Answer:
No
.3 = .30
.32 > .30
Step-by-step explanation:
Chef plans to serve 4.5 oz. steaks to 180 Guests. How many POUNDS only of raw steak should Chef order if Waste (trimming and cooking) 25% a. 1080 oz. b. 50.625 lbs. C. 67.5 lbs. d. 1080 lbs. e. NONE of the above answers are correct
Answer: Option 'c' is correct.
Step-by-step explanation:
Since we have given that
Number of ounces steaks = 4.5 oz
Number of guests = 180
Percentage of waste = 25%
Number of ounces of fresh steak is given by
[tex]4.5\times 180\\\\=810\ oz[/tex]
Let the number of ounces of raw steak be x.
According to question,
[tex]\dfrac{100-25}{100}\times x=810\\\\\dfrac{75}{100}\times x=810\\\\0.75\times x=810\\\\x=\dfrac{810}{0.75}\\\\x=1080\ oz[/tex]
As we need in pounds, and we know that
1 pound = 16 ounces
So, Number of pounds of raw steak that Chef could order is given by
[tex]\dfrac{1080}{16}=67.5\ lbs[/tex]
Hence, Option 'c' is correct.
To determine how many pounds of raw steak Chef should order for 180 guests with 25% waste, calculate the weight per guest and multiply by the number of guests. Convert the weight to pounds by dividing by 16. The answer is approximately B. 63.3 lbs.
Explanation:To determine how many pounds of raw steak Chef should order, we need to calculate the total weight of steak required for 180 guests, taking into account the 25% waste.
First, we calculate the weight of steak for one guest by multiplying 4.5 oz by 1.25 (to account for the waste), which equals 5.625 oz.
Then, we multiply the weight per guest by the number of guests to get the total weight of steak required: 5.625 oz x 180 = 1012.5 oz.
Finally, we convert the ounces to pounds by dividing by 16, since there are 16 ounces in a pound.
Thus, Chef should order 1012.5 oz ÷ 16 = 63.28125 lbs, which rounds to approximately 63.3 lbs (option B).
Aminophylline is infusing at 30 mL/hr. The drug is mixed 250 mg in 500 ml D5W. How many mg/hr is the patient receiving?
Answer:
the patient receiving is 15 mg/hr
Step-by-step explanation:
given data
infusing = 30 mL/hr
drug mixed = 250 mg
D5W = 500ml
to find out
How many mg/hr is the patient receiving
solution
we know infusing rate is = 30 mL/hr ...................1
and we know mixed is = [tex]\frac{250mg}{500ml}[/tex] ..............2
so
now we multiply both equation 1 and 2
patient receiving = [tex]\frac{250mg}{500ml}[/tex] × [tex]\frac{30ml}{hr}[/tex]
solve we get here
patient receiving = [tex]\frac{15mg}{hr}[/tex]
so the patient receiving is 15 mg/hr
The patient is receiving 15 mg/hr of aminophylline.
Explanation:To calculate the mg/hr that the patient is receiving, we need to find the concentration of aminophylline in the infusion and multiply it by the infusion rate. The concentration of aminophylline is 250 mg in 500 ml of D5W. This means that there are 250 mg of aminophylline in 500 ml, or 0.5 mg/ml. The infusion rate is 30 ml/hr. So, to find the mg/hr, we multiply the concentration by the infusion rate: 0.5 mg/ml * 30 ml/hr = 15 mg/hr. Therefore, the patient is receiving 15 mg/hr of aminophylline.
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You manage an ice cream factory that makes two flavors: Creamy Vanilla and Continental Mocha. Into each quart of Creamy Vanilla go 2 eggs and 3 cups of cream. Into each quart of Continental Mocha go 1 egg and 3 cups of cream. You have in stock 350 eggs and 600 cups of cream. You make a profit of $3 on each quart of Creamy Vanilla and $2 on each quart of Continental Mocha. How many quarts of each flavor should you make to earn the largest profit? HINT (See Example 2.] (If an answer does not exist, enter DNE.) Creamy Vanilla quarts Continental Mocha quarts
Answer:
You should make 150 quarts of Creamy Vanilla and 50 quarts of Continental Mocha.
Step-by-step explanation:
This problem can be solved by a system of equations.
The largest profit is going to earned when all the eggs and cups of cream in stock are used.
I am going to call x the number of quarts of Creamy Vanilla and y the number of quarts of Continental Mocha.
The problem states that each quart of Creamy Vanilla uses 2 eggs and each quart of Continental Mocha uses 1 egg. There are 350 eggs in stock, so:
[tex]2x + y = 350[/tex]
Each quart of Creamy Vanilla uses 3 cups of cream, as does each quart of Continental Mocha. There are 600 cups of cream in stock.
So:
[tex]3x + 3y = 600[/tex] Simplifying by 3.
[tex]x + y = 200[/tex]
We have the following system
[tex]1)2x + y = 350[/tex]
[tex]2)x + y = 200[/tex]
I am going to multiply 2) by (-1) and then add with 1), so i can eliminate y
[tex]1)2x + y = 350[/tex]
[tex]2)-x - y = -200[/tex]
[tex]2x - x + y - y = 350 - 200[/tex]
[tex]x = 150[/tex]
[tex]x + y = 200[/tex]
[tex]y = 200 - 150 = 50[/tex]
You should make 150 quarts of Creamy Vanilla and 50 quarts of Continental Mocha.
