tenth and DO NOT include units
Answer:
3600 mg of drug per day.
[tex]\text{The amount of drug per dosage}=\text{1200 mg}[/tex]
Step-by-step explanation:
We have been given that a patient is ordered 60 mg/kg/day of an antibiotic.
Since the weight of the patient is 60 kg, so the dosage of drug per day would be 60 mg times 60.
[tex]\text{The dosage of drug per day}=\frac{\text{60 mg}}{\text{ kg}}\times 60\text{ kg}[/tex]
[tex]\text{The dosage of drug per day}=\text{60 mg}\times 60[/tex]
[tex]\text{The dosage of drug per day}=\text{3600 mg}[/tex]
Therefore, the patient should receive 3600 mg of drug per day.
Since the patient gets 3 doses per day, so dosage of drug per dosage would be amount of drug per day divided by 3.
[tex]\text{The amount of drug per dosage}=\frac{\text{3600 mg}}{3}[/tex]
[tex]\text{The amount of drug per dosage}=\text{1200 mg}[/tex]
Therefore, the amount of drug per dosage would be 1200 mg.
State whether the following statement is true or false, and explain why. If the statement is false, state the true change. A high school reports that its students' SAT scores were down by 12% in one year. The next year, however, the test scores rose by 20%. The high school principal announces, "Overall, test scores have improved by 8% over the past two years."
False. The high school principal's claim of an overall improvement of 8% is incorrect; the actual improvement is 5.6% over the two years.
Initial SAT scores were down by 12%.
Next year, the scores rose by 20%.
Let's assume the initial SAT score was 100 (just for simplicity):
After the 12% decrease: 100 - 12% of 100
= 88.
After the 20% increase on 88:
88 + 20% of 88
= 105.6.
So, in this scenario, the scores have actually increased from 100 to 105.6, which is an increase of 5.6%, not 8%.
Hence, the high school principal's claim of an overall improvement of 8% is incorrect. The true change is an increase of 5.6% over the two years.
To learn more on Percentage click:
https://brainly.com/question/24159063
#SPJ12
The principal's statement is false because percentage change does not simply add and subtract, but rather it compounds. After the changes, the scores have increased by 5.6% of the original scores, not 8%.
Explanation:The statement given by the principal is false. It's a common misconception that you can add and subtract percentages like usual numbers.
Let's say the original score was 100%. A 12% decrease would leave the scores at 88% of their original. The next year, when the scores rose by 20%, it didn't rise by 20% of the original 100%, but 20% of the new 88%, that is, 17.6%, making the new score 105.6% of the original scores.
So, over the two years, the scores did not increase by 8%, but rather, they increased by 5.6% of the original scores. The principal does not correctly understand the mathematical concept of percentage change.
Learn more about Percentage Change here:https://brainly.com/question/35855541
#SPJ11
use completeling the square to solve for x in the equation (x-12)(x+4)=9
Answer:
x= [tex]4\pm \sqrt{73}[/tex]
Step-by-step explanation:
First step. Solve the binomial product from the left side of the equation:
[tex](x-12)(x+4)=9[/tex]
[tex]x^2+4x-12x-48=9[/tex]
Second step. Simplify and move independent terms to the right side of the equation:
[tex]x^2-8x=57[/tex]
Third step. Find a number that multiplied by two gives -8, then square this number and sum it on both sides of the equation:
[tex]x^2-8x+16=57+16[/tex]
Fourth step. Write the left side of the equation as a squared binomial:
[tex](x-4)^2=73[/tex]
Fifth step. First, take the square root and then add 4 to both sides of the equation to solve for x:
[tex](x-4)=\pm \sqrt{73}[/tex]
[tex]x= 4\pm \sqrt{73}[/tex]
One milligram of streptomycin sulfate contains the antibiotic activity of 650 μg of streptomycin base. How many grams streptomycin sulfate would be the equivalent of 1 g of streptomycin base?
Answer:
1.538 g of streptomycin sulfate
Step-by-step explanation:
As we know, we have 650 μg of streptomycin base in 1 milligram of streptomycin sulfate.
If we convert everithing to grams:
650 μg= 0.00065 g of Streptomycin base for every 0.001 grams of Streptomycin Sulfate so we have :0.001 grs Streptomycin Sulfate/0.00065 gr Streptomycin base=1.538 gr Streptomycin Sulfate/Streptomycin base
Now if we want 1 gram of Streptomycin base we will need:
1 g of Streptomycin base*1.538 gr Streptomycin Sulfate/Streptomycin base= 1.538 gr Streptomycin Sulfate
If P and Q are predicates over some domain, and if it is true that Vx(P(x)V Q(x)), must VxP(x) v VæQ(x) also be true? Explain.
