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vichka [17]
2 years ago
10

A machine can stamp 40 envelopes in 8 mins. How many of these machines working simultaneously are required or needed to stamp 12

0 envelopes per minutes.​
Mathematics
2 answers:
zzz [600]2 years ago
8 0

Answer:  The required number of machines is 24.

Step-by-step explanation:  Given that a machine can stamp 40 envelopes in 8 mins.

We are to find the number of machines that are working simultaneously are required or needed to stamp 120 envelopes per minutes.​

We will be using the UNITARY method to solve the given problem.

In 8 minutes, number of envelopes stamped by a machine = 40.

So, in 1 minute, the number of envelopes stamped by the machine is given by

\dfrac{40}{8}=5.

Now, number of machines required to stamp 5 envelopes in 1 minute = 1.

So, number of machines required to stamp 1 envelope in 1 minute is

\dfrac{1}{5}.

Therefore, the number of machines required to stamp 120 envelopes in 1 minute is given by

\dfrac{1}{5}\times120=24.

Thus, the required number of machines is 24.

kati45 [8]2 years ago
3 0

Answer:

24 minutes

Step-by-step explanation:

let y be the 120 envelope per minutes.

40/8 = 120/y

40y= 8 x 120

40x =960

x = 960/40

= 24 minutes

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N 2004, the General Social Survey (which uses a method similar to simple random sampling) asked, "Do you consider yourself athle
zmey [24]

<u>Answer-</u>

The standard error of the confidence interval is 0.63%

<u>Solution-</u>

Given,

n = 2373 (sample size)

x = 255 (number of people who bought)

The mean of the sample M will be,

M=\frac{x}{n} =\frac{255}{2373} =0.1075

Then the standard error SE will be,

SE=\sqrt{\frac{M\times (1-M)}{n}}

SE=\sqrt{\frac{0.1075\times (1-0.1075)}{2373}}=\sqrt{\frac{0.0959}{2373}}=0.0063=0.63\%

Therefore, the standard error of the confidence interval is 0.63%




6 0
2 years ago
△XYZ was reflected over a vertical line, then dilated by a scale factor of , resulting in △X'Y'Z'. Which must be true of the two
Igoryamba

Answer:

correct option is △XYZ ~ △X'Y'Z'.

Step-by-step explanation:

since ΔXYZ is dilated by some scale factor so, the resulting triangle can not be congruent to the ΔXYZ. so option 2 is wrong.

as we have explained that the two triangles are not congruent then it's sides and angles also can't be congruent so, option 3 is also incorrect.

As we don't know by what factor the triangle XYZ is dilated so we can't say anything about correctness of option 4 and 5.

ΔXYZ  was reflected over a vertical line and then dilated so the resulting ΔX'Y'Z' is similar to ΔXYZ.

i.e., △XYZ ~ △X'Y'Z'.

5 0
2 years ago
Read 2 more answers
Is it possible to draw three lines in two planes such that all three lines are skew?
nata0808 [166]
Yes; it is possible 

<span>If the planes are parallel to each other then they don't intersect. </span>

<span>There can also be two or more lines on different planes which do not intersect each other. These lines are called skew lines.</span>
8 0
2 years ago
Read 2 more answers
James works as a waiter. He served meals with bills of $23.59, $40.65, $30.50, and $15.68. If his total for tips was $17.67, wha
mr_godi [17]
<span><span>1.       </span>James works as a waiter.
He served food with the following price
=> 23.59 dollars
=> 40.65 dollars
=> 30.50 dollars
=> 15.68 dollars
Let’s add this altogether:
=> 23.59 + 40.65 + 30.50 + 15.68
=> 110.42 the total of the ordered food.
Now, James received a tip of 17.67 dollars. Find the interest rate of his tip.
=> 110.42 * .50 = 55.21, thus the interest rate of tip is less than 50%
=> You can do trial and error to get the correct rate,
=> 110.42 * .16 = 17.67, thus he received 16% tip</span>



3 0
2 years ago
Read 2 more answers
Find the moments of inertia Ix, Iy, I0 for a lamina in the shape of an isosceles right triangle with equal sides of length a if
Katyanochek1 [597]

Answer:

Ix = Iy =  a^4 / 12

Ixy = a^4 / 24

Step-by-step explanation:

Solution:-

- Sketch the right angled isosceles triangle as shown in the attachment.

- The density (ρ) is a multivariable function of both coordinates (x and y).

                             ρ ( x, y ) = k*(x^2 + y^2)

Where,  k: coefficient of proportionality.

- The lamina and density are symmetrical across the line y = x.  (see attachment). The center of mass must lie on this line.

- The coordinates of centroid ( xcm and ycm) are given by:

                            x_c_m = \frac{M_y}{m} = \frac{\int \int {x*p(x,y)} \, dA }{\int \int {p(x,y)} \, dA } \\\\y_c_m = \frac{M_x}{m} = \frac{\int \int {y*p(x,y)} \, dA }{\int \int {p(x,y)} \, dA } \\\\m = mass = \int \int {p(x,y)} \, dA

- The coordinates of xcm = ycm, they lie on line y = x.

- Calculate the mass of lamina (m):

                            m = mass = \int \int {p(x,y)} \, dA = \int\limits^0_a \int\limits_0 {k*(x^2 + y^2)} \, dy.dx \\\\m = k\int\limits^a_0 {(x^2y + \frac{y^3}{3}) } \, dx|\limits^a^-^x_0 = k\int\limits^a_0 {(\frac{-4x^3}{3}  + 2ax^2-ax^2+\frac{a^3}{3}) } \, dx\\\\m = k* [ \frac{-x^4}{3} + \frac{2ax^3}{3} - \frac{a^2x^2}{2} +\frac{a^3x}{3}] | \limits^a_0   \\\\m = k* [ \frac{-a^4}{3} + \frac{2a^4}{3} - \frac{a^4}{2} +\frac{a^4}{3}] = \frac{ka^4}{6}

- Calculate the Moment (My):

                     M_y = \int \int {x*p(x,y)} \, dA = \int\limits^0_a \int\limits_0 {k*x*(x^2 + y^2)} \, dy.dx \\\\M_y = k\int\limits^a_0 x*{(x^2y + \frac{y^3}{3}) } \, dx|\limits^a^-^x_0 = k\int\limits^a_0 x*{(\frac{-4x^3}{3}  + 2ax^2-ax^2+\frac{a^3}{3}) } \, dx\\\\M_y = k* [ \frac{-4x^5}{15} + \frac{ax^4}{2} - \frac{a^2x^3}{3} +\frac{a^3x^2}{6}] | \limits^a_0   \\\\M_y = k* [ \frac{-4a^5}{15} + \frac{a^5}{2} - \frac{a^5}{3} +\frac{a^5}{6}] = \frac{ka^5}{15}

- Calculate ( xcm = ycm ):

               xcm = ycm = ( ka^5 /15 ) / ( ka^4/6) = 2a/5        

- Now using the relations for Ix, Iy and I: We have:

                   Ix = bh^3 / 12

                   Iy = hb^3 / 12

Where,        h = b = a .... (Right angle isosceles)

                   Ix = Iy =  a^4 / 12

                   Ixy = b^2h^2 / 24

                   Ixy = a^4 / 24

                                           

8 0
2 years ago
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