A concrete mix is designed to withstand 3000 pounds per square inch​ (psi) of pressure. The following data represent the strength of nine randomly selected casts​ (in psi). 3970​, 4100​, 3100​, 3200​, 2950​, 3830​, 4100​, 4050​, 3460 Compute the​ mean, median and mode strength of the concrete​ (in psi). Compute the mean strength of the concrete. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. A. The mean strength of the concrete is nothing psi of pressure. ​(Round to the nearest tenth as​ needed.) B. The mean does not exist.

Answer:

R2 = 43.03 ohms

Step-by-step explanation:

If R2=R4-15, then R4 = R2+15

According to the Wheastone bridge equation we have:

[tex]\frac{R1}{R2} =\frac{R3}{R4}\\\frac{10}{R2} =\frac{6470}{R2+15}\\\\10*(R2+15) = 6470*R2\\10*R2+150 = 6470*R2\\150=(6470-15)*R2\\R2=\frac{6455}{150}= 43.03333[/tex]

Answer:

(a) The probability is 0.4696

(b) The probability is 0.5304

(c) The probability is 0.0929

Step-by-step explanation:

The total number of ways in which we can select k elements from a group n elements is calculate as:

[tex]nCx=\frac{n!}{x!(n-x)!}[/tex]

So, the number of ways in which we can select four students from a group of 19 students is:

[tex]19C4=\frac{19!}{4!(19-4)!}=3,876[/tex]

On the other hand, the number of ways in which we can select four students with no math majors is:

[tex](16C4)*(3C0)=(\frac{16!}{4!(16-4)!})*(\frac{3!}{0!(3-0)!})=1820[/tex]

Because, we are going to select 4 students form the 16 students that aren't math majors and select 0 students from the 3 students that are majors.

At the same way, the number of ways in which we can select four students with one, two and three math majors are 1680, 360 and 16 respectively, and they are calculated as:

[tex](16C3)*(3C1)=(\frac{16!}{3!(16-3)!})*(\frac{3!}{1!(3-1)!})=1680[/tex]

[tex](16C2)*(3C2)=(\frac{16!}{2!(16-2)!})*(\frac{3!}{2!(3-1)!})=360[/tex]

[tex](16C1)*(3C3)=(\frac{16!}{1!(16-1)!})*(\frac{3!}{3!(3-3)!})=16[/tex]

Then, the probability that the group has no math majors is:

[tex]P=\frac{1820}{3876} =0.4696[/tex]

The probability that the group has at least one math major is:

[tex]P=\frac{1680+360+16}{3876} =0.5304[/tex]

The probability that the group has exactly two math majors is:

[tex]P=\frac{360}{3876} =0.0929[/tex]

In short, to calculate the probability of certain events in a group selection, you would identify the total possible groups, and then calculate how many of these groups satisfy your desired conditions. The probability is then calculated as the favorable events over the total possibilities.

This problem is a classic example of combinatorics and probability. The total number of ways to select four students from a total of 19 is given by the combination function: 19 choose 4. The denominator for all our probability calculations will be this total number of possible groups.

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Answer:

2,777.8 km or 1,729.2 mi

Step-by-step explanation:

first think about how many degrees would you travel if you wanted to do a whole circunference always going south: it would take 360 degress to complete a circunference.

Then you can use a rule of three to find the answer:

If the whole circunference is 40,000km and in degrees is 360, then how much 25 degrees would be?

[tex]x= \frac{25}{360}*40,000[/tex]

[tex]x= 2777.8[/tex]

Answer:

25

Step-by-step explanation:

Given:

Volume of Prochlorperazine injection available = 10 mL

Dose per vial = 5 mg/mL

Now,

The total mass of dose present in 10 mL = Volume × Dose

or

The total mass of dose present in 10 mL = 10 × 5 = 50 mg

Thus,

The number of 2 mg dose that can be withdrawn = [tex]\frac{\textup{50 mg}}{\textup{2 mg}}[/tex]

or

The number of 2 mg dose that can be withdrawn = 25

Answer: 25 doses of 2 mg each from the 10-mL vial

Step-by-step explanation:

To determine how many 2-mg doses can be withdrawn from a 10-mL vial containing Prochlorperazine at a concentration of 5 mg/mL, you can use the following calculation:

1. Calculate the total amount of Prochlorperazine in the vial:

  Total amount = Concentration × Volume

  Total amount = 5 mg/mL × 10 mL

  Total amount = 50 mg

2. Now, calculate how many 2-mg doses can be withdrawn:

  Number of 2-mg doses = Total amount / Dose per patient

  Number of 2-mg doses = 50 mg / 2 mg/dose

  Number of 2-mg doses = 25 doses

So, you can withdraw 25 doses of 2 mg each from the 10-mL vial of Prochlorperazine.