Answer:
It is not true
Step-by-step explanation:
Suppose your domain is the integer numbers. Define
P(x)="x is even"
Q(x)="x is odd"
So we have that the predicate [tex]\forall x(P(x) \vee Q(x))[/tex] is always true because the integers are always even or odd. But the predicate [tex]\forall x P(x) \vee \forall x Q(x)[/tex] means that all the integer numbers are even or all the integer numbers are odd, which is false. So we can't deduce [tex]\forall x P(x) \vee \forall x Q(x)[/tex] from [tex]\forall x(P(x) \vee Q(x))[/tex].
You are asked to bake muffins for a breakfast meeting. Just as you are about to start making them, you get a call saying that the number of people coming to the meeting has doubled! Your original recipe called for three eggs. How many eggs do you need to make twice as many muffins?
Answer: 6
Step-by-step explanation:
Given : The number of eggs required to make the current muffin receipe = 3
If the number of people coming to the meeting has doubled, then we need twice of as many eggs .
i.e. the number of eggs required to make twice of muffin =[tex]2\times3=6[/tex]
Hence, the number of people coming to the meeting has doubled = 6
A vial contains 80 mg of drug in 2 mL of injection. How many milliliters of the injection should be administered to obtain 0.02 g of the drug?
Answer:
0.5mL of the injection should be administered to obtain 0.02 g of the drug.
Step-by-step explanation:
First step: The first step of this problem is the conversion of 0.02g to mg.
Each gram has 1000 miligrams. So:
1g - 1000mg
0.02g - xmg
x = 1000*0.02
x = 20mg
Final step:
A vial contains 80 mg of drug in 2 mL of injection. How many mL should be administered to obtain 0.02 g = 20mg of the drug.
This can be solved as a rule of three problem.
In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.
When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.
When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.
In this step, as the dose of the injection increases, so the quantity of the drug. So the relationship between the measures is direct. So:
80 mg - 2mL
20 mg - xmL
80x = 40
[tex]x = \frac{40}{80}[/tex]
x = 0.5mL
0.5mL of the injection should be administered to obtain 0.02 g of the drug.
To obtain 0.02 grams of the drug, the required volume to administer from the vial containing 80 mg in 2 mL is 0.5 mL.
Explanation:To solve this problem, we need to determine how many milliliters contain 0.02 grams of the drug. First, convert the amount we want from grams to milligrams since our given concentration is in milligrams: 0.02 grams is equal to 20 milligrams.
Given that the vial contains 80 mg of drug in 2 mL, we can calculate the volume required for 20 mg. The formula we will use is:
(desired dose / concentration of vial) × volume of vial = required volume
(20 mg / 80 mg) × 2 mL = 0.5 mL
Therefore, to obtain 0.02 grams (20 mg) of the drug, the required volume to administer would be 0.5 mL.
One side of a triangular lot is 150 ft and the angel oppiste this side is 55 degrees. Another angel is 63 degrees. Determine how much fencing is needed to enclose it.
Answer:
474.84 ft of fencing is needed
Step-by-step explanation:
We know that the angles of a triangle sum up 180º. We already know 2 of the triangle's angles (55º and 63º). Therefore the third angle measures:
180 - 55 - 63 = 62.
To know how much fencing is needed, we need the perimeter of the triangle, so we need to find out how much the other sides of the lot measure.
We will use law of sins to solve this problem.
First we solve for y:
[tex]\frac{150}{sin55}= \frac{y}{sin63} \\y=150 (sin63)/(sin55)\\y=133.65/.8191\\y=163.16[/tex]
Now we solve for the other side of the lot, x:
[tex]\frac{150}{sin55}=\frac{x}{sin62}\\ x=150(sin62)/(sin55)\\x=150(.8829)/.8191\\x=132.435/.8191\\x=161.68[/tex]
Now that we have the measures of all the sides we sum them up
total fencing needed= 150 + 163.16 + 161.68 = 474.84
Rewrite using necessary or sufficient condition, as appropriate: "if the oven is hot then you may put the cake in."
Answer:
Answered
Step-by-step explanation:
"if the oven is hot then you may put the cake in". This the premise.
Now, The necessary condition will be "If the oven is hot" meaning the cake can be put inside the oven only when the oven is hot and not otherwise. So, here the hotness of the oven becomes the necessary condition.
