Answer:
30 million.
Step-by-step explanation:
Let x be the number of accountants who e-filed income tax returns in 2003.
We have been given that in 2004, 34.2 million accountants e-filed income tax returns. That was 114% of the number who e-filed in 2003.
We can represent our given information in an equation as:
[tex]\frac{114}{100}\cdot x=34.2[/tex]
[tex]1.14x=34.2[/tex]
[tex]\frac{1.14x}{1.14}=\frac{34.2}{1.14}[/tex]
[tex]x=30[/tex]
Since the number is given in millions, therefore, 30 million accountants e-filed income tax returns in 2003.
How many grams of Total Sugars in one serving of this food can be attributed to naturally occurring sugar (i.e., NOT added sugars)?
22g
2g
12g
10g
Answer:
Taking into account the information presented in the image attached for this food.
Naturally sugar are
2 g
Step-by-step explanation:
Information presented regarding to the composition of this food is attached.
One serving has a weight of 55 g.
Total carbohydrates in one serving are 37 g. This value includes fiber and sugars.
If we detail sugars, it has 12 g of total sugars (St), where 10 g are added sugars (Sa) and all the rest Natural sugars (Sn).
Expressing sugars as an equation
St = Sa + Sn
Isolating Sn value
Sn = St - Sa
Sn = 12 g - 10 g
Sn = 2 g
Finally, Natural sugars (Sn) is 2 g.
All 22 grams of total sugars per serving in the food are naturally occurring, as the label specifies there are 0g of added sugars.
Explanation:To determine the grams of naturally occurring sugar in a serving of food, we need to examine the nutritional information provided.
According to the label, it includes 0g added sugars, which implies that all of the sugars listed are naturally occurring.
Therefore, if there are 22g total sugars in one serving of the food, and none of these are added sugars, then all 22 grams can be attributed to naturally occurring sugars.
(7)-0, at the points x 71, 72, 73, 74, and 7.5 Use Euler's method with step size 0.1 to approximate the solution to the initial value pro oblemy - 2x+y The approximate solution to y'=2x-y?.y(7)=0, at the point x = 71 is (Round to five decimal places as needed.)
Answer:
2.68
Step-by-step explanation:
We are given that [tex]x_0=7,x_1=7.1,x_2=7.2,x_3=7.3,x_4=7.4,x_5=7.5[/tex]
h=0.1
y'=2x-y
y(7)=0,f(x,y)=2x-y
[tex]x_0=7,y_0=0[/tex]
We have to find the approximate solution to the initial problem at x=7.1
[tex]y_1=y_0+hf(x_0,y_0)[/tex]
Substitute the value then, we get
[tex]y_1=0+(0.1)(2(7)-0)=0+(0.1)(14)=1.4[/tex]
[tex]y_1=1.4[/tex]
[tex]x_1=x_0+h=7+0.1=7.1[/tex]
[tex]y_2=y_1+hf(x_1,y_1)[/tex]
Substitute the values then, we get
[tex]y_2=1.4+(0.1)(2(7.1)-1.4)=1.4+(0.1)(14.2-1.4)=1.4+(0.1)(12.8)=1.4+1.28[/tex]
[tex]y_2=1.4+1.28=2.68[/tex]
Hence, the approximation solution to the initial problem at x=7.1 is =2.68
Use the power series method to solve the given initial-value problem. (Format your final answer as an elementary function.)
(x − 1)y'' − xy' + y = 0, y(0) = −3, y'(0) = 2
[tex]y=\displaystyle\sum_{n\ge0}a_nx^n=a_0+\sum_{n\ge1}a_nx^n[/tex]
[tex]y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n=a_1+\sum_{n\ge0}(n+1)a_{n+1}x^n[/tex]
[tex]y''=\displaystyle\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^n[/tex]
Notice that [tex]y(0)=-3=a_0[/tex], and [tex]y'(0)=2=a_1[/tex].
Substitute these series into the ODE:
[tex](x-1)y''-xy'+y=0[/tex]
[tex]\displaystyle\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^{n+1}-\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^n-\sum_{n\ge0}(n+1)a_{n+1}x^{n+1}+\sum_{n\ge0}a_nx^n=0[/tex]
Shift the indices to get each series to include a [tex]x^n[/tex] term.