Answer:  The proof is given below.

Step-by-step explanation:  We are given to show that the following equality is true :

[tex]^{n+1}C_r=^nC_{r-1}+^nC_r.[/tex]

We know that

the number of combinations of n different things taken r at a time is given by

[tex]^nC_r=\dfrac{n!}{r!(n-r)!}.[/tex]

Therefore, we have

[tex]R.H.S.\\\\=^nC_{r-1}+^nC_r\\\\\\=\dfrac{n!}{(r-1)!(n-(r-1))!}+\dfrac{n!}{(r)!(n-r)!}\\\\\\=\dfrac{n!}{(r-1)!(n-r+1)!}+\dfrac{n!}{(r)!(n-r)!}\\\\\\=\dfrac{n!}{(r-1)!(n-r+1)(n-r)!}+\dfrac{n!}{r(r-1)!(n-r)!}\\\\\\=\dfrac{n!}{(r-1)!(n-r)!}\left(\dfrac{1}{n-r+1}+\dfrac{1}{r}\right)\\\\\\=\dfrac{n!}{(r-1)!(n-r)!}\left(\dfrac{r+n-r+1}{(n-r+1)r}\right)\\\\\\=\dfrac{n!}{(r-1)!(n-r)!}\times\dfrac{n+1}{(n-r+1)r}\\\\\\=\dfrac{(n+1)!}{r!(n-r+1)!}\\\\\\=\dfrac{(n+1)!}{r!((n+1)-r)!}\\\\\\=^{n+1}C_r\\\\=L.H.S.[/tex]

Thus, [tex]^{n+1}C_r=^nC_{r-1}+^nC_r.[/tex]

Hence proved.

Answer:

The statement is true for every n between 0 and 77 and it is false for [tex]n\geq 78[/tex]

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: [tex]\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4[/tex]

For n=1: [tex]\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4[/tex]

From this point we will assume that [tex]n\geq 2[/tex]

As we can see, [tex]\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4[/tex] and [tex](4n)^4=256n^4[/tex]. Then,

[tex]\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4[/tex]

Now, we will use the formula for the sum of the first 4th powers:

[tex]\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}[/tex]

Therefore:

[tex]\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0[/tex]

and, because [tex]n \geq 0[/tex],

[tex]465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1[/tex]

Observe that, because [tex]n \geq 2[/tex] and is an integer,

[tex]n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5[/tex]

In concusion, the statement is true if and only if n is a non negative integer such that [tex]n\leq 77[/tex]

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  [tex](4n)^4- \sum^{n}_{i=0} (2i)^4[/tex] for 77 and 78 you will obtain:

[tex](4n)^4- \sum^{n}_{i=0} (2i)^4=53810064[/tex]

[tex](4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992[/tex]

We will use mathematical induction to prove this inequality.

We will prove this inequality using mathematical induction. First, let's check the base case when n = 0. The left-hand side (LHS) of the inequality is 0 and the right-hand side (RHS) is (4*0)^4 = 0. So, the inequality holds for n = 0.

Next, assume that the inequality holds for some positive integer k, i.e.,

∑([2i]^4) ≤ (4k)^4 (where the sum is taken from i = 0 to k)

We will prove that it also holds for k + 1. Adding the (k+1)th term to both sides of the inequality:

∑([2i]^4) + ([2(k+1)]^4) ≤ (4k)^4 + ([2(k+1)]^4)

Now, simplifying the LHS and RHS:

(∑([2i]^4)) + ([2(k+1)]^4) ≤ (4k)^4 + ([2(k+1)]^4)

Answer:

False

Step-by-step explanation:

Correlation measures the strength of the relation between two variables.

Further, Correlation is said to be positive if increasing/decreasing the one variable, also increases/decreases the values of another variable.