"Oven is hot" is the necessary condition "to put the cake in"
Consider the quadratic function f(x) = -x^2 + 1x + 12. Determine the following: (enter all numerical answers as integers, fractions, or decimals): The smallest -intercept is 2 = The largest x-intercept is z = The y-intercept is y = The vertex is ( The line of symmetry has the equation Preview
Answer and Explanation:
Given : The quadratic function [tex]f(x)=-x^2+x+12[/tex]
To find : Determine the following ?
Solution :
The x -intercept are where f(x)=0,
So, [tex]-x^2+x+12=0[/tex]
Applying middle term split,
[tex]-x^2+4x-3x+12=0[/tex]
[tex]-x(x-4)-3(x-4)=0[/tex]
[tex](x-4)(-x-3)=0[/tex]
[tex]x=4,-3[/tex]
The x-intercepts are (4,0) and (-3,0).
The smallest x-intercept is x=-3
The largest x-intercept is x=4
The y -intercept are where x=0,
So, [tex]f(0)=-(0)^2+0+12[/tex]
[tex]f(0)=12[/tex]
The y-intercept is y=12.
The quadratic function is in the form, [tex]y=ax^2+bx+c[/tex]
On comparing, a=-1 , b=1 and c=1 2
The vertex of the graph is denote by (h,k) and the formula to find the vertex is
For h, The x-coordinate of the vertex is given by,
[tex]h=-\frac{b}{2a}[/tex]
[tex]h=-\frac{1}{2(-1)}[/tex]
[tex]h=\frac{1}{2}[/tex]
For k, The y-coordinate of the vertex is given by,
[tex]k=f(h)[/tex]
[tex]k=-h^2+h+12[/tex]
[tex]k=-(\frac{1}{2})^2+\frac{1}{2}+12[/tex]
[tex]k=-\frac{1}{4}+\frac{1}{2}+12[/tex]
[tex]k=\frac{-1+2+48}{4}[/tex]
[tex]k=\frac{49}{4}[/tex]
The vertex of the function is [tex](h,k)=(\frac{1}{2},\frac{49}{4})[/tex]
The x-coordinate of the vertex i.e. [tex]x=-\frac{b}{2a}[/tex] is the axis of symmetry,
So, [tex]x=-\frac{b}{2a}=\frac{1}{2}[/tex] (solved above)
The axis of symmetry is [tex]x=\frac{1}{2}[/tex].
Assume that 155 students are surveyed and every student takes at least one of the following languages. The results of the survey are as follows:
90 take French.
83 take German.
42 take French and German.
41 take German and Russian.
22 take French as their only foreign language.
22 take French, Russian, and German.
(1) How many take Russian?
(2) How many take French and Russian but not German?
Answer:
91 people take Russian
26 people take French and Russian but not German
Step-by-step explanation:
To solve this problem, we must build the Venn's Diagram of this set.
I am going to say that:
-The set A represents the students that take French.
-The set B represents the students that take German
-The set C represents the students that take Russian.
We have that:
[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]
In which a is the number of students that take only Franch, A \cap B is the number of students that take both French and German , A \cap C is the number of students that take both French and Russian and A \cap B \cap C is the number of students that take French, German and Russian.
By the same logic, we have:
[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]
[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]
This diagram has the following subsets:
[tex]a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)[/tex]
There are 155 people in my school. This means that:
[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 155[/tex]
The problem states that:
90 take Franch, so:
[tex]A = 90[/tex]
83 take German, so:
[tex]B = 83[/tex]
22 take French, Russian, and German, so:
[tex]A \cap B \cap C = 22[/tex]
42 take French and German, so:
[tex]A \cap B = 42 - (A \cap B \cap C) = 42 - 22 = 20[/tex]
41 take German and Russian, so:
[tex]B \cap C = 41 - (A \cap B \cap C) = 41 - 22 = 19[/tex]
22 take French as their only foreign language, so:
[tex]a = 22[/tex]
Solution:
(1) How many take Russian?
[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]
[tex]C = c + (A \cap C) + 19 + 22[/tex]
[tex]C = c + (A \cap C) + 41[/tex]
First we need to find [tex]A \cap C[/tex], that is the number of students that take French and Russian but not German. For this, we have to go to the following equation:
[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]
[tex]90 = 22 + 20 + (A \cap C) + 22[/tex]
[tex](A \cap C) + 64 = 90[/tex].