The question regarding the number of quarts of Creamy Vanilla and Continental Mocha for maximum profit can be solved using a method in mathematics called linear programming. It involves setting up inequalities to represent the constraints (number of eggs and cups of cream) and a profit function which we want to maximize. The specific solution will depend on the particular set of constraints and profit per flavor.
Explanation:This problem can be approached as a linear programming problem. Let x represent the number of quarts of Creamy Vanilla and y represent the number of quarts of Continental Mocha. The constraints from the problem can be represented with the following inequalities, considering the number of eggs and cups of cream:
2x + y ≤ 350 (eggs)
3x + 3y ≤ 600 (cups of cream)
We want to maximize the profit function P which is given by P = 3x + 2y. This can be solved graphically by plotting the constraints on a graph, finding the feasible region and then determining which point in that region provides the maximum value of P.
The solution to this problem can vary depending on the specific set of supplies and the profit of each flavor.
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Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G
[tex]\Rightarrow[/tex]
Suppose first that [tex]H\subset G[/tex] is a normal subgroup. Then by definition we must have for all [tex]a\in H[/tex], [tex]xax^{-1} \in H[/tex] for every [tex]x\in G[/tex]. Let [tex]a\in G[/tex] and choose [tex](ab)\in aH[/tex] ([tex]b\in H[/tex]). By hypothesis we have [tex]aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H[/tex], i.e. [tex]aba^{-1}=c[/tex] for some [tex]c\in H[/tex], thus [tex]ab=ca \in Ha[/tex]. So we have [tex]aH\subset Ha[/tex]. You can prove [tex]Ha\subset aH[/tex] in the same way.
[tex]\Leftarrow[/tex]
Suppose [tex]aH=Ha[/tex] for all [tex]a\in G[/tex]. Let [tex]h\in H[/tex], we have to prove [tex]aha^{-1} \in H[/tex] for every [tex]a\in G[/tex]. So, let [tex]a\in G[/tex]. We have that [tex]ha^{-1} =a^{-1}h'[/tex] for some [tex]h'\in H[/tex] (by the hypothesis). hence we have [tex]aha^{-1}=h' \in H[/tex]. Because [tex]a[/tex] was chosen arbitrarily we have the desired .
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:
Mean = 3640
Mode = 4100
Median = 3830.
Step-by-step explanation:
We are given the following data in the question:
Strength of casts (in psi):
3970,4100,3100,3200,2950,3830,4100,4050,3460
Formula:
[tex]Mean = \displaystyle\frac{\text{Sum of all observation}}{\text{Total number of observations}}[/tex]
[tex]\displaystyle\frac{3970+4100+3100+ 3200+ 2950+ 3830+4100+ 4050+ 3460}{9} = \displaystyle\frac{32760} {9} = 3640[/tex]
Mode is the entry with most frequency. Thus, for the given sample mode = 4100.
Median
Since n = 9 is odd,
Formula:
[tex]Median = \displaystyle\frac{n+1}{2}th~term[/tex]
Data in ascending order:
2950,3100,3200,3460,3830,3970,4050,4100,4100
Median = 5th term = 3830.
The mean strength of the concrete is 3640.
The mode strength of the concrete is 4100.
The median strength of the concrete is 3830.
Given
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
What formula is used to calculate the mean value?The mean value of the data is given by;
[tex]\rm Mean = \dfrac{Sum \ of \ all \ observation}{Total \ number \ of \ observation}[/tex]
The mean of the strength of the concrete is;
[tex]\rm Mean = \dfrac{Sum \ of \ all \ observation}{Total \ number \ of \ observation}\\\\Mean =\dfrac{3970+4100+3100+3200+2950+3830+ 4100+4050+3460 }{9}\\\\Mean =\dfrac{32760}{9}\\\\Mean=3640[/tex]
The mean strength of the concrete is 3640.
Mode is the entry with the most frequency.
Thus, for the given sample mode = 4100.
The mode strength of the concrete is 4100.
The mid-value of the data is called the median.
The median strength of the concrete is 3830.
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A restaurant serves a luncheon special with a choice of two appetizers, three choices of entr´ee, 4 choices of salad, and 5 choices of desert. How many different meals are possible?
Answer: 80
Step-by-step explanation:
Given : A restaurant serves a luncheon special .
The number of choices for appetizers= 4
The number of choices for salad = 4
The number of choices for desert = 5
Then, the total number of different possible meals are given by :_
[tex]4\times4\times5=80[/tex]
Therefore, the total number of different possible meals are =80
FOCUS ON HIGHER ORDER THINKING Work Area Katie said, "Negative numbers are integers." What was Explain the Error her error?
Step-by-step explanation:
Consider the provided information.
Integers: Integers are the set of whole number and the negatives of the natural numbers, i.e, … ,-2, -1, 0, 1, 2,...
Rational number: A number is said to be rational, if it is in the form of p/q. Where p and q are integer and denominator is not equal to 0.The decimal expansion of rational numbers may terminate or become periodic.
Irrational number: A number is irrational if it cannot be expressed be expressed by dividing two integers. The decimal expansion of Irrational numbers are neither terminate nor periodic.
Now consider the provided statement "Negative numbers are integers."
The statement is incorrect as all negative numbers are not integers they might be rational number or irrational numbers.