Correlation is said to be negative if increasing/decreasing the one variable, also decreases/increases the values of another variable.

Since we don't know here exists a positive correlation or negative correlation.

So here are two possible conditions:

The person who has Gum disease also has heart disease.

And, the person has Gum disease can never have heart disease.

Thus, the given statement is false.

The statement is false because a correlation found in a study does not necessarily mean one factor (gum disease) is the cause of the other (heart disease). The cause and effect relationship must be established through further studies.

The statement 'A recent study found a correlation between gum disease and heart disease. This result indicates that gum disease causes people to develop heart disease.' is False. A correlation implies a relationship between two elements, but it does not indicate a cause and effect relationship.

This means although the study shows a link or association between gum disease and heart disease, it does not mean gum disease causes heart disease. It could be that people with poor gum health also tend to have poor overall health including heart health. Alternatively, there could be a third underlying factor that leads to both conditions. Therefore, the cause and effect relationship must be established through further studies.

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Answer:

Solution: [tex]B=5+25e^{4-4t}[/tex]

Step-by-step explanation:

Given: [tex]\dfrac{dB}{dt}+4B=20[/tex]  

with B(1)=30

The differential equation in form of linear differential equation,

[tex]\dfrac{dy}{dt}+Py=Q[/tex]

Integral factor, IF: [tex]e^{\int Pdt}[/tex]

General Solution:

[tex]y\cdot IF=\int Q\cdot IFdt[/tex]

[tex]\dfrac{dB}{dt}+4B=20[/tex]  

P=4, Q=20

IF= [tex]e^{\int 4dt}=e^{4t}[/tex]

Solution:

[tex]Be^{4t}=\int 20e^{4t}dt[/tex]

[tex]Be^{4t}=5e^{4t}+C[/tex]

[tex]B=5+Ce^{-4t}[/tex]

B(1)=30 , Put t=1, B=30

[tex]30=5+Ce^{-4}[/tex]

[tex]C=25e^4[/tex]

[tex]B=5+25e^{4-4t}[/tex]

Answer:

[tex]\frac{16}{195}[/tex]

Step-by-step explanation:

To obtain the result of a fractions multiplication we need to multiply both numerators and the divide by the multiplication of the denomitators. In general, given a,b,c,d real numbers with b and d not zero, we have that

[tex]\frac{a}{b}*\frac{c}{d}=\frac{a*c}{b*d}[/tex]

Substituting a,b,c and d for 8,15,2 and 13 we obtain that

[tex]\frac{8}{15}* \frac{2}{13} =\frac{16}{195}[/tex]

To find the value of 8/15 x 2/13, multiply the numerators together and multiply the denominators together. The fraction 16/195 is the final answer.

To find the value of 8/15 x 2/13, we multiply the numerators together (8 x 2) and multiply the denominators together (15 x 13). This gives us 16 in the numerator and 195 in the denominator.

The fraction 16/195 cannot be simplified further, so that is the final answer.

Calculation:

We have 8/15 x 2/13 = (8 x 2)/(15 x 13) = 16/195.

Answer:

[tex]f'(0)=0[/tex]

Step-by-step explanation:

Applying the chain rule

[tex]\frac{d}{dx} (f(-x))=-\frac{df}{dx}[/tex]

Then it becomes

[tex]\frac{df}{dx} =-\frac{df}{dx}[/tex]

In x=0

[tex]\frac{d[tex]f'(0)=-f'(0)\\f'(0)+f'(0)=0\\2f'(0)=0\\[/tex]f}{dx} =-\frac{df}{dx}[/tex]

Then

[tex]f'(0)=0[/tex]

Answer:

False

Step-by-step explanation:

In the above situation where the professor took a random sample of size 10 from each, conducted a test and found out that the variances are equal.  should not use a t test with related samples. The professor should use the t test for the difference in means testing for independence. Hence, the statement is false.

According to the hypothesis tested, it is found that it is true that the professor should use a t test with related samples, hence option A is correct.

A t-test should be used when we do not have the standard deviation for the population, which is the case in this problem, as we have it for the sample.

Related samples are used when comparisons are made between two samples, which is the case here for the samples of upper and lower classmen.

Hence, option A is correct.

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Answer:

$114

Step-by-step explanation:

make a calender and count every 4 days for dogs and every 6 days for the lawn. Then add all the money up.