[tex](A \cap C) = 26[/tex]
----------------------------
The number of students that take Russian is:
[tex]C = c + 26 + 41[/tex]
[tex]C = c + 67[/tex]
------------------------------
Now we have to find c, that we can find in the equation that sums all the subsets:
[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 155[/tex]
[tex]22 + b + c + 20 + 26 + 19 + 22 = 155[/tex]
[tex]b + c + 109= 155[/tex]
[tex]b + c = 46[/tex]
For this, we have to find b, that is the number of students that take only German. Then we go to this eqaution:
[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]
[tex]B = b + 19 + 20 + 22[/tex]
[tex]B = b + 61[/tex]
[tex]b + 61 = 83[/tex]
[tex]b = 22[/tex]
-------
[tex]b + c = 46[/tex]
[tex]c = 46 - b[/tex]
[tex]c = 24[/tex]
The number of people that take Russian is:
[tex]C = c + 67[/tex]
[tex]C = 24 + 67[/tex]
[tex]C = 91[/tex]
91 people take Russian
(2) How many take French and Russian but not German?
[tex](A \cap C) = 26[/tex]
26 people take French and Russian but not German
A number multiplied by a negative three is 12.
(Can somebody please help me write this into an expression??) :(
In order to find the answer to this question you will have to find the missing number in the given numbers we have then fill in the blank to make your expression and your answer.
[tex]-3 = 12[/tex]
[tex]-4 \times -3 = 12[/tex]
[tex]= -4[/tex]
Therefore your answer is "-4 * -3 = 12."
Hope this helps.
Evaluate 4P3
24
7
10
12
Answer:
24
Step-by-step explanation:
4P3 = 4!/(4-3)! = 4·3·2 = 24
Determine whether p ∧ (p → q) and p ∧ q are logically equivalent.
Answer:
Yes they are logically equivalent
p∧(p→q)≡p∧q
Step-by-step explanation:
The statements are logically equivalent if they have the same truth tables. So let´s use truth tables in order to determine if they are logically equivalent or not:
The picture that I attached you shows the truth table for each case. As you can see, p∧(p→q) has exactly the same truth values as p∧q. So we can conclude that they are logically equivalent.
Calculate the potential energy associated with 1 m^3 of water at 607 feet tall taking the mass of 1 m^3 of water to be 1000 kg
Answer:
The potential energy associated with the given mass equals 1814.98 kilo Joules.
Step-by-step explanation:
We know that for a object of mass 'm' standing at a height of 'h' meters above the surface of earth the potential energy associated with the object is given by
[tex]P.E=mass\times g\times h[/tex]
where
'g' is acceleration due to gravity.
Since it is given that mass of 1 cubic meter of water is 1000 kilograms that the mass associated with given quantity of water is also 1000 kilograms since the volume is 1 cubic meter.
The height is given as 607 feet = [tex]{607}\times 0.3048=185.0136[/tex]meters
Applying the values in the above equation we get
[tex]P.E=1000\times 9.81\times 185.0136=1814.98kJ[/tex]
The slope f′(x) at each point (x,y) on a curve y=f(x) is given, along with a point (a,b) on the curve. Use this information to find f(x). f′(x) = 4x/(1 + 7x^2) (0,10) NOTE: OF absolute value symbols, | | , are needed for the answer, then use abs(expression). For example, ln|x| must be entered as ln(abs(x))
[tex]f'(x)=\dfrac{4x}{1+7x^2}[/tex]
Integrating gives
[tex]f(x)=\displaystyle\int\frac{4x}{1+7x^2}\,\mathrm dx[/tex]
To compute the integral, substitute [tex]u=1+7x^2[/tex], so that [tex]\frac27\,\mathrm du=4x\,\mathrm dx[/tex]. Then
[tex]f(x)=\displaystyle\frac27\int\frac{\mathrm du}u=\frac27\ln|u|+C[/tex]
Since [tex]u=1+7x^2>0[/tex] for all [tex]x[/tex], we can drop the absolute value, so we end up with
[tex]f(x)=\dfrac27\ln(1+7x^2)+C[/tex]
Given that [tex]f(0)=10[/tex], we have
[tex]10=\dfrac27\ln1+C\implies C=10[/tex]
so that
[tex]\boxed{f(x)=\dfrac27\ln(1+7x^2)+10}[/tex]
Keith has 40-pound bags of mulch in his truck that weigh a total of 3600 pounds. His Owner’s Manual lists the truck’s capacity as at most 3000 pounds. How many bags does Keith need to remove in order to meet the weight requirements?
Please show step by step explanation
Keith is 600 pounds over the weight limit of his truck. Given that each bag of mulch weighs 40 pounds, Keith needs to remove 15 bags of mulch to be within the truck's weight capacity.
Explanation:This is a straightforward math problem involving subtraction and division. First, let's find out how much weight is over the truck's capacity. Keith's truck is currently carrying 3600 pounds of mulch, but his truck's capacity is only 3000 pounds. So, he is over by 3600 - 3000 = 600 pounds.