[tex]\displaystyle\sum_{n\ge1}n(n+1)a_{n+1}x^n-\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^n-\sum_{n\ge1}na_nx^n+\sum_{n\ge0}a_nx^n=0[/tex]
Remove the first term from both series starting at [tex]n=0[/tex] to get all the series starting on the same index [tex]n=1[/tex]:
[tex]\displaystyle-2a_2+a_0+\sum_{n\ge1}n(n+1)a_{n+1}x^n-\sum_{n\ge1}(n+1)(n+2)a_{n+2}x^n-\sum_{n\ge1}na_nx^n+\sum_{n\ge1}a_nx^n=0[/tex]
[tex]\displaystyle-2a_2+a_0+\sum_{n\ge1}\bigg[n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}-(n-1)a_n\bigg]x^n=0[/tex]
The coefficients are given recursively by
[tex]\begin{cases}a_0=-3\\a_1=2\\\\a_n=\dfrac{(n-2)(n-1)a_{n-1}-(n-3)a_{n-2}}{n(n-1)}&\text{for }n>1\end{cases}[/tex]
Let's see if we can find a pattern to these coefficients.
[tex]a_2=\dfrac{a_0}2=-\dfrac32=-\dfrac3{2!}[/tex]
[tex]a_3=\dfrac{2a_2}{3\cdot2}=-\dfrac12=-\dfrac3{3!}[/tex]
[tex]a_4=\dfrac{2\cdot3a_3-a_2}{4\cdot3}=-\dfrac18=-\dfrac3{4!}[/tex]
[tex]a_5=\dfrac{3\cdot4a_4-2a_3}{5\cdot4}=-\dfrac1{40}=-\dfrac3{5!}[/tex]
[tex]a_6=\dfrac{4\cdot5a_5-3a_4}{6\cdot5}=-\dfrac1{240}=-\dfrac3{6!}[/tex]
and so on, suggesting that
[tex]a_n=-\dfrac3{n!}[/tex]
which is also consistent with [tex]a_0=3[/tex]. However,
[tex]a_1=2\neq-\dfrac3{1!}=-3[/tex]
but we can adjust for this easily:
[tex]y(x)=-3+2x-\dfrac3{2!}x^2-\dfrac3{3!}x^3-\dfrac3{4!}x^4+\cdots[/tex]
[tex]y(x)=5x-3-3x-\dfrac3{2!}x^2-\dfrac3{3!}x^3-\dfrac3{4!}x^4+\cdots[/tex]
Now all the terms following [tex]5x[/tex] resemble an exponential series:
[tex]y(x)=5x-3\displaystyle\sum_{n\ge0}\frac{x^n}{n!}[/tex]
[tex]\implies\boxed{y(x)=5x-3e^x}[/tex]
The given differential equation is a second-order homogeneous differential equation. The power series method may not straightforwardly work for this equation due to the x dependence in the coefficients. Even using advanced techniques like the Frobenius method, the solution cannot be expressed as an elementary function.
Explanation:The given differential equation (x − 1)y'' - xy' + y = 0 is an example of a second order homogeneous differential equation. To solve the equation using the power series method, let's assume a solution of the form y = ∑(from n=0 to ∞) c_n*x^n. Substituting this into the equation and comparing coefficients, we can find a relationship for the c_n's and thus the power series representation of the solution.
However, for this particular differential equation, the power series method is not straightforward because of the x dependence in the coefficients of y'' and y'. Therefore, the standard power series approach would not work, and we would need more advanced techniques like the Frobenius method which allows for non-constant coefficients.
Unfortunately, even with the Frobenius method, the solution isn't an elementary function, meaning the solution cannot be expressed in terms of a finite combination of basic arithmetic operations, exponentials, logarithms, constants, and solutions to algebraic equations.
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Given:
An = [6 n/(-4 n + 9)]
For both of the following answer blanks, decide whether the given sequence or series is convergent or divergent. If convergent, enter the limit (for a sequence) or the sum (for a series). If divergent, enter INF if it diverges to infinity, MINF if it diverges to minus infinity, or DIV otherwise.
(a) The sequence {An }._________________
(b) The series ∑n=1[infinity]( An )________________
The given sequence is convergent with a limit of -3/2, while the series is divergent since its terms do not approach zero
Explanation:The sequence in question is An = [6*n/(-4*n + 9)]. To find out if this sequence is convergent or divergent, we need to take the limit as n approaches infinity. As n approaches infinity, the 'n' in the numerator and the 'n' in the denominator will dominate, making the sequence asymptotically equivalent to -6/4 = -3/2. Thus, the sequence is convergent, and its limit is -3/2.
On the other hand, the series ∑n=1[infinity]( An ) is the sum of the terms in the sequence. We can see that as n approaches infinity, the terms of this series do not approach zero, which is a necessary condition for a series to be convergent (using the nth term test). Therefore, the series is divergent.