Answer:

Pigs = 20 and Chickens = 50

Step-by-step explanation:

Let the number of pigs be x

and let the number of chicken be y

Thus, x + y = 70

Since Chicken has 2 legs and pigs has 4 legs.

⇒ 4x + 2y = 180

Solving both equations,

We get, x = 20 and y = 50

Thus number of pigs = 20

and, number of chickens = 50.

Answer:

[tex]\frac{20,410 \text{ grams}}{254\text{ mm}}[/tex]

Step-by-step explanation:

We have been given that adhesive tape made from fabric has a tensile strength of not less than 20.41 kg/2.54 cm of width. We are asked to reduce these quantities to grams and millimeters.

We know 1 kg equals 1000 grams and 1 cm equals 10 mm.

[tex]\frac{20.41\text{ kg}}{\text{2.54 cm}}[/tex]

[tex]\frac{20.41\text{ kg}}{\text{2.54 cm}}\times \frac{\text{1 cm}}{\text{10 mm}}[/tex]

[tex]\frac{20.41\text{ kg}}{2.54\times\text{10 mm}}[/tex]

[tex]\frac{20.41\text{ kg}}{254\text{ mm}}[/tex]

[tex]\frac{20.41\text{ kg}}{254\text{ mm}}\times \frac{\text{1000 grams}}{\text{1 kg}}[/tex]

[tex]\frac{20.41\times \text{1000 grams}}{254\text{ mm}}[/tex]

[tex]\frac{20,410 \text{ grams}}{254\text{ mm}}[/tex]

Therefore, our required measurement would be [tex]\frac{20,410 \text{ grams}}{254\text{ mm}}[/tex].

Answer:

The answers are given below:

Step-by-step explanation:

1. a. (─13)³⁵

1. b. The product will be negative. The expanded form shows 34 negative factors plus one more negative factor. Any even number of negative factor yields a positive product. The remaining 35th negative factor negates the resulting product.

2. 4 times.

3. Arnie is not correct. The base, ─3.1, should be in parentheses to prevent ambiguity. At present the notation is not correct.

Answer:

The value of x is 16 and the value of y is 8.

Step-by-step explanation:

Consider the provided equation.

[tex]\frac{3}{4}x -\frac{1}{2}y= 8\ and\ 2x +y=40[/tex]

Isolate x for [tex]\:\frac{3}{4}x-\frac{1}{2}y=8[/tex]

[tex]\frac{3}{4}x-\frac{1}{2}y+\frac{1}{2}y=8+\frac{1}{2}y[/tex]

[tex]\frac{3}{4}x=8+\frac{1}{2}y[/tex]

Multiply both side by 4 and simplify.

[tex]3x=32+2y[/tex]

[tex]x=\frac{32+2y}{3}[/tex]

Substitute the value of x in [tex]2x +y=40[/tex]

[tex]2\cdot \frac{32+2y}{3}+y=40[/tex]

[tex]\frac{64}{3}+\frac{7y}{3}=40[/tex]

[tex]64+7y=120[/tex]

[tex]7y=56[/tex]

[tex]y=8[/tex]

Now substitute the value of y in [tex]x=\frac{32+2y}{3}[/tex]

[tex]x=\frac{32+2\cdot \:8}{3}[/tex]

[tex]x=16[/tex]

Hence, the value of x is 16 and the value of y is 8.

Answer:

  {3, 4, 5, 6, 7, 8}

Step-by-step explanation:

Integers that are less than 9 and greater than 2 include the integers 3 through 8.

The correct set equal to {x | x < 9 and x > 2} is {3, 4, 5, 6, 7, 8}, as it includes all the integers that satisfy the given condition.

The given set is {x | x < 9 and x > 2}, which translates to all numbers greater than 2 and less than 9. When comparing this to the options provided, we need to ensure that the numbers within the set are all and only the integers that satisfy these conditions, regardless of their order. The set {3, 4, 5, 6, 7, 8} matches this description exactly, as it includes all the integers that are greater than 2 and less than 9. Sets in mathematics do not consider the order of elements; they only consider the presence of elements. Therefore, the correct option that is equal to the given set is {3, 4, 5, 6, 7, 8}.