Each bag of mulch weighs 40 pounds, so to find out how many bags Keith needs to remove, we simply divide the total excess weight by the weight of each bag: 600 / 40 = 15 bags. Therefore, Keith needs to remove 15 bags of mulch from his truck to meet the weight requirements.
Learn more about Subtraction and Division here:https://brainly.com/question/1292482
#SPJ12
Keith needs to remove 15 bags of mulch to meet the weight requirements.
Explanation:To find out how many bags Keith needs to remove, we need to determine the weight of one bag of mulch. If he has 40-pound bags and a total weight of 3600 pounds, we can divide the total weight by the weight of one bag:
Number of bags = Total weight / weight of one bag = 3600 pounds / 40 pounds = 90 bags
Since the truck's capacity is at most 3000 pounds, Keith needs to remove the excess weight:
Excess weight = Total weight - Truck's capacity = 3600 pounds - 3000 pounds = 600 pounds
Now, we can calculate how many bags he needs to remove using the weight of one bag:
Bags to remove = Excess weight / weight of one bag = 600 pounds / 40 pounds = 15 bags
Learn more about Mulch bags here:https://brainly.com/question/15368655
#SPJ12
Suppose you pick 6 different numbers in [10]. Prove that 2 of the numbers are next to each other. (Hint: use the pigeonhole principle. What are the pigeons and what are the holes?)
Step-by-step explanation:
We are picking 6 numbers from the numbers 1,2,3,4,5,6,7,8,9,10. Since we care about numbers being next to each other, we might think of the 10 numbers as being distributed in 5 boxes (which you can think of as the holes):
| 1 2 | 3 4 | 5 6 | 7 8 | 9 10 |
So on the first box we have the numbers 1 and 2, on the second box we have the numbers 3 and 4, and so on. Since we are picking 6 numbers from those 10 numbers, that means we'll have to pick 6 boxes (and inside each box we pick a number), but we only have 5 available boxes, so by the pigeonhole principle, we'll have to pick 1 same box at least two times. Since on each picked box we'll need to pick a number, on this box which was picked two times, we will have to pick both of its numbers. And so those 2 numbers inside that box will be next to each other (meaning they're consecutive numbers).
Use Gaussian Elimination to find an equation of a polynomial that passes through points A(-5,-3), B(-2,3). C(3,3), D(6,19). Indicates row operations with the R notation. Leave coefficients in fraction form, do not report in decimals.
Answer:
The polynomial equation that passes through the points is [tex]2-\frac{2}{3}x+\frac{1}{12}x^{2}+\frac{1}{12}x^{3}[/tex]
Step-by-step explanation:
Suppose you have a function y = f(x) which goes through these points
A(-5,-3), B(-2,3). C(3,3), D(6,19)
there is a polynomial P(x) of degree 3 which goes through these point.
We use the fact that four distinct points will determine a cubic function.
P(x) is the degree 3 polynomial through the 4 points, a standard way to write it is
[tex]P(x) = a+bx+cx^2+dx^3[/tex]
Next replace the given points one by one, which leads to a system of 4 equations and 4 variables (namely a,b,c,d)
[tex]-3=a+b\cdot-5+c\cdot -5^2+d\cdot -5^3\\3=a+b\cdot-2+c\cdot -2^2+d\cdot -2^3\\3=a+b\cdot 3+c\cdot 3^2+d\cdot 3^3\\19=a+b\cdot 6+c\cdot 6^2+d\cdot 6^3[/tex]
We can rewrite this system as follows:
[tex]-3=a-5\cdot b+25\cdot c-125\cdot d\\3=a-2\cdot b+4\cdot c-8\cdot d\\3=a+3\cdot b+9\cdot c+27\cdot d\\19=a+6\cdot b+36\cdot c+216\cdot d[/tex]
To use the Gaussian Elimination we need to express the system of linear equations in matrix form (the matrix equation Ax=b).
The coefficient matrix (A) for the above system is
[tex]\left[\begin{array}{cccc}1&-5&25&-125\\1&-2&4&-8\\1&3&9&27\\1&6&36&216\end{array}\right][/tex]
the variable matrix (x) is
[tex]\left[\begin{array}{c}a&b&c&d\end{array}\right][/tex]
and the constant matrix (b) is
[tex]\left[\begin{array}{c}-3&3&3&19\end{array}\right][/tex]
We also need the augmented matrix, it is obtained by appending the columns of the coefficient matrix and the constant matrix.