For example: -2.5 is a negative number but it is not an integer. The number is a rational number.
Hence, her statement is incorrect.
She can say that all negative whole numbers are integers.
Solve the initial value problem dydx=(x−2)(y−10),y(0)=5dydx=(x−2)(y−10),y(0)=5.
y=
Answer:
The solution for the initial value problem is:
[tex]y(x) = -5e^{\frac{x^{2}}{2} - 2x} + 10[/tex]
Step-by-step explanation:
We have the following initial value problem:
[tex]\frac{dy}{dx} = (x-2)(y-10)[/tex]
The first step is solving the differential equation. We can do this by the variable separation method. It means that every term with y in on one side of the equality, every term with x on the other side. So:
[tex]\frac{dy}{dx} = (x-2)(y-10)[/tex]
[tex]\frac{dy}{y-10} = (x-2)dx[/tex]
To find y in function of x, we integrate both sides.
[tex]\frac{dy}{y-10} = (x-2)dx[/tex]
[tex]\int {\frac{1}{y-10}} \, dy = \int {(x-2)} \, dx[/tex]
Solving each integral separately
[tex]\int {\frac{1}{y-10}} \, dy[/tex]
This one we solve by substitution
[tex]u = y-10, du = dy[/tex]
[tex]\int {\frac{1}{y-10}} \, dy = \int {\frac{1}{u}} \, du = \ln{u} = \ln{(y-10)}[/tex]
[tex] \int {(x-2)} \, dx = \frac{x^{2}}{2} - 2x + K[/tex]
Now we have that:
[tex]\int {\frac{1}{y-10}} \, dy = \int {(x-2)} \, dx[/tex]
[tex]\ln{(y-10) = \frac{x^{2}}{2} - 2x + K}[/tex]
To solve for y, we apply the exponential to both sides, since the exponential and ln are inverse operations:
[tex]e^{\ln{(y-10)} = e^{\frac{x^{2}}{2} - 2x + K}[/tex]
[tex]y - 10 = Ke^{\frac{x^{2}}{2} - 2x}[/tex]
[tex]y(x) = Ke^{\frac{x^{2}}{2} - 2x} + 10[/tex]
[tex]y(0) = 5[/tex] means that when [tex]x = 0, y(x) = 5[/tex]. So:
[tex]5 = Ke^{\frac{0^{2}}{2} - 2*0} + 10[/tex]
[tex]Ke^{0} = -5[/tex]
[tex]K = -5[/tex]
So, the solution for the initial value problem is:
[tex]y(x) = -5e^{\frac{x^{2}}{2} - 2x} + 10[/tex]
An baseball player’s batting average decreases from 0.312 to 0.278. What is the percent decreased?
Answer:
[tex]10.897[/tex]
Step-by-step explanation:
An baseball player’s batting average decreases from 0.312 to 0.278 .
Let [tex]x_0[/tex] be the initial baseball player’s batting average and [tex]x_1[/tex] be the final baseball player’s batting average .
Initial value [tex]\left ( x_0 \right )[/tex] = 0.312
Final value [tex]\left ( x_1 \right )[/tex] = 0.278
So, change in value =Final value - Initial value = [tex]x_1-x_0[/tex] = [tex]0.278-0.312=-0.034[/tex]
Therefore , decrease in value = 0.034
We know that percent decreased = ( decrease in value × 100 ) ÷ Initial value
i.e. , [tex]\frac{0.034}{0.312}\times 100=\frac{3400}{312}=10.897[/tex]
Or we can say percentage change in baseball player’s batting average = [tex]-10.897 \%[/tex]
Find the linear trend forecast for period 8 given the following data: Period Sales 1 19 2 18 3 15 4 20 5 18 6 22 7 20
Answer:
The equation of trend line is [tex]y = 0.500000x + 16.857143[/tex].
The linear trend forecast for period 8 is about 20.86.
Step-by-step explanation:
The given data table is
Period Sales
1 19
2 18
3 15
4 20
5 18
6 22
7 20
We need to find the linear trend forecast for period 8.
The general form of linear regression is
[tex]y=a+bx[/tex] .... (1)
where, a is y-intercept and b is slope.
[tex]b=\frac{\sum_{i=1}^nx_iy_i-n\overline{x}\overline{y}}{\sum_{i=1}^nx_i^2-n\overline{x}^2}[/tex]
[tex]a=\overline{y}-b\overline{x}[/tex]
Using the graphing calculator we get
[tex]a=16.857143[/tex]
[tex]b=0.500000[/tex]
Substitute these values in equation (1).
[tex]y = 0.500000x + 16.857143[/tex]
The equation of trend line is [tex]y = 0.500000x + 16.857143[/tex].
Substitute x=8 to find the linear trend forecast for period 8.
[tex]y = 0.500000(8) + 16.857143[/tex]
[tex]y =20.857143[/tex]
[tex]y \approx 20.86[/tex]
Therefore the linear trend forecast for period 8 is about 20.86.
There are 14 juniors and 16 seniors in a chess club. a) From the 30 members, how many ways are there to arrange 5 members of the club in a line? b) How many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line? 0 c) If the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there? d) If the club sends either 4 juniors or 4 seniors, how many possible groupings are there?
Answer:
No. of juniors = 14
No. of seniors = 16
Total students = 30
A) From the 30 members, how many ways are there to arrange 5 members of the club in a line?