Answer:

The answer is no: the set of all 2x2 matrices such that det(A)=0 is not a subspace of the vector space of all 2x2 matrices.

Step-by-step explanation:

In order for a set of matrices to be a subspace of all 2x2 matrices, three conditions must be satisfied:

1) That the set is not empty.

2) If A and B are both 2x2 matrices with zero determinant then the matrix A+B should also be a matrix with zero determinant.

3) The determinant of c*A, where "c" is any complex number and A is any matrix of the set, should be zero.

1)

The first condition is satisfied by the set of all 2x2 matrices such that det(A)=0, since there are plenty 2x2 matrices with zero determinant.

2)

The second condition is not satisfied, since from the determinant properties, we know that:

[tex]det(A+B)\geq det(a)+det(B)[/tex]

The equality might hold, but it is not a general characteristic. For example, if we consider the following matrices:

[tex]A = \left[\begin{array}{cc}1&0\\0&0\end{array}\right], \quad B = \left[\begin{array}{cc}0&0\\0&1\end{array}\right][/tex]

We can easily check that the determinant of both matrices is zero, nevertheless the determinant of the sum is different than zero.

Therefore, the set of all 2x2 matrices such that det(A)=0 is not a subspace of the vector space of all 2x2 matrices.

The set of all 2x2 matrices such that det(A) = 0 is not a subspace of the vector space of all 2x2 matrices.

To determine if a set is a subspace of a vector space, it must satisfy three conditions:

1. The set must contain the zero vector (in this case, the zero matrix).

2. The set must be closed under addition, meaning that if A and B are matrices in the set, then A + B must also be in the set.

3. The set must be closed under scalar multiplication, meaning that if A is a matrix in the set and c is a scalar, then cA must also be in the set.

Let's examine each condition:

1. The zero matrix, denoted by O, has a determinant of det(O) = 0, so it is included in the set. This condition is satisfied.

2. Consider two 2x2 matrices A and B with determinant 0:

[tex]\[ A = \begin{pmatrix} a b \\ c d \end{pmatrix}, \quad \text{det}(A) = ad - bc = 0 \][/tex]

 [tex]\[ B = \begin{pmatrix} e f \\ g h \end{pmatrix}, \quad \text{det}(B) = eh - fg = 0 \][/tex]

 The sum of A and B is:

 [tex]\[ A + B = \begin{pmatrix} a+e b+f \\ c+g d+h \end{pmatrix} \][/tex]

 The determinant of A + B is:

 [tex]\[ \text{det}(A + B) = (a+e)(d+h) - (b+f)(c+g) \][/tex]

 [tex]\[ \text{det}(A + B) = ad + ah + ea + eh - bc - bg - cf - fg \][/tex]

 Since det(A) = 0 and det(B) = 0, we have:

 [tex]\[ ad - bc = 0 \][/tex]

 [tex]\[ eh - fg = 0 \][/tex]

 However, this does not imply that det(A + B) = 0. The cross terms ah, ea, bg, and cf may not sum to zero, and thus det(A + B) may not be zero. Therefore, the set is not closed under addition, and this condition is not satisfied.

3. Consider a scalar c and a matrix A with determinant 0:

 [tex]\[ cA = c \begin{pmatrix} a b \\ c d \end{pmatrix} = \begin{pmatrix} ca cb \\ cc cd \end{pmatrix} \][/tex]

 The determinant of cA is:

 [tex]\[ \text{det}(cA) = (ca)(cd) - (cb)(cc) = c^2(ad - bc) \][/tex]

 Since det(A) = 0, we have ad - bc = 0, and thus det[tex](cA) = c^2(0) = 0[/tex]for any scalar c. This means that the set is closed under scalar multiplication, and this condition is satisfied.

Answer:

1880 Milliliters(mL) = 3.97 pints

Step-by-step explanation:

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.

Unit conversion problems, like this one, is an example of a direct relationship between measures.

Each ml has 0.002 pints. How many pints are there in 1880mL. So:

1ml - 0.002 pints

1880ml - x pints

[tex]x = 1800*0.002[/tex]

[tex]x = 3.97[/tex] pints

1880 Milliliters(mL) = 3.97 pints

Answer:

1) yes

2) no, (fog)(x)=3-x/2

3) no, (f o g)(x)= -2x+6/3x-5

The first and second pair of functions are not inverses of each other, while the third pair of functions are inverses.