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Listed below are amounts (in millions of dollars) collected from parking meters by a security service company and other companies during similar time periods. Do the limited data listed here show evidence of stealing by the security service company's employees? Security Service Company: 1.4 1.6 1.7 1.6 1.5 1.6 1.6 1.5 1.4 1.8 Other Companies: 1.5 1.8 1.6 1.9 1.7 1.9 1.8 1.7 1.7 1.6 Find the coefficient of variation for each of the two samples, then compare the variation.
Answer:
[tex]CV=\frac{0.1252}{1.57}=0.07975[/tex]
[tex]CV=\frac{0.1317}{1.72}=0.07657[/tex]
The relative variability is almost equal in both samples a slight greater variability can be noticed in the first sample.
Step-by-step explanation:
The coefficient of variation of a sample is defined as the ratio between the mean standard deviation and the sample mean. And it represents the percentage relation of the variation of the data with respect to the average.
[tex]CV=\frac{S}{\bar X}[/tex]
In the case of the first sample you have:
[tex]CV=\frac{0.1252}{1.57}=0.07975[/tex]
In the case of the second sample you have:
[tex]CV=\frac{0.1317}{1.72}=0.07657[/tex]
The relative variability is almost equal in both samples a slight greater variability can be noticed in the first sample.
Suppose you are planning to sample cat owners to determine the average number of cans of cat food they purchase monthly. The following standards have been set: a confidence level of 99 percent and an error of less than 5 units. Past research has indicated that the standard deviation should be 6 units. What is the final sample required?
Answer: 10
Step-by-step explanation:
The formula to find the sample size is given by :-
[tex]n=(\dfrac{z_{\alpha/2}\ \sigma}{E})^2[/tex]
Given : Significance level : [tex]\alpha=1-0.99=0.1[/tex]
Critical z-value=[tex]z_{\alpha/2}=2.576[/tex]
Margin of error : E=5
Standard deviation : [tex]\sigma=6[/tex]
Now, the required sample size will be :_
[tex]n=(\dfrac{(2.576)\ 6}{5})^2=9.55551744\approx10[/tex]
Hence, the final sample required to be of 10 .
If the probability that a bird will lay an egg is 75%, the probability that the egg will hatch is 50%, and the probability that the chick will be eaten by a snake before it fledges is 20%, what is the probability that a parent will have progeny that survive to adulthood?
(A) 30%
(B) 14.5%
(C) 7.5%
(D) 2%
Answer: (A) 30%
Step-by-step explanation:
Given : The probability that a bird will lay an egg =0.75
The probability that the egg will hatch =0.50
Now, the probability that the egg will lay an egg and hatch = [tex]0.75\times0.50[/tex]
Also,The probability that the chick will be eaten by a snake before it fledges =0.20
Then, the probability that the chick will not be eaten by a snake before it fledges : 1-0.20=0.80
Now, the probability that a parent will have progeny that survive to adulthood will be :-
[tex]0.75\times0.50\times0.80=0.30=30\%[/tex]
Hence, the probability that a parent will have progeny that survive to adulthood = 30%
Consider a fair coin which when tossed results in either heads (H) or tails (T). If the coin is tossed TWO times 1. List all possible outcomes. (Order matters here. So, HT and TH are not the same outcome.) 2. Write the sample space. 3. List ALL possible events and compute the probability of each event, assuming that the probability of each possible outcome from part (a) is equal. (Keep in mind that there should be many more events than outcomes and not all events will have the same probability.)
Answer:
Sample space = {(T,T), (T,H), (HT), (HH)}
Step-by-step explanation:
We are given a fair coin which when tossed one times either gives heads(H) or tails(T).
Now, the same coin is tossed two times.
1) All the possible outcomes
Tails followed by tails
Rails followed by heads
Heads followed by a tail
Heads followed by heads
2) Sample space
{(T,T), (T,H), (HT), (HH)}
3) Formula:
[tex]Probability = \displaystyle\frac{\text{Favourable outcome}}{\text{Total number of outcome}}[/tex]
Using the above formula, we can compute the following probabilities.