Since we are asked about arrangement so we will use permutation
Formula : [tex]^nP_r=\frac{n!}{(n-r)!}[/tex]
n = 30
r = 5
[tex]^{30}P_5=\frac{30!}{(30-5)!}[/tex]
[tex]^{30}P_5=17100720[/tex]
So, From the 30 members, there are 17100720 ways to arrange 5 members of the club in a line?
B) How many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line?
Out of 16 seniors 2 will be selected
So, 3 places are vacant
Remaining students = 30-2 = 28
So, out of 28 students 3 students will be selected
No. of ways = [tex]^{16}P_2 \times ^{28}P_3[/tex]
No. of ways = [tex]\frac{16!}{(16-2)!}\times\frac{28!}{(28-3)!}[/tex]
= [tex]4717440[/tex]
There are 4717440 ways to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line.
C)If the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there?
Since we are not asked about arrangement so we will use combination
Out of 16 seniors 2 will be selected
Out of 14 juniors 2 will be selected
Formula : [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]
So, No. of possible groupings = [tex]^{16}C_2 \times ^{14}C_2[/tex]
= [tex]\frac{16!}{2!(16-2)!} \times \frac{14!}{2!(14-2)!}[/tex]
= [tex]10920[/tex]
If the club sends 2 juniors and 2 seniors to the tournament, there are 10920 possible groupings
D) If the club sends either 4 juniors or 4 seniors, how many possible groupings are there?
Out of 16 seniors 4 will be selected
or
Out of 14 juniors 4 will be selected
So, No. of possible groupings = [tex]^{16}C_4 + ^{14}C_4[/tex]
= [tex]\frac{16!}{4!(16-4)!} + \frac{14!}{4!(14-4)!}[/tex]
= [tex]2821[/tex]
So,If the club sends either 4 juniors or 4 seniors, there are 2821 possible groupings .
If eggs in the basket are removed two at a time, one egg will remain. If the eggs are removed three at a time, 2 eggs will remain. If the eggs are removed 4, 5, or 6 at a time 3, 4, and 5 eggs will remain, respectfully. If they are taken out seven at a time, however, no eggs will be left over. Find the smallest number of eggs that could be in the basket.
Smallest number of eggs in the basket = 119
Step-by-step explanation:In the question,
We know that the number of eggs in the basket should be a a multiple of 7.
But it can not be a multiple of 2, 3, 4, 5 and 6 because every time we pick up the eggs 2, 3, 4, 5 or 6 at a time we are left with some eggs with us.
Therefore, the number of eggs can not be a multiple of these numbers.
Now,
Let us say the number of eggs in the basket be 7x.
So,
Let us take the LCM of 2, 3, 4, 5 and 6.
So,
LCM = 60
Now, the number would be greater than 60 and the multiple of 7.
So, checking on all the multiples of 7 above 60 and checking the condition that, the remainder left on dividing by,
2 is 1. (2x + 1)
by 3 is 2. (3x + 2)
by 4 is 3. (4x + 3)
by 5 is 4. (5x + 4)
by 6 is 5. (6x + 5)
So,
On Checking the multiples of 60 which are divisible by 7 are ,
60 + 1 , 120 + 1, 60 - 1, 120 - 1.
So,
For 61 it is not satisfying all the conditions.
121 is also not satisfying all the conditions.
59 is also not satisfying all the conditions.
But,
The number 119 on checking its divisibility by 2. Leaves remainder 1 as, (2(59) + 1).
Divisibility by 3 leaves remainder 2 as, {3(39) + 2}.
Divisibility by 4 leaves remainder 3 as, {4(29) + 3}.
Divisibility by 5 leaves remainder 4 as, {5(23) + 4}.
Divisibility by 6 leaves remainder 5 as, {6(19) + 5}.
Divisibility by 7 leaves remainder 0 as, {7(17) + 0}.
Therefore, the minimum number of eggs in the basket are 119.
Answer: 119 is the smallest number of eggs in the basket
Step-by-step explanation:
let k be the number
If eggs in the basket are removed two at a time, one egg will remainso k is less than a multiple of 2 by 1 so we have k = 2A - 1 for a positive integer A
If the eggs are removed three at a time, 2 eggs will remainso k is 2 more and thus less with 1, for a multiple of 3, k = 3B - 1, for some positive integer B
If the eggs are removed 4 at a time, 3 eggs will remainso k is 3 more and thus less with 1, with a multiple of 4, so k = 4C -1 for some positive integer
If the eggs are removed 5 at a time, 4 eggs will remainso k is 4 more and thus less with 1, with a multiple of 5, so k = 5D -1 for some positive integer
If the eggs are removed 6 at a time, 5 eggs will remainso k is 5 more and thus less with 1, with a multiple of 6, so k = 6E -1 for some positive integer
If they are taken out seven at a time, however, no eggs will be left overthus k is a multiple of 7, so k = 7F for some positive integer F
THUS K = 2A - 1 = 3B - 1 = 4C -1 = 5D -1 = 6E -1 = 7F
ADD 1 to k above gives k + 1 = 2A = 3B = 4C = 5D = 6E = 7F + 1
of k + 1 = 7F + 1 has to be a common multiple of 2, 3, 4, 5, 6. The least common multiple of 2, 3, 4, 5, 6 is 60
thus k + 1 is a multiple of 60 so k = 60 -1 = 59
again find the least multiple of 60 that is 1 greater than a multiple of 60
the first multiple of 60 is 60 and 60 - 1 = 59, but 59 is not a multiple of 7
the second multiple of 60 is 120 and 120 -1 = 119
and 7 is a multiple of 119 because 7 * 17 = 119
Mercoxit Energy Inc. has a retention ratio of 0.79, an ROA of 5.94%, and an EM of 1.11. This means that they will have a sustainable growth rate of ____%
Final answer:
To calculate the sustainable growth rate of Mercoxit Energy Inc., the retention ratio is multiplied by the calculated Return on Equity (ROE), which is derived from the ROA and EM figures given. The result yields a sustainable growth rate of 5.21% for Mercoxit Energy Inc.