[tex]\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\1&-2&4&-8&3\\1&3&9&27&3\\1&6&36&216&19\end{array}\right][/tex]
To transform the augmented matrix to the reduced row echelon form we need to follow these steps:
Subtract row 1 from row 2 [tex]\left(R_2=R_2-R_1\right)[/tex][tex]\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&3&-21&117&6\\1&3&9&27&3\\1&6&36&216&19\end{array}\right][/tex]
Subtract row 1 from row 3 [tex]\left(R_3=R_3-R_1\right)[/tex][tex]\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&3&-21&117&6\\0&8&-16&152&6\\1&6&36&216&19\end{array}\right][/tex]
Subtract row 1 from row 4 [tex]\left(R_4=R_4-R_1\right)[/tex][tex]\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&3&-21&117&6\\0&8&-16&152&6\\0&11&11&341&22\end{array}\right][/tex]
Divide row 2 by 3 [tex]\left(R_2=\frac{R_2}{3}\right)[/tex][tex]\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&1&-7&39&2\\0&8&-16&152&6\\0&11&11&341&22\end{array}\right][/tex]
Add row 2 multiplied by 5 to row 1 [tex]\left(R_1=R_1+\left(5\right)R_2\right)[/tex][tex]\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&8&-16&152&6\\0&11&11&341&22\end{array}\right][/tex]
Subtract row 2 multiplied by 8 from row 3 [tex]\left(R_3=R_3-\left(8\right)R_2\right)[/tex][tex]\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&0&40&-160&-10\\0&11&11&341&22\end{array}\right][/tex]
Subtract row 2 multiplied by 11 from row 4 [tex]\left(R_4=R_4-\left(11\right)R_2\right)[/tex][tex]\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&0&40&-160&-10\\0&0&88&-88&0\end{array}\right][/tex]
Divide row 3 by 40 [tex]\left(R_3=\frac{R_3}{40}\right)[/tex][tex]\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&0&1&-4&-1/4\\0&0&88&-88&0\end{array}\right][/tex]
Add row 3 multiplied by 10 to row 1 [tex]\left(R_1=R_1+\left(10\right)R_3\right)[/tex][tex]\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&-7&39&2\\0&0&1&-4&-1/4\\0&0&88&-88&0\end{array}\right][/tex]
Add row 3 multiplied by 7 to row 2 [tex]\left(R_2=R_2+\left(7\right)R_3\right)[/tex][tex]\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&88&-88&0\end{array}\right][/tex]
Subtract row 3 multiplied by 88 from row 4 [tex]\left(R_4=R_4-\left(88\right)R_3\right)[/tex][tex]\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&0&264&22\end{array}\right][/tex]
Divide row 4 by 264 [tex]\left(R_4=\frac{R_4}{264}\right)[/tex][tex]\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&0&1&1/12\end{array}\right][/tex]
Subtract row 4 multiplied by 30 from row 1 [tex]\left(R_1=R_1-\left(30\right)R_4\right)[/tex][tex]\left[\begin{array}{cccc|c}1&0&0&0&2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&0&1&1/12\end{array}\right][/tex]
Subtract row 4 multiplied by 11 from row 2 [tex]\left(R_2=R_2-\left(11\right)R_4\right)[/tex][tex]\left[\begin{array}{cccc|c}1&0&0&0&2\\0&1&0&0&-2/3\\0&0&1&-4&-1/4\\0&0&0&1&1/12\end{array}\right][/tex]
Add row 4 multiplied by 4 to row 3 [tex]\left(R_3=R_3+\left(4\right)R_4\right)[/tex][tex]\left[\begin{array}{cccc|c}1&0&0&0&2\\0&1&0&0&-2/3\\0&0&1&0&1/12\\0&0&0&1&1/12\end{array}\right][/tex]
From the reduced row-echelon form the solutions are:
[tex]\left[\begin{array}{c}a=2&b=-2/3&c=1/12&d=1/12\end{array}\right][/tex]
The polynomial P(x) is:
[tex]2-\frac{2}{3}x+\frac{1}{12}x^{2}+\frac{1}{12}x^{3}[/tex]
We can check our solution plotting the polynomial and checking that it passes through the points.
Assistance in understanding and solving this example on Elementary Number Theory with the details of the solution to better understand, thanks.
a) Find all solutions to 3x+4y=60 in positive integers.
b) A roadside stand bought 11 large baskets of eggs from a farmer and sold 39 small baskets of eggs, which hold fewer than a dozen. There were 19 eggs left over. How many eggs does a large basket hold?
Answer:
a)
x | y
==========
16 | 3
12 | 6
8 | 9
4 | 12
b) A large basket contains 23 eggs
Step-by-step explanation:
a) Isolating 3x, we get
3x = 60-4y
This means that 60-4y is a multiple of 3.