The first pair of functions, f(x) = 2x + 3/(4x - 3) and g(x) = 3x + 3/(4x - 2), are not inverses of each other. To determine this algebraically, we need to calculate (f o g)(x) and (g o f)(x) and check if they equal to x. In this case, (f o g)(x) is x + 2/3, which is not equal to x, therefore f and g are not inverses of each other.

The second pair of functions, f(x) = -x^3 + 2 and g(x) = 3(cubedroot)x - 2/(2), are also not inverses. By calculating (f o g)(x) and (g o f)(x), we found that (f o g)(x) = x - 14/8, which is not equal to x.

The third pair of functions, f(x) = -2x + 4/(2 - 5x) and g(x) = (4 - 2x)/(5 - 2x), is indeed inverses of each other. By calculating (f o g)(x) and (g o f)(x), we found that (f o g)(x) = x and (g o f)(x) = x, which means f and g are inverses of each other.

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Answer:

The exact interest on $5,870 at 12% is $58.70.

Step-by-step explanation:

Given information:

Principal = $5870

Interest rate  = 12% = 0.12

Time = June-December = 7 months.

We know that

1 year = 12 months

1/12 year = 1 month

7/12 year = 7 month

Time = 7/12 year

Formula for simple interest:

[tex]I=P\times r\times t[/tex]

where, P is principal, r is rate of interest and t is time in years.

Substitute P=5870, r=0.12 and t=7/12 in the above formula.

[tex]I=5870\times 0.12\times \frac{1}{12}[/tex]

[tex]I=5870\times 0.01[/tex]

[tex]I=58.70[/tex]

Therefore the exact interest on $5,870 at 12% is $58.70.

Answer:

Considering the fire point at (0,h), x-direction positive to the right (→) and y-direction positive to up (↑) and the only force acting after fire is the projectile weight = -mg in the y-direction.

[tex]\\ x(t)=Vo*cos(e)*t\\ v_x(t)=Vo*cos(e)\\ a_y(t)=0\\ y(t)=h+Vo*sin(e)*t-\frac{g}{2}t^{2}\\ v_x(t)=Vo*sin(e)-gt\\ a_y(t)=-g[/tex]

Step-by-step explanation:

First, we apply the Second Newton's Law in both x and y directions:

x-direction:

[tex]\sum F_x= m\frac{dv_x}{dt} =0[/tex]

Integrating we have

[tex]\int\limits^{V_x} _{V_{0x}}{}\, dV_x =\int\limits^{t} _0{0}\, dt\\ V_{0x}=Vo*cos(e)\\ V_x(t)=Vo*cos(e)[/tex]

Taking into account that a=(dv/dt) and v=(dx/dt):

[tex]a_x(t)=\frac{dV_x(t)}{dt}=0\\V_x(t)=\frac{dx(t)}{dt}-->\int\limits^x_0 {} dx = \int\limits^t_0 {Vo*cos(t)} \, dt \\x(t)=Vo*cos(e)*t[/tex]

y-direction:

[tex]\sum F_y= m\frac{dv_x}{dt} =-mg[/tex]

Integrating we have

[tex]\int\limits^{V_y} _{V_{0y}}{}\, dV_y =\int\limits^{t} _0 {-g} \, dt\\ V_{0y}=Vo*sin(e)\\ V_y(t)=Vo*sin(e)-g*t[/tex]

Taking into account that a=(dv/dt) and v=(dy/dt):

[tex]a_y(t)=\frac{dV_y(t)}{dt}=-g\\V_y(t)=\frac{dy(t)}{dt}-->\int\limits^y_h {} dy = \int\limits^t_0 {(Vo*sin(t)-g*t)} \, dt \\y(t)=h+Vo*sin(e)*t-\frac{g}{2}t^{2}[/tex]

Answer:

Step-by-step explanation:

given number,

247₁₀ to be converted into

a) base 7

divide the number by 7 and write the remainder on the left side

solution is (502)₇

b) base 2

divide the number by 2 and write the remainder on the left side and write in the direction from down to up as shown in the diagram attached below.

solution is (11110111)₂

c) base 8

divide the number by 8 and write the remainder on the left side

solution is (367)₈

d) base 16

divide the number by 16 and write the remainder on the left side

solution is (F 7)₈

15 - F

diagram is attached below.