Explanation:
The sustainable growth rate (SGR) is a company's maximum growth rate in sales using internal financial resources while maintaining a constant debt to equity ratio. To calculate the SGR for Mercoxit Energy Inc., one needs to use the following formula:
SGR = Retention Ratio x Return on Equity (ROE)
While we are given the retention ratio and the Return on Assets (ROA), we are not directly given the ROE. However, we can calculate ROE using the provided Equity Multiplier (EM), since ROE can be written as:
ROE = ROA x EM
Considering the given figures:
Retention Ratio = 0.79
ROA = 5.94%
EM = 1.11
Now calculate ROE:
ROE = 0.0594 x 1.11 = 0.065934 (or 6.5934%)
Finally, we calculate the SGR:
SGR = 0.79 x 0.065934 = 0.05208786 (or 5.208786%)
If we round to two decimal places, Mercoxit Energy Inc. has a sustainable growth rate of 5.21%.
Write the equation in slope-intercept form of the line below
A. Y= 2
B. Y= -2
C. X= 2
D. X= -2
Answer:
y = 2
Step-by-step explanation:
Answer:
y=2
Step-by-step explanation:
Define fn : [0,1] --> R by the
equationfn(x) = xn. Show that the
sequence(fn(x)) converges for each x belongs to [0,1],
but thatthe sequence (fn) does not converge uniformly.
Answer:
The sequence of functions [tex]\{x^{n}\}_{n\in \mathbb{N}}[/tex] converges to the function
[tex]f(x)=\begin{cases}0&0\leq x<1\\1&x=1\end{cases}[/tex].
Step-by-step explanation:
The limit [tex]\lim_{n\to \infty }c^{n}[/tex] exists and converges to zero whenever [tex]\lvert c \rvert <1 [/tex]. But, if [tex]c=1[/tex] the sequence [tex]\{c^{n}\}[/tex] is constant and all its terms are equal to [tex]1[/tex], then converges to [tex]1[/tex]. Using this result, consider the sequence of functions [tex]\{f_{n}\}[/tex] defined on the interval [tex][0,1][/tex] by [tex]f_{n}(x)=x^{n}[/tex]. Then, for all [tex]0\leq x<1[/tex] we have that [tex]\lim_{n\to \infty}x^{n}=0[/tex]. Now, if [tex]x=1[/tex], then [tex]\lim_{n\to \infty }x^{n}=1[/tex]. Therefore, the limit function of the sequence of functions is
[tex]f(x)=\begin{cases}0&0\leq x<1\\1&x=1\end{cases}[/tex].
To show that the convergence is not uniform consider [tex]0<\varepsilon<1[/tex]. For any [tex]n>1[/tex] choose [tex]x\in (0,1)[/tex] such that [tex]\varepsilon^{1/n}<x<1[/tex]. Then
[tex]\varepsilon <x^{n}=\lvert f(x)-f_{n}(x)\rvert[/tex]
This implies that the convergence is not uniform.
inverse laplace transform of H(s) = 1/(s+4)^2
Answer:
Inverse Laplace of [tex]\frac{1}{(S+4)^2}[/tex] will be [tex]te^{-4t}[/tex]
Step-by-step explanation:
We have to find the inverse Laplace transform of [tex]H(S)=\frac{1}{(S+4)^2}[/tex]
We know that of [tex]\frac{1}{s+4}[/tex] is [tex]e^{-4t}[/tex]
As in H(s) there is square of [tex]s+4[/tex]
So i inverse Laplace there will be multiplication of t
So the inverse Laplace of [tex]\frac{1}{(s+4)^2}[/tex] will be [tex]te^{-4t}[/tex]
[tex]L^{-1}\frac{1}{(S+4)^2}=te^{-4t}[/tex]
Calculate the IRR if the power company gets a fixed feed-in tariff of $0.25/kWh, for 20 years and the salvage value of the plant after 20 years is $20 M.
Answer:
0.2449 or 24.49%
Step-by-step explanation:
Calculate the IRR if the power company gets a fixed feed-in tariff of $0.25/kWh.
