At a grocery store, Daniel wants to buy 3 lb of ham.
What decimal should the digital scale show?
Write 3 as a fraction and then divide.
The scale should read

Answer:

The remaining interior angles of this triangle are 140º and 10º

Step-by-step explanation:

The sum of the interior angles of a triangle is always 180º.

A triangle has 3 angles. In this problem, we have one of them, that i am going to call A1 = 30º.

The sum of a interior angle with it's respective exterior angle is also always 180º.

We have that one of the exterior angles is equal to 40°. So it's respective interior angle is

40º + A2 = 180º

A2 = 180º - 40º

A2 = 140º

Now we have two interior angles, and we know that the sum of the 3 interior angles is 180º. So:

A1 + A2 + A3 = 180º

A3 = 180º - A1 - A2

A3 = 180º - 30º - 140º

A3 = 180º - 170º

A3 = 10º

Answer:

140 and 10

Step-by-step explanation:

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

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

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

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]

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.

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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%.

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.

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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.

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.

Answer:

A proof can be as follows:

Step-by-step explanation:

Remember that an odd interger is of the form [tex]2p+1[/tex] where [tex]p[/tex] is a integer and remember that two consecutive integer are two numbers of the form [tex]p, p+1[/tex]

[tex](\Rightarrow)[/tex] Suppose the [tex]n[/tex] is an odd integer.

Then [tex]n-1[/tex] must be an even integer and hence divisible by 2. Then we define

[tex]p=\dfrac{n-1}{2}\\q=\dfrac{n-1}{2}+1[/tex]

Then we have that

[tex]p+q=\dfrac{n-1}{2}+\dfrac{n-1}{2}+1=\frac{(n-1)+(n-1)}{2}+1=\frac{2(n-1)}{2}+1=n-1+1=n[/tex]

The converse is as follows:

[tex](\Leftarrow)[/tex] Let [tex]p[/tex] an integer, then[tex]p,p+1[/tex]  are two consecutive integers. Then

[tex]n=p+(p+1)=2p+1[/tex] is an odd integer.

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.

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].

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.

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

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:

[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]

[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]

[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]

[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]

[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]

[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]

[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]

[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]

[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]

[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]

[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]

[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]

[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]

[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]

[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.

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.

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.

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Keith needs to remove 15 bags of mulch to meet the weight requirements.

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

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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.

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.

Answer:

  24

Step-by-step explanation:

4P3 = 4!/(4-3)! = 4·3·2 = 24

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

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

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]

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]

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.

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.

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 :

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

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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

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"

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].

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%.

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.

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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.

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.

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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

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]