Answer:

[tex]\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}[/tex]

Step-by-step explanation:

Given equation of ellipsoids,

[tex]u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}[/tex]

The vector normal to the given equation of ellipsoid will be given by

[tex]\vec{n}\ =\textrm{gradient of u}[/tex]

            [tex]=\bigtriangledown u[/tex]

[tex]=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})[/tex]

[tex]=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}[/tex]

[tex]=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}[/tex]

Hence, the unit normal vector can be given by,

[tex]\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}[/tex]

             [tex]=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}[/tex]

[tex]=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}[/tex]

Hence, the unit vector normal to each point of the given ellipsoid surface is

[tex]\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}[/tex]

Answer: Our required values would be -10x+5, 2x+5 and -25.

Step-by-step explanation:

Since we have given that

g(x) = -4x+5

and

h(x) = 6x

We need to find  (g-h)(x) and (g+h)(x).

So, (g-h)(x) is given by

[tex]g(x)-h(x)\\\\=-4x+5-6x\\\\=-10x+5[/tex]

and (g+h)(x) is given by

[tex]g(x)+h(x)\\\\=-4x+5+6x\\\\=2x+5[/tex]

and (g-h)(3) is given by

[tex]-10(3)+5\\\\=-30+5\\\\=-25[/tex]

Hence, our required values would be -10x+5, 2x+5 and -25.

Answer:

a) 3 inch pulley: 11,309.7 radians/min

6) 6 inch pulley: 5654.7 radians/min

b) 900 RPM (revolutions per minute)

Step-by-step explanation:

Hi!

When a pulley wirh radius R rotantes an angle θ, the arc length travelled by a point on its rim is Rθ.  Then the tangential speed V is related to angular speed  ω as:

[tex]V=R\omega[/tex]

When you connect two pulleys with a belt, if the belt doesn't slip, each point of the belt has the same speed as each point in the rim of both pulleys: Then, both pulleys have the same tangential speed:

[tex]\omega_1 R_1 = \omega_2 R_2\\[/tex]

[tex]\omega_2 = \omega_1 \frac{R_1}{R_2} =1800RPM* \frac{3}{6}= 900RPM[/tex]

We need to convert RPM to radias per minute. One revolution is 2π radians, then:

[tex]\omega_1 = 1800*2\pi \frac{radians}{min} = 11,309.7\frac{radians}{min}[/tex]

[tex]\omega_2 = 5654.7 \frac{radians}{min}[/tex]

The saw rotates with the same angular speed as the 6 inch pulley: 900RPM

a. The angular speed of the 3 inch pulley is 3600π radians/min and the angular speed of the 6 inch pulley is 7200π radians/min. b. The revolutions per minute of the saw is 900.

a. To find the angular speed in radians per minute, we need to convert the revolutions per minute to radians per minute. Since 1 revolution is equal to 2π radians, we can calculate the angular speed of the 3 inch pulley as follows:

Angular speed = (Revolutions per minute) x (2π radians per revolution)

Angular speed = (1800 rev/min) x (2π radians/rev) = 3600π radians/min

Similarly, for the 6 inch pulley:

Angular speed = (Revolutions per minute) x (2π radians per revolution)

Angular speed = (1800 rev/min) x (2π radians/rev) = 7200π radians/min

b. To find the revolutions per minute of the saw, we need to use the ratio of the diameters of the two pulleys. Since the diameter of the 6 inch pulley is twice the diameter of the 3 inch pulley, the revolutions per minute of the saw will be half of the revolutions per minute of the motor. Therefore, the revolutions per minute of the saw is 900.

Answer: The number is 40.

Step-by-step explanation:

Since we have given that

Let the number be 'x'.

If a number is decreased by 40%.

So, number becomes,

[tex]\dfrac{100-40}{100}\times x\\\\=\dfrac{60}{100}\times x\\\\=0.6x[/tex]

According to question, the result becomes 24.

So, our equation becomes,

[tex]0.6x=24\\\\x=\dfrac{24}{0.6}\\\\x=\dfrac{240}{6}\\\\x=40[/tex]

Hence, the number is 40.