IRR means Internal Rate of Return is given by
[tex]IRR=(\frac{FV}{PV})^{\frac{1}{n}}-1[/tex]
Where, FV = Final value ($20)
PV = Present value ($0.25)
N = 20 years
Now put the values
[tex]IRR=(\frac{20}{0.25})^{\frac{1}{20}}-1[/tex]
= [tex]IRR=(80)^{\frac{1}{20}}-1[/tex]
= 1.24495742 - 1
= 0.24495742
Converting in percentage :
0.24495742 × 100 = 24.49%
IRR = 0.2449 or 24.49%
Answer:
IRR = 0.24495
Step-by-step explanation:
Given data:
Tariff =$0.25
So, present value = $0.25
N = 20 year
salvage value after 20 year is $20 M
final value is $20 M
IRR means internal rate of return and it is given as
[tex]IRR =[\frac{FV}{PV}]^{1/n} -1[/tex]
Where FV is final value and PV is present value
[tex]IRR = [\frac{20}{0.25}]^{1/20} -1[/tex]
IRR = 0.24495
1. What is the negation of the following: "If this triangle has two 45 degree angles then it is a right triangle."
A. this triangle does not have two 45 degree angles and it is a right triangle.
B. this triangle has two 45 degree angles or it is not a right triangle.
C. this triangle has two 45 degree angles and it is not a right triangle.
D. this triangle does not have two 45 degree angles or it is a right triangle.
E. this triangle has two 45 degree angles or it is a right triangle.
F. If this triangle does not have two 45 degree angles then it is a right triangle.
G. this triangle has two 45 degree angles and it is a right triangle.
H. If it is not a right triangle then this triangle does not have two 45 degree angles.
I. If this triangle does not have two 45 degree angles then it is not a right triangle.
J. If this triangle has two 45 degree angles then it is a right triangle.
K. If it is a right triangle then this triangle has two 45 degree angles.
Final answer:
The correct negation of the statement is option C: 'This triangle has two 45 degree angles and it is not a right triangle,' which follows the 'P and not Q' format for negating 'If P, then Q' statements.
Explanation:
The negation of the statement 'If this triangle has two 45 degree angles then it is a right triangle' is 'This triangle has two 45 degree angles and it is not a right triangle.' To negate a conditional statement like the one given, you would state that while the condition holds true, the conclusion does not. This is represented in option C. When negating an 'if-then' statement, the format usually follows 'If P, then Q' to 'P and not Q.' Thus, the correct answer is C, 'this triangle has two 45 degree angles and it is not a right triangle.'
The amount of red blood cells in a blood sample is equal to the total amount in the sample minus the amount of plasma. What is the total amount of blood drawn?
Answer:
Step-by-step explanation:
The question is:
The amount of red blood cells in a blood sample is equal to the total amount in the sample minus the amount of plasma. What is the total amount of blood drawn? Red blood cells = 45% Plasma = 5.5 ml
Solution:
We have given:
The amount of red blood cells in a blood sample is equal to the total amount in the sample minus the amount of plasma. We need to find how many ml is 1% of blood.
The equation we get is:
Red blood cells = total sample - amount of plasma
45% = 100% - 5.5 ml
Combine the percentages:
5.5 ml = 100%-45%
5.5 ml = 55%
0.1 ml = 1%
1% = 0.1 ml
Now you can find out red blood cells volume or total sample volume.
Red blood cells volume = 45% x 0.1ml/%
= 4.5ml
Total sample volume =100% x 0.1ml/%
= 10ml .
Formulate the situation as a system of two linear equations in two variables. Be sure to state clearly the meaning of your x- and y-variables. Solve the system by the elimination method. Be sure to state your final answer in terms of the original question. A jar contains 70 nickels and dimes worth $6.10. How many of each kind of coin are in the jar? nickels dimes
Answer:
There are 52 dimes and 18 nickles
Step-by-step explanation:
Lets call x= number of dimes and y= number of nickles
then we have the first equation
(1) x + y = 70
As a nickel is worth 0.05 US$ and a dime is worth 0.10 US$, we have the second equation
(2) 0.10x + 0.05y = 6.10
We then have a linear system of 2 equations and 2 unknowns
(1) x +y = 70
(2) 0.10x + 0.05y = 6.10
In order to solve the system by the elimination method, we have to multiply on of the equations by a suitable number to eliminate one unknown when adding the two equations.
There are several ways of doing this. We could, for example, multiply (1) by -0.05 and then add it to (2)
(1) -0.05x – 0.05y = (-0.05)70
(2) 0.10x + 0.05y = 6.10
That is to say,
(1) -0.05x – 0.05y = -3.5
(2) 0.10x + 0.05y = 6.10
Adding (1) and (2) we get
-0.05x = -2.6 => x = (-2.6)/(-0.05) = 52 => x = 52
So we have 52 dimes.
Substituting this value in equation 1, we obtain
y = 70 - x = 70 – 52 = 18
Then we have 18 nickels
Show that the sum of any two odd numbers is even.
Hi here´s a way to solve it
Let m and n be odd integers. Then, we can express m as 2r + 1 and n as 2s + 1, where r and s are integers.
This means that any odd number can be written as the sum of some even integer and one.
Substituting, we have that m + n = (2r + 1) + 2s + 1 = 2r + 2s + 2.
As we defined r and s as integers, 2r + 2s + 2 is also an integer.
Now It is clear that 2r + 2s + 2 is an integer divisible by 2 becasue we have 2 in each of the integers.
Therefore, 2r + 2s + 2 = m + n is even.
So, the sum of two odd integers is even.
Total cost to stay one night at sleepyhead Motel is $119 without breakfast and $142 with breakfast included what is the difference in the cost for a one-night stay with and without breakfast
The difference in the cost for a one-night stay with and without breakfast is:
$ 23
Step-by-step explanation:Total cost to stay one night at sleepyhead Motel without breakfast is: $119.
and cost to stay one night at sleepyhead Motel with breakfast is: $142.