Probability((T,T)) =[tex]\frac{1}{4}[/tex]
Probability((T,H)) =[tex]\frac{1}{4}[/tex]
Probability((H,T)) =[tex]\frac{1}{4}[/tex]
Probability((H, H)) =[tex]\frac{1}{4}[/tex]
Probability(Atleast one tails) = [tex]\frac{3}{4}[/tex]
Probability(Atleast one heads) = [tex]\frac{3}{4}[/tex]
Probability(Exactly one tails) = [tex]\frac{2}{4}[/tex]
Probability(Exactly one heads) = [tex]\frac{2}{4}[/tex]
a,b,c,d are integers and GCD(a,b)=1. if c divides a and d divides b, prove that GCD(c,d) = 1.
Answer:
One proof can be as follows:
Step-by-step explanation:
We have that [tex]g.c.d(a,b)=1[/tex] and [tex]a=cp, b=dq[/tex] for some integers [tex]p, q[/tex], since [tex]c[/tex] divides [tex]a[/tex] and [tex]d[/tex] divides [tex]b[/tex]. By the Bezout identity two numbers [tex]a,b[/tex] are relatively primes if and only if there exists integers [tex]x,y[/tex] such that
[tex]ax+by=1[/tex]
Then, we can write
[tex]1=ax+by=(cp)x+(dq)y=c(px)+d(qy)=cx'+dy'[/tex]
Then [tex]c[/tex] and [tex]d[/tex] are relatively primes, that is to say,
[tex]g.c.d(c,d)=1[/tex]
Ben earns $9 per hour and $6 for each delivery he makes.He wants to earn more than $155 in an 8 hour work day.What is the least number of deliveries he must make to reach his goal?
Answer:
Ben must make at least 14 deliveries to reach his goal.
Step-by-step explanation:
The problem states that Ben earns $9 per hour and $6 for each delivery he makes. So his daily earnings can be modeled by the following function.
[tex]E(h,d) = 9h + 6d[/tex],
in which h is the number of hours he works and d is the number of deliveries he makes.
He wants to earn more than $155 in an 8 hour work day.What is the least number of deliveries he must make to reach his goal?
This question asks what is the value of d, when E = $156 and h = 8. So:
[tex]E(h,d) = 9h + 6d[/tex]
[tex]156 = 9*8 + 6d[/tex]
[tex]156 = 72 + 6d[/tex]
[tex]6d = 84[/tex]
[tex]d = \frac{84}{6}[/tex]
d = 14
Ben must make at least 14 deliveries to reach his goal.
1. The volume of a cube is increasing at a rate of 1200 cm/min at the moment when the lengths of the sides are 20cm. How fast are the lengths of the sides increasing at that [10] moment?
Answer:
[tex]1\,\,cm/min[/tex]
Step-by-step explanation:
Let V be the volume of cube and x be it's side .
We know that volume of cube is [tex]\left ( side \right )^{3}[/tex] i.e., [tex]x^3[/tex]
Given :
[tex]\frac{\mathrm{d} V}{\mathrm{d} t}=1200\,\,cm^3/min[/tex]
[tex]x=20\,\,cm[/tex]
To find : [tex]\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
Solution :
Consider equation [tex]V=x^3[/tex]
On differentiating both sides with respect to t , we get
[tex]\frac{\mathrm{d} V}{\mathrm{d} t}=3x^2\left ( \frac{\mathrm{d} x}{\mathrm{d} t} \right )\\1200=3(20)^2\left ( \frac{\mathrm{d} x}{\mathrm{d} t} \right )\\\frac{\mathrm{d} x}{\mathrm{d} t} =\frac{1200}{3(20)^2}=\frac{1200}{3\times 400}=\frac{1200}{1200}=1\,\,cm/min[/tex]
So,
Length of the side is increasing at the rate of [tex]1\,\,cm/min[/tex]
Last year, Anthony's grandmother gave him 33 silver coins and 16 gold coins to start a coin collection. Now Anthony has six times as many coins in his collection. How many coins does Anthony have in his collection?
Answer:
59 coins
Step-by-step explanation:
Add 16 gold coins and 33 silver coinsMultiply by 6The number of coins in Anthony's collection is 294
Using the parameters given for our Calculation;
silver coins = 33gold coins = 16Total number of coins now :
33 + 16 = 49Six times as many coins can be calculated thus :
6(49) = 294Therefore, the number of coins in Anthony's collection is 294
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Bob, the proprietor of Midland Lumber, believes that the odds in favor of a business deal going through are 9 to 5. What is the (subjective) probability that this deal will not materialize? (Round your answer to three decimal places.)
Answer:
There is a 35.7% probability that this deal will not materialize.
Step-by-step explanation:
This problem can be solved by a simple system of equations.
-I am going to say that x is the probability that this deal materializes and y is the probability that this deal does not materialize.
