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lianna [129]
2 years ago
15

A piece of buttered toast falls to the floor 17 times. The toast landed buttered side up 6 times. What is the probability that t

he toast lands buttered side down?
Mathematics
1 answer:
kirza4 [7]2 years ago
8 0

Step-by-step explanation:

Given that,

A piece of buttered toast falls to the floor 17 times. The toast landed buttered side up 6 times.

It means that the total number of outcomes are 17

We need to find the probability that the toast lands buttered side down. Favourable oucome is 17-6 = 11

So, probability is given by :

P(E)=\dfrac{\text{favourable outcomes}}{\text{total no of outcomes}}

P(E)=\dfrac{11}{17}

So, the  probability that the toast lands buttered side down is 11/17.

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Each year a town holds a winter carnival this year 40% of the attendees were children under the age of 10 if 304 children under
tatuchka [14]

Answer:

760 attendees

Step-by-step explanation:

40% of the attendees is 304. That means you can add 304 to 304 (304 x 2) to get 608. To get the last 20%, divide 304 by 2, because 40(%) divided by 2 is 20(%). The answer to that is 152. Now, add it all up. 608 + 152 = 760.

In conclusion, there were 760 attendees at the carnival.

4 0
2 years ago
Find the area of equilateral triangle with side a.
marishachu [46]

Answer:

\frac{\sqrt{3} }{4} a^2

Step-by-step explanation:

To find the area of an equilateral triangle, we can apply a formula.

A=\frac{\sqrt{3} }{4} s^2

A= area\\s=side \: length

The side length is given a.

Plug a in the formula as the side length.

A=\frac{\sqrt{3} }{4} a^2

8 0
2 years ago
Read the proof. Given: m∠H = 30°, m∠J = 50°, m∠P = 50°, m∠N = 100° Prove: △HKJ ~ △LNP Statement Reason 1. m∠H = 30°, m∠J = 50°,
motikmotik

Answer:

 2. m\angle H+m\angle J +m\angle K =180^{\circ}

Reason: the sum of all interior angles of any triangle is equal to 180º.

Step-by-step explanation:

1) Organizing

Statement  1. m∠H = 30°, m∠J = 50°, m∠P = 50°, m∠N = 100° 1. Reason: given

2) Since the sum of all internal angles of any triangle is equal to 180º, just like the formula. N the number of sides of a triangle:

S_{i}=180^{\circ}(n-2)\rightarrow 180(3-2) \therefore S_{i}=180^{\circ}

3) We can also say

m\angle K=180^{\circ}-(m\angle H+m\angle J)\\m\angle  K=180^{\circ}-\left ( 30^{\circ}+50^{\circ} \right )\\m\angle K=100^{\circ}

Similarly to the other triangle:

m\angle L=180^{\circ}-(m\angle N+m\angle P)\\m\angle L=180^{\circ}-\left ( 100^{\circ}+50^{\circ} \right )\\m\angle L=30^{\circ}

4) Hence,

2. m\angle H+m\angle J +m\angle K =180

Reason: The sum of interior angles is equal to 180º.

8 0
2 years ago
Discuss what some of those rules are, and how they get applied in your analysis. If an engineering challenge includes "more than
ExtremeBDS [4]

Answer:

The engineer must verify and verify through a statistical inference that estimates and possible parameters may emerge, as well as determine what hypothesis tests should be performed to draw the most accurate conclusion.

Step-by-step explanation:

The engineer must assume that there may be more than one reasonable estimator for a different event or experiment; Something that could help you would be to perform an estimation of parameters, in that estimation it is required to know the properties of the estimators; that is to say that the closer the value of an estimator is to the real value of the parameter, it could be said that it is the most efficient or exact extimator.

6 0
2 years ago
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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