Besides, as x is a positive integer, then x = 60-4y has to be greater than zero
60-4y>0
Solving the inequality for y
60>4y
60/4 > y
15>y (or y<15, which is the same)
The multiples of 3 which are smaller than 15 are 3,6,9 and 12
Now we draw a table for these values of y and found the value of x by replacing in the equation 3x = 60-4y---> x = (60-4y)/3
x | y
=======
16 | 3
12 | 6
8 | 9
4 | 12
And these are all the solutions in positive integers.
b) This problem is pretty much like problem a)
Let's call x the number of eggs contained in the large baskets and y the number of eggs contained in the small baskets.
Then
Number of eggs bought - number of eggs sold = number of eggs left over.
In this case
11x-39y = 19
But we also know that 0<y<12 because the small baskets hols fewer than a dozen.
Isolating x in the equation 11x-39y + 19, we get
x=(19+39y)/11
Now we make a table with these values of x and y remembering that y is a positive integer smaller than 12
x | y
=======
5.2 | 1
8.8 | 2
12.4 | 3
15.9 | 4
19.5 | 5
23 | 6
26.6 | 7
30.1 | 8
33.6 | 9
37.2 |10
40.7 |11
From this table we see that the only integer solution for x is 23, so a large basket contains 23 eggs.
Solve the following circle graph problems.
If savings are represented by 45° on a circle graph showing all expenses, and the total expenses are $1440, how much go into savings?
Select one:
a. $240
b. $120
c. $180
d. $300
Answer:
$180 go into savings
Step-by-step explanation:
Given : Savings are represented by 45° on a circle graph showing all expenses.
Total expenses are $1440
To Find : How much go into savings?
Solution:
Total angle of expense = 360°
Savings = [tex]\frac{\text{Saving angle}}{\text{Total angle}} \times 1440[/tex]
= [tex]\frac{45}{360} \times 1440[/tex]
= [tex]180[/tex]
Hence $180 go into savings
Find all values of m so that the function y = xm is a solution of the given differential equation. (Enter your answers as a comma-separated list.)
x^2y'' − 14xy' + 54y = 0
m =
Answer:
m = 6,9
Step-by-step explanation:
We are given that [tex]y = x^m[/tex] is a solution to given differential equation.
[tex]x^2y'' - 14xy' + 54y = 0[/tex]
First we 3evaluate the value of:
[tex]y'' = m(m-1)x^{(m-2)}[/tex]
[tex]y' = mx^{(m-1)}[/tex]
Putting these value in the above differential equation, we get,
[tex]x^2m(m-1)x^{(m-2)} + 14xmx^{(m-1)} + 54x^m = 0[/tex]
[tex]x^m[m(m-1) -14m + 54] = 0[/tex]
[tex]x^m(m^2 - 15m + 54) = 0[/tex]
[tex](m^2 - 15m + 54) = 0[/tex]
[tex](m-9)(m-6) = 0[/tex]
[tex]m = 9,6[/tex]
Thus, for m = 9,6 , the function [tex]y = x^m[/tex] is a solution of given differential equation.
Values of m so that the function [tex]y = x^m[/tex] is a solution of the given differential equation are 6 and 9 and this can be determined by differentiating the [tex]y = x^m[/tex].
Given :
[tex]y = x^m[/tex][tex]x^2y"-14xy'+54y = 0[/tex] ------ (1)The following calculations can be used to determine the value of 'm':
[tex]y = x^m[/tex] ---- (2)
Now, differentiate the above equation with respect to 'x'.
[tex]\dfrac{dy}{dx}=\dfrac{d}{dx}(x^m)[/tex]
[tex]y' = mx^{m-1}[/tex] ----- (3)
Now, differentiate equation (2) with respect to 'x'.
[tex]\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(mx^{m-1})[/tex]
[tex]y" = m(m-1)x^{m-2}[/tex] ----- (4)
Now, put the values of y, y', and y" in equation (1).
[tex]x^2\times m(m-1)x^{m-2}-14x(mx^{m-1})+54x^m=0[/tex]
[tex]x^m(m(m-1)-14m+54)=0[/tex]
[tex]x^m(m^2-15m+54)=0[/tex]
(m - 9)(m - 6) = 0
m = 9 , 6
For more information, refer to the link given below:
https://brainly.com/question/13077606
Replacement times for CD players are normally distributed
with=5.5
years (66 months) and
= 1.5 years (18 months). Find the probabilitythat a randomly
selected CD player will have a replacement time orMORE THAN 7
YEARS.