The sum of all probabilities is always 100%. So
[tex]1) x + y = 100[/tex].
Bob, the proprietor of Midland Lumber, believes that the odds in favor of a business deal going through are 9 to 5.
Mathematically, this means that:
[tex]2) \frac{x}{y} = \frac{9}{5}[/tex]
We want to find the value of y. So, we can write x as a function of y in equation 2), and replace it in equation 1).
Solution:
[tex]\frac{x}{y} = \frac{9}{5}[/tex]
[tex]x = \frac{9y}{5}[/tex]
[tex]x + y = 100[/tex]
[tex]\frac{9y}{5} + y = 100[/tex]
[tex]\frac{14y}{5} = 100[/tex]
[tex]14y = 500[/tex]
[tex]y = \frac{500}{14}[/tex]
[tex]y = 35.7[/tex]
There is a 35.7% probability that this deal will not materialize.
A man in a maze makes three consecutive displacements. His first displacement is 6.70 m westward, and the second is 11.0 m northward. At the end of his third displacement he is back to where he started. Use the graphical method to find the magnitude and direction of his third displacement.
Answer:
The man had a displacement of 12.88 m southeastward
Step-by-step explanation:
The path of man forms a right triangle. The first two magnitudes given in the problem form the legs and the displacement that we must calculate forms the hypotenuse of the triangle. To do this we will use the equation of the pythagorean theorem.
H = magnitude of displacement
[tex]H^2 = \sqrt{L_1^2 + L_2^2} = \sqrt{6.70^2 + 11.0^2} = \sqrt{165.89} =12.88 m[/tex]
using the graphic method, we will realize that the displacement is oriented towards the southeast
Find q, r in \mathbb{Z} so that 105 = 11q + r
with 0 \leq r < 11 as in the division algorithm
Answer:
[tex]q=9\,,\,r=6[/tex]
Step-by-step explanation:
Division Algorithm :
As per division algorithm , for numbers a and b , there exist numbers q and r such that [tex]a=bq+r\,\,,0\leq r< b[/tex]
Here ,
a = Dividend
b = Divisor
q = quotient
r = remainder
Given : 105 = 11q + r such that [tex]0 \leq r < 11[/tex]
Here, clearly a = 105 , b = 11
To find : q and r
Solution : On dividing 105 by 11 , we get [tex]105=11\times 9+6[/tex]
On comparing [tex]105=11\times 9+6[/tex] with [tex]a=bq+r\,\,,0\leq r< b[/tex] , we get [tex]q=9\,,\,r=6[/tex]
Cheese costs $4.40 per pound. Find the cost per kilogram. (1kg = 2.2lb)
Answer:
The cost is $9.70 per kilogram.
Step-by-step explanation:
This can be solved by a rule of three.
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. In this case, the rule of three is a cross multiplication.
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 case, the rule of three is a line multiplication.
In this problem, the measures are the weight of the cheese and the price. As the weight increases, so does the price. It means that this is a direct rule of three.
Solution:
The problem states that cheese costs $4.40 per pound. Each kg has 2.2 pounds. How many kg are there in 1 pound. So:
1 pound - xkg
2.2 pound - 1 kg
[tex]2.2x = 1[/tex]
[tex]x = \frac{1}{2.2}[/tex]
[tex]x = 0.45[/tex]kg
Since cheese costs $4.40 per pound, and each pound has 0.45kg, cheese costs $4.40 per 0.45kg. How much does is cost for 1kg?
$4.40 - 0.45kg
$x - 1kg
[tex]0.45x = 4.40[/tex]
[tex]x = \frac{4.40}{0.45}[/tex]
[tex]x = 9.70[/tex]
The cost is $9.70 per kilogram.
"The cost per kilogram of cheese is approximately $2.00.
To find the cost per kilogram, we need to convert the cost from dollars per pound to dollars per kilogram using the conversion factor between pounds and kilograms. Given that 1 kilogram is equal to 2.2 pounds, we can set up the following conversion:
Cost per pound of cheese = $4.40
Conversion factor = 2.2 pounds/kilogram
Now, to find the cost per kilogram, we divide the cost per pound by the conversion factor:
Cost per kilogram = Cost per pound / Conversion factor
Cost per kilogram = $4.40 / 2.2 pounds/kilogram
Performing the division, we get:
Cost per kilogram ≈ $2.00
Before the industrial revolution in 1800 the concentration of carbon in Earth’s atmo- sphere was 280 ppm. The concentration in 2015 was 399 ppm. What is the percent increase in the amount of carbon in the atmosphere?