Answer:

110,000 base 2

Step-by-step explanation:

column 1 [the first position in the number]:

1+1=0, (carry 1)

column 2:

0+1 +1 (carried)=0, (carry 1)

column 3:

1+0+1 (carried)=0, (carry 1)

column 4:

0+1+1 (carried)=0, (carry 1)

column 5:

1+1+1=1, (carry 1)

then you write the last 1 'cause there is n number to add with:

[tex]10,101_{2}+11,011_{2}=110,000_{2}[/tex]

In binary system the highest number to write is 1, if you add 1+1, it jumps to 0, and you have to carry 1 to the next position.

If you are not sure about the sum, you can convert the numbers in base 2, to base 10, so you can know if it is correct:

[tex]10,101_{2}=21_{10}\\11,011_{2}=27_{10}\\110,000_{2}=48_{10}[/tex]

So 21+27=48.

In decimal system when you add 9+1, it jumps to 0 and then you have to carry 1 to the next position, because the the highest number you can write is 9.

Answer:

a. [tex]p \wedge q[/tex]

b. [tex]\neg p \wedge q[/tex]

c. [tex]p\Rightarrow q[/tex]

d. [tex]p \Leftrightarrow q[/tex]

e. [tex]\neg p \Rightarrow \neg q[/tex]

Step-by-step explanation:

a. 4 is an even integer and -5 is a negative prime number, can be represented by: [tex]p \wedge q[/tex]

b. 4 is not an even integer and-5 is a negative prime number, can be represented by: [tex]\neg p \wedge q[/tex]

c. If 4 is an even integer, then-5 is a negative prime number, can be represented by: [tex]p\Rightarrow q[/tex]

d. 4 is an even integer if and only if-5 is a negative prime number, can be represented by: [tex]p \Leftrightarrow q[/tex]

e. If 4 is not an even integer, then-5 is not a negative prime number,  can be represented by: [tex]\neg p \Rightarrow \neg q[/tex]

The statements a-e are translated into the language of logic as p ∧ q, ¬p ∧ q, p → q, p ↔ q, and ¬p → ¬q, respectively. These represent different logical relationships between the statements '4 is an even integer' and '-5 is a negative prime number'.

To rewrite the provided statements using logical connectors and the given identifiers (p: 4 is an even integer, q: -5 is a negative prime number), we proceed as follows:

a. p ∧ q: This reads "p and q", representing the statement "4 is an even integer and -5 is a negative prime number".

b. ¬p ∧ q: The symbol ¬ stands for "not", so this reads "not p and q", representing "4 is not an even integer and -5 is a negative prime number".

c. p → q: This reads "p implies q", representing "If 4 is an even integer, then -5 is a negative prime number".

d. p ↔ q: This reads "p if and only if q", representing "4 is an even integer if and only if -5 is a negative prime number".

e. ¬p → ¬q: This reads "not p implies not q", representing "If 4 is not an even integer, then -5 is not a negative prime number".

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The maximum potential error is 92% as per the concept of percentage.

The pharmacist weighed 0.050 g of a substance on a balance insensitive to quantities smaller than 0.004 g.

To find the maximum potential error in terms of percentage, we need to determine the difference between the actual weight of the substance and the closest value that the balance can measure, which is 0.004 g.

The difference is 0.050 g - 0.004 g = 0.046 g.

The maximum potential error is the difference between the actual weight and the closest value that the balance can measure, divided by the actual weight, multiplied by 100%.

Therefore, the maximum potential error in terms of percentage is (0.046 g / 0.050 g) x 100% = 92%.

To learn more about the percentage;

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The question deals with the calculation of the maximum potential error in a measurement. Given the insensitivity of the balance to 0.004 g and the actual measurement of substance of 0.050 g, the maximum potential error by calculation comes out to be 8%.

The question is asking about the potential error in a measurement made by a pharmacist. The error is the difference between the smallest measurable quantity by the balance and the actual measurement. In this case, we have a balance that is insensitive to quantities smaller than 0.004 g, and the pharmacist is measuring 0.050 g of a substance.

To find the potential error percentage, we take the maximum potential error (which is defined by the sensitivity of the balance, 0.004 g), divide it by the actual measurement (0.050 g) and multiply by 100 to make it a percentage.

Maximum potential error percentage = (0.004 g / 0.050 g) * 100% = 8%

So the maximum potential error in this measurement is 8%.

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