Answer:
0.04
Step-by-step explanation:
X~N(μ=14.2; σ=0.9), a=7
[tex]P(X>7)=1-P(X\leq 7)[/tex]
[tex]P(0\leq X\leq a)=P(\frac{0-\mu}{\sigma} \leq \frac{X-\mu}{\sigma}\leq \frac{a-\mu}{\sigma}), Z= \frac{X-\mu}{\sigma}=P(0\leq X\leq a)=P(\frac{0-\mu}{\sigma} \leq \frac{X-\mu}{\sigma}\leq \frac{a-\mu}{\sigma}), Z= \frac{X-\mu}{\sigma}[/tex]
[tex]P(0\leq X\leq 7) = P(\frac{0-5.5}{1.5} \leq Z\leq \frac{7-5.5}{1.5}) = \Phi(1.66)-\Phi(-3.66) = \Phi(1.66) -(1-\Phi(3.66)) = 0.95-(1-0.99)=0.96[/tex]
P(X>7)=1-0.96=0.04
You have a large jar that initially contains 30 red marbles and 20 blue marbles. We also have a large supply of extra marbles of each color. Draw a marble out of the jar. If it's red, put it back in the jar, and add three red marbles to the jar from the supply of extras. If it's blue, put it back into the jar, and add five blue marbles to the jar from the supply of extras. Do this two more times. Now, pull a marble from the jar, at random. What's the probability that this last marble is red? What's the probability that we actually drew the same marble all four times?
Answer:
There is a 57.68% probability that this last marble is red.
There is a 20.78% probability that we actually drew the same marble all four times.
Step-by-step explanation:
Initially, there are 50 marbles, of which:
30 are red
20 are blue
Any time a red marble is drawn:
The marble is placed back, and another three red marbles are added
Any time a blue marble is drawn
The marble is placed back, and another five blue marbles are added.
The first three marbles can have the following combinations:
R - R - R
R - R - B
R - B - R
R - B - B
B - R - R
B - R - B
B - B - R
B - B - B
Now, for each case, we have to find the probability that the last marble is red. So
[tex]P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}[/tex]
[tex]P_{1}[/tex] is the probability that we go R - R - R - R
There are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is [tex]\frac{30}{50} = \frac{3}{5}[/tex].
Now the red marble is returned to the bag, and another 3 red marbles are added.
Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is [tex]\frac{33}{53}[/tex]
Again, the red marble is returned to the bag, and another 3 red marbles are added
Now there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is [tex]\frac{36}{56}[/tex]
Again, the red marble sorted is returned, and another 3 are added.
Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is [tex]\frac{39}{59}[/tex]. So
[tex]P_{1} = \frac{3}{5}*\frac{33}{53}*\frac{36}{56}*\frac{39}{59} = \frac{138996}{875560} = 0.1588[/tex]
[tex]P_{2}[/tex] is the probability that we go R - R - B - R
[tex]P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788[/tex]
[tex]P_{3}[/tex] is the probability that we go R - B - R - R
[tex]P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076[/tex]
[tex]P_{4}[/tex] is the probability that we go R - B - B - R
[tex]P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511[/tex]
[tex]P_{5}[/tex] is the probability that we go B - R - R - R
[tex]P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733[/tex]
[tex]P_{6}[/tex] is the probability that we go B - R - B - R
[tex]P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493[/tex]
[tex]P_{7}[/tex] is the probability that we go B - B - R - R
[tex]P_{7} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{33}{63} = \frac{825}{17325} = 0.0476[/tex]
[tex]P_{8}[/tex] is the probability that we go B - B - B - R
[tex]P_{8} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{30}{65} = \frac{750}{17875} = 0.0419[/tex]
So, the probability that this last marble is red is:
[tex]P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8} = 0.1588 + 0.0788 + 0.076 + 0.0511 + 0.0733 + 0.0493 + 0.0476 + 0.0419 = 0.5768[/tex]
There is a 57.68% probability that this last marble is red.
What's the probability that we actually drew the same marble all four times?
[tex]P = P_{1} + P_{2}[/tex]
[tex]P_{1}[/tex] is the probability that we go R-R-R-R. It is the same [tex]P_{1}[/tex] from the previous item(the last marble being red). So [tex]P_{1} = 0.1588[/tex]
[tex]P_{2}[/tex] is the probability that we go B-B-B-B. It is almost the same as [tex]P_{8}[/tex] in the previous exercise. The lone difference is that for the last marble we want it to be blue. There are 65 marbles, 35 of which are blue.
[tex]P_{2} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{35}{65} = \frac{875}{17875} = 0.0490[/tex]
[tex]P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078[/tex]
There is a 20.78% probability that we actually drew the same marble all four times