Hence, the difference in the cost for a one-night stay with and without breakfast is calculated by:
$ (142-119)
= $ 23
Hence, the answer is: $ 23
Prove that every line in R^3 is a regular curve.
Answer:
Every line in [tex]\mathbb{R}^{3}[/tex] is a function of the form [tex]\gamma (t)={\bf p}+t {\bf v} [/tex], where [tex]{\bf p}[/tex] is point where the line passes and [tex]{\bf v}[/tex] is a nonzero vector which is called the direction vector of the line. Then, if we derive the function [tex]\gamma[/tex] we obtain [tex]\gamma'(t)={\bf v} \neq (0,0,0)[/tex], so [tex]\gamma(t)={\bf p}+t {\bf v}[/tex] is a regular curve.
Step-by-step explanation:
Every line in [tex]\mathbb{R}^{3}[/tex] can be parametrized by
[tex]\gamma (t)={\bf p}+t{\bf v}=(p_{1},p_{2},p_{3})+t(v_{1},v_{2},v_{3})=(p_{1}+tv_{1},p_{2}+tv_{2},p_{3}+tp_{3})[/tex], where [tex]t\in \mathbb{R}[/tex]. To derivate the function [tex]\gamma [/tex] we only need to derive each component. Then we have that
[tex]\gamma'(t)=(\frac{d}{dt}(p_{1}+tv_{1}),\frac{d}{dt}(p_{2}+tv_{2}),\frac{d}{dt}(p_{3}+tv_{3}))=(v_{1},v_{2},v_{3})={\bf v}\neq (0,0,0).[/tex]
Now, remember that a a parametrized curve is said to be regular if [tex]\gamma'\neq 0[/tex] for all [tex]t[/tex].
A company makes greeting cards and their research shows that that price and demand are related linearly: p = mx + b. They know that for every additional card they wish to sell they need to drop the price by $0.05. They also know that in order to sell 360 cards they need to set the price at $4. Find the linear equation relating price to demand. p= 0 Preview
Answer:
The linear equation relating price to demand is [tex]p=-0.05x+360.2[/tex]
Step-by-step explanation:
A Linear Demand Function expresses demand p (the number of items demanded) as a function of the unit price x (the price per item).
From the information given we know two facts:
For every additional card sold, the price decrease $0.05To sell 360 cards they need to set the price at $4Let x be the price and p the number of cards sold.
The fact 1. is the slope of the function because is the change in demand per unit change in price.
We can use fact 2. to find the value of b in the equation
[tex]p=-0.05x+b\\b=p+0.05x[/tex]
[tex]b=360+0.05\cdot 4\\b=360+0.2\\b=360.2[/tex]
The linear equation relating price to demand is [tex]p=-0.05x+360.2[/tex]
In an arithmetic sequence, the nth term an is given by the formula An=a1+(n−1)d, where a1
is the first term and d is the common difference. Similarly, in a geometric sequence, the nth term is given by an=a1•rn−1.
Here r is the common ratio. Use these formulas to determine the indicated term in the given sequence.
The 30th term of 1, 4, 7, 10,...
Answer:
88
Step-by-step explanation:
We are given that in arithmetic sequence , the nth term [tex]a_n[/tex] is given by the formula
[tex]A_n=a_1+(n-1)d[/tex]
Where [tex]a_1=first term[/tex]
d=Common difference
In an geometric sequence, the nth term is given by
[tex]a_n=a_1r^{n-1}[/tex]
Where r= Common ratio
1,4,7,10,..
We have to find 30th term.
[tex]a_1=1,a_2=4,a_3=7,a_4=10[/tex]
[tex]d=a_2-a_1=4-1=3[/tex]
[tex]d=a_3-a_2=7-4=3[/tex]
[tex]d=a_4-a_3=10-7=3[/tex]
[tex]r_1=\frac{a_2}{a_1}=\frac{4}{1}=4[/tex]
[tex]r_2=\frac{a_3}{a_2}=\frac{7}{4}[/tex]
[tex]r_1\neq r_2[/tex]
Therefore, given sequence is an arithmetic sequence because the difference between consecutive terms is constant.
Substitute n=30 , d=3 a=1 in the given formula of arithmetic sequence
Then, we get
[tex]a_{30}=1+(30-1)(3)=1+29(3)=1+87=88[/tex]
Hence, the 30th term of sequence is 88.
Find the area of the shaded region. All angles are right angles.
Area = Length time width.
The area of the shaded area is Length x width of the entire shape, minus the length x width of the unshaded area.
Full area: 27 x 10 = 270 square m.
Unshaded area: 17 x 5 = 85 square m.
Area of shaded region: 270 - 85 = 185 square m.
(From an actuarial exam) A company offers a health insurance plan, a life insurance plan, and an investment insurance plan. An employee can have 0, 1, or 2 plans, but cannot have both life insurance and investment plans. You are given the following information: • 450 employees have at least one plan. • 330 employees have only one plan. • 320 employees have the health insurance plan. • 45 employees have only the life insurance plan. • There are 20 more employees that have both health and life plans than those that have both health and investment plans. How many people have the investment plan
Answer:
Step-by-step explanation:
We can work with these values as a set value, and build a Venn Diagram from them.