Answer: There is increase of 4.255 in the amount of carbon in the atmosphere.
Step-by-step explanation:
Since we have given that
Concentration of carbon in Earth's atmosphere in 1800 = 280 ppm
Concentration of carbon in Earth's atmosphere in 2015 = 399 ppm
We need to find the percentage increase in the amount of carbon in the atmosphere.
So, Difference = 399-280 = 119 ppm
so, percentage increase in the amount of carbon is given by
[tex]\dfrac{Difference}{Original}\times 100\\\\=\dfrac{119}{280}\times 100\\\\=\dfrac{11900}{280}\\\\=42.5\%[/tex]
Hence, there is increase of 4.255 in the amount of carbon in the atmosphere.
Which nutrient category on the Nutrition Facts panel does NOT have a Daily Value or % Daily Value ascribed to it?
Dietary Fiber
Protein
Total Fat
Sodium
Answer:
Protein
Step-by-step explanation:
According to U.S. Food and Drug Administration (FDA) information, the current scientific evidence indicates that protein intake is not a matter of public concern, this is why the FDA does not request to add a % Daily Value on the Nutrition Facts panel, unless if the product is made for protein such as 'high protein' products or if this food is meant for use by childrend under 4 years old.
Answer:Protein
Step-by-step explanation: took quiz :)
Justin deposited $2,000 into an account 5 years ago. Simple interest was paid on the account. He has just withdrawn $2,876. What interest rate did he earn on the account?
Answer: [tex]8.76\%[/tex]
Step-by-step explanation:
The formula to find the final amount after getting simple interest :
[tex]A=P(1+rt)[/tex], where P is the principal amount , r is rate of interest ( in decimal )and t is time(years).
Given : Justin deposited $2,000 into an account 5 years ago.
i.e. P = $2,000 and t= 5 years
He has just withdrawn $2,876.
i.e. we assume that A = $2876
Now, Put all the values in the formula , we get
[tex](2876)=(2000)(1+r(5))\\\\\Rightarrow\ 1+5r=\dfrac{2876}{2000}\\\\\Rightarrow\ 1+5r=1.438\\\\\Righhtarrow\ 5r=0.438\\\\\Rightarrow\ r=\dfrac{0.438}{5}=0.0876[/tex]
In percent, [tex]r=0.0876\times100=8.76\%[/tex]
hence, He earned [tex]8.76\%[/tex] of interest on account.
A recipe calls for 1 cup of ground almonds. How many ounces of ground almonds should you use for this recipe if 1 pint of ground almonds weighs 0.42 pounds?
Answer: There are 3.36 ounces of ground almonds in 1 cup.
Step-by-step explanation:
Since we have given that
1 pint = 0.42 pounds
As we know that
1 cup = 0.5 pints
1 pound = 16 ounces
So, We need to find the number of ounces.
As 1 pint = 0.42 pounds = 0.42 × 16 = 6.72 ounces
0.5 pints is given by
[tex]6.72\times 0.5\\\\=3.36\ ounces[/tex]
Hence, there are 3.36 ounces of ground almonds in 1 cup.
To find out how many ounces are in one cup of ground almonds if 1 pint (2 cups) weighs 0.42 pounds, you divide 0.42 pounds by 2 to get the weight per cup, then multiply by 16 to convert pounds to ounces, resulting in 3.36 ounces per cup.
The question involves converting weight measurements from one unit to another, specifically from pounds to ounces. To do this, you use the conversion factor of 16, because there are 16 ounces in 1 pound. Applying this to the recipe question:
1 pint of ground almonds weighs 0.42 pounds. Since 1 pint equals 2 cups, this weight corresponds to 2 cups of ground almonds.
To find out how many ounces 1 cup of ground almonds weighs, first divide the total weight by the number of cups:
0.42 pounds ÷ 2 cups = 0.21 pounds per cup.
Then, convert pounds to ounces:
0.21 pounds × 16 ounces/pound = 3.36 ounces.
So, for the recipe, you should use 3.36 ounces of ground almonds.
Let p stand for "This statement is false." What can be said about the truth value of p. (Hint: Did we really assign a truth value to p? See Example 5 for a discussion of truth value assignment.)
Answer: P means "This statement is false"
then, P is a "function" of some statement,
if i write P( 3> 1932) this could be read as:
3>1932, this statement is false.
You could see that 3> 1932 is false, so P( 3>1932) is true.
Then you could se P(x) at something that is false if x is true, and true if x is false, so p is a negation.
Find the probability that Z is to the right of 3.05.
Answer: 0.0011
Step-by-step explanation:
By using the standard normal distribution table , the probability that Z is to the left of 3.05 is [tex]P(z<3.05)= 0.9989[/tex]
We know that the probability that Z is to the right of z is given by :-
[tex]P(Z>z)=1-P(Z<z)[/tex]
Similarly, the probability that Z is to the right of 3.05 will be :-
[tex]P(Z>3.05)=1-P(Z<3.05)=1-0.9989=0.0011[/tex]
Hence, the probability that Z is to the right of 3.05 = 0.0011
Round the following number to the indicated place. 66.1086 to hundredths
Answer:
66.11
Step-by-step explanation:
We are given that a number
66.1086
We have to round the number to hundredths
Place of 6=One;s
Place of second 6=Tens
Place of 1=Tenths
Place of 0=Hundredths
Place of 8=Thousandths
Place of 6=Ten thousandths
Thousandths place is 8 which is greater than 5 therefore, one will be added to hundredth place and other number on the left side of hundredth place remain same and the numbers on the right side of hundredth place will be replace by zero.
Therefore, the given number round to hundredths=66.11
Show your work:
Express 160 pounds (lbs) in kilograms (kg). Round to the nearest hundredths.
Answer:
160 lbs = 72.57kg
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 lb has 0.45kg. How many kg are there in 160lbs. So:
1lb - 0.45kg
160 lbs - xkg
[tex]x = 0.45*160[/tex]
[tex]x = 72.57[/tex] kg
160 lbs = 72.57kg
Simplify the following expressions
X2 X-2/3
X1/4/ X-5/2
(4/5)x-2/5 y3/2 / (2/3) x3/5y1/2
(2/3)x2/3y2/3 / (1/3)x-1/3y-1/3
(z2/3 x2/3y-2/3 ) y + (z2/3 x-1/3y-1/3 ) x
(4x3/5y3 z2 ) 1/3
Step-by-step explanation:
For each case we have the next step by step solution.
[tex]x^2(\dfrac{x-2}{3})=\dfrac{x^3-2x^2}{3}[/tex][tex]\dfrac{x^{1/4}}{x-\dfrac{5}{2}}=\dfrac{x^{1/4}}{\dfrac{2x}{2}-\dfrac{5}{2}}=\dfrac{x^{1/4}}{\dfrac{2x-5}{2}}=\dfrac{2x^{1/4}}{{2x-5}}[/tex][tex]\dfrac{\dfrac{4}{5}x-\dfrac{2}{5}y^{3/2}}{\dfrac{\dfrac{2}{3}x^3}{5y^{1/2}}}=\dfrac{\dfrac{4x}{5}-\dfrac{2y^{3/2}}{5}}{\dfrac{\dfrac{2x^3}{3}}{5y^{1/2}}}=\dfrac{\dfrac{4x-2y^{3/2}}{5}}{\dfrac{2x^3}{15y^{1/2}}}={\dfrac{(4x-2y^{3/2})\cdot 15y^{1/2}}{5\cdot 2x^3}}[/tex] [tex]{\dfrac{(4x-2y^{3/2})\cdot 15y^{1/2}}{5\cdot 2x^3}}={\dfrac{(60xy^{1/2}-30y^{3/2}y^{1/2})}{10x^3}}={\dfrac{(60xy^{1/2}-30y^{4/2})}{10x^3}}={\dfrac{(60xy^{1/2}-30y^{2})}{10x^3}}[/tex][tex]\dfrac{\dfrac{\dfrac{2}{3}x^2}{3y^{2/3}}}{\dfrac{1}{3}x-\dfrac{1}{3}y-\dfrac{1}{3}}=\dfrac{\dfrac{\dfrac{2x^2}{3}}{3y^{2/3}}}{\dfrac{x}{3}-\dfrac{y}{3}-\dfrac{1}{3}}=\dfrac{\dfrac{2x^2}{9y^{2/3}}}{\dfrac{x-y-1}{3}}=\dfrac{2x^2\cdot 3}{(x-y-1)\cdot 9y^{2/3}}}=\dfrac{6x^2}{(9xy^{2/3}-9yy^{2/3}-9y^{2/3})}}=\dfrac{6x^2}{(9xy^{2/3}-9y^{5/3}-9y^{2/3})}}[/tex][tex](z^{2/3}x^{2/3}y+\dfrac{2}{3})y+(z^{2/3}x-\dfrac{1}{3}y-\dfrac{1}{3})x=(z^{2/3}x^{2/3}y^2+\dfrac{2}{3}y)+(z^{2/3}x^2-\dfrac{1}{3}yx-\dfrac{x}{3})=z^{2/3}x^{2/3}y^2+z^{2/3}x^2-\dfrac{1}{3}yx+\dfrac{2}{3}y-\dfrac{x}{3}[/tex][tex](\dfrac{4x^3}{5y^2}z^2)^{1/3}=\dfrac{(4x^3)^{1/3}}{(5y^2)^{1/3}}(z^2)^{1/3}=\dfrac{4^{1/3}x}{5^{1/3}y^{2/3}}z^{2/3}[/tex]Prove that the following two sets are the same. S1 = {a + bx : a, b ∈ R} = all polynomials which can expressed as a linear combination of 1 and x; S2 = {ax + b(2 + x) : a, b ∈ R} = all polynomials which can expressed as a linear combination of x and 2 + x.