I am going to say the set A are those that have the health insurance plan.
Set B are those that have the life insurance plan
Set C are those that have the investment plan.
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 employees that only have the health insurance plan, [tex]A \cap B[/tex] is the number of employees that have both the health and the life insurance plans, [tex]A \cap C[/tex] is the number of employees that have both the health insurance and the investment plans. and [tex]A \cap B \cap C[/tex] is the number of employees that have all three of those plans.
By the same logic, we have that:
[tex]B = b + (A \cap B) + (B \cap C) + (A \cap B \cap C)[/tex]
[tex]C = c + (B \cap C) + (A \cap C) + (A \cap B \cap C)[/tex]
The problem states that:
An employee cannot have both life insurance and investment plans. So:
[tex]B \cap C = 0, A \cap B \cap C = 0[/tex]
45 employees have only the life insurance plan. So:
[tex]b = 45[/tex]
There are 20 more employees that have both health and life plans than those that have both health and investment plans
[tex]A \cap B = A \cap C + 20[/tex]
320 employees have the health insurance plan.
[tex]A = 320[/tex]
450 employees have at least one plan
[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 450[/tex]
330 employees have only one plan
[tex]a + b + c = 330[/tex]
How many people have the investment plan?
We have to find the value of C.
Now we solve:
[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 450[/tex]
Applying what we have
-----------
[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 450[/tex]
[tex]330 + A \cap C + 20 + A \cap C = 450[/tex]
[tex]2(A\capC) = 100[/tex]
[tex]A \cap C = 50[/tex]
[tex]A \cap B = A \cap C + 20 = 50 + 20 = 70[/tex]
----------------
[tex]A = 320[/tex]
[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]
[tex]a + 70 + 50 = 320[/tex]
[tex]a = 200[/tex]
------------------
[tex]b = 45[/tex]
[tex]a + b + c = 330[/tex]
[tex]245 + c = 330[/tex]
[tex]c = 85[/tex]
The number of people that have the investment plan is:
[tex]C = 85 + 50 = 135[/tex]
135 people have the investment plan
The population mean annual salary for environmental compliance specialists is about $62,000. A random sample of 32 specialists is drawn from this population. What is the probability that the mean salary of the sample is less than $59,000? Assume σ=$6,200.
Answer: 0.002718
Step-by-step explanation:
Given : The population mean annual salary for environmental compliance specialists is about $62,000.
i.e. [tex]\mu=62000[/tex]
Sample size : n= 32
[tex]\sigma=6200[/tex]
Let x be the random variable that represents the annual salary for environmental compliance specialists.
Using formula [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex], the z-value corresponds to x= 59000 will be :
[tex]z=\dfrac{59000-62000}{\dfrac{6200}{\sqrt{32}}}\approx\dfrac{-3000}{\dfrac{6200}{5.6568}}=-2.73716129032\approx-2.78[/tex]
Now, by using the standard normal z-table , the probability that the mean salary of the sample is less than $59,000 :-
[tex]P(z<-2.78)=0.002718[/tex]
Hence, the probability that the mean salary of the sample is less than $59,000= 0.002718
For the following sequence, determine if it is an arithmeticsequence, a geometric sequence, or neither. If it is either arithmetic or geometric, give the next term in the sequence.
5, 8,13, 21, 34, 55,...
Answer:
Neither
Step-by-step explanation:
We are given that a sequence
5,8,13,21,34,55,....
We have to determine the sequence is an arithmetic , geometric or neither.
[tex]a_1=5,a_2=8,a_3=13,...a_6=55[/tex]
We know that if a sequence is an arithmetic then the difference between consecutive terms are equal.
If a sequence is geometric then the ratio of consecutive terms is constant.
[tex]d_1=a_2-a_1=8-5=3[/tex]
[tex]d_2=a_3-a_2=13-8=5[/tex]
[tex]d_1\neq d_2[/tex]
Hence, the difference between the consecutive terms is not equal .Therefore, sequence is not an arithmetic sequence.
[tex]r_1=\frac{a_2}{a_1}=\frac{8}{5}[/tex]
[tex]r_2=\frac{a_3}{a_2}=\frac{13}{8}[/tex]
[tex]r_1\neq r_2[/tex]
Hence, the ratio of consecutive terms is not equal .Therefore, given sequence is not geometric sequence.
Therefore, given sequence is not an arithmetic nor geometric sequence.
Solve for x in the expression using the quadratic formula. 2x2+31x−4.9=0
Answer:
[tex]x=\frac{-31\pm \sqrt{1000.2}}{4}[/tex]
Step-by-step explanation:
Given quadratic equation,
[tex]2x^2+31x-4.9=0[/tex]
Since, by the quadratic formula,
The solution of a quadratic equation [tex]ax^2+bx+c=0[/tex] is,
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Here, a = 2, b = 31, c = -4.9,
Thus, by the quadratic formula,
[tex]x=\frac{-31\pm \sqrt{31^2-4\times 2\times -4.9}}{2\times 2}[/tex]
[tex]=\frac{-31\pm \sqrt{961+39.2}}{4}[/tex]
[tex]=\frac{-31\pm \sqrt{1000.2}}{4}[/tex]
[tex]\implies x = \frac{-31+\sqrt{1000.2}}{4}\text{ or }x=\frac{-31- \sqrt{1000.2}}{4}[/tex]