Answer with Step-by-step explanation:
We are given that twos sets
[tex]S_1[/tex]={a+bx:[tex]a,b\in R[/tex]}=All polynomials which can expressed as a linear combination of 1 and x.
[tex]S_2[/tex]={ax+b(2+x):[tex]a,b\in R[/tex]}=All polynomials which can be expressed as a linear combination of x and 2+x.
We have to prove that given two sets are same.
[tex]S_2[/tex]={ax+2b+bx}={(a+b)x+2b}={cx+d}
[tex]S_2[/tex]={cx+d}=All polynomials which can be expressed as a linear combination of 1 and x.
Because a+b=c=Constant
2b= Constant=d
Hence, the two sets are same .
slope-intercept formula y=mx+b
1. Suppose that A , B and C are sets. Show that A \ (B U C) (A \ B) n (A \ C).
Step-by-step explanation:
We want to show that
[tex]=A \setminus (B \cup C) = (A\setminus B) \cap (A\setminus C)[/tex]
To prove it we just use the definition of [tex]X\setminus Y = X \cap Y^c[/tex]
So, we start from the left hand side:
[tex]=A \setminus (B \cup C) = A \cap (B \cup C)^c[/tex] (by definition)
[tex]=A \cap (B^c \cap C^c)[/tex] (by DeMorgan's laws)
[tex]=A \cap B^c \cap C^c[/tex] (since intersection is associative)
[tex]=A \cap B^c \cap A \cap C^c[/tex] (since intersecting once or twice A doesn't make any difference)
[tex]=(A \cap B^c) \cap (A \cap C^c)[/tex] (since again intersection is associative)
[tex]=(A\setminus B) \cap (A \setminus C)[/tex] (by definition)
And so we have reached our right hand side.
Calculate (a) the number of milligrams of metoclopramide HCl in each milliliter of the prescription:
Metoclopramide HCl 10 g
Methylparaben 50 mg
Propylparaben 20 mg
NaCl 800 mg
Purifed water qs ad 100 mL
Answer:
There are 100 milligrams of metoclopramide HCl in each milliliter of the prescription
Step-by-step explanation:
When the prescription says Purified water qs ad 100 mL means that if we were to make this, we should add the quantities given and then, fill it up with water until we have 100 mL of solution, being the key words qs ad, meaning sufficient quantity to get the amount of mixture given.
Then, knowing there is 10 grams of metoclopramide HCl per 100 mL of prescription, that means there is (1 gram = 1000 milligrams) 10000 milligrams of metoclopramide HCl per 100 mL of prescription. That is a concentration given in a mass/volume way.
Knowing the concentration, we can calculate it per mL instead of per 100 mL
[tex]Concentration_{metoclopramide HCL}= \frac{10000mg}{100mL} =100 \frac{mg}{mL}[/tex]
What is the answer to (n+4) +7 =
(n+4) +7 remove the parenthesis
n+4+7 add the same number answer is n +11
DECISION SCIENCE Assume more restrictions. The books pass through two departments: Graphics and Printing, 1. before they can be sold, X requires 3 hours in Graphics and 2 hours in printing. Y requires 1 hour in graphics and 1 hour in printing. There are 21 hours available in graphics and 19 hours in printing respectively. Solve for x and y.
Answer:
Step-by-step explanation:
The given information can be tabulated as follows:
Graphics G Printing P Total
X 3 2 5
Y 1 1 2
Available 21 19
We have the constraints as
[tex]3x+y\leq 21\\2x+y\leq 19\\x\leq 2\\y\leq 15[/tex]
Thus we have solutions as
[tex]0\leq x\leq 2\\0\leq y\leq 15[/tex]