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3 years ago

Suppose that a randomly generated list of numbers from 0 to 9 is being used

Mathematics

2 answers:

sdas [7]3 years ago

5 0

Answer: D

Step-by-step explanation:

counting 0 there are 7 numbers which represent 70% since there are 10 total numbers

Leni [432]3 years ago

3 0

Answer:

Step-by-step explanation:

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Convert 123000000 milligram into gram

velikii [3]

Answer:

123,000 grams

Step-by-step explanation:

1000 mg is equivalent to 1 gram.

3 0

2 years ago

Read 2 more answers

A photography studio charges $50 that includes a sitting fee and 6 prints. Luigi increased his order to 11 prints and paid $65.

Effectus [21]

Answer:

Sitting fee - 32$

Step-by-step explanation:

This is a system of equations(let x represent the sitting fee)

x+6y=50

x+11y=65

You want to isolate the x variable - x+6y-6y=50-6y ; x = 50-6y

Input this into the 2nd equation: 50-6y+11y=65 ; 50+5y=65

Subtract 50 from both sides. 5y=15 (Divided 5y on both sides) ; y=3

Now that y = 3 input this into any equation I choose the 1st one.

x+6(3) = 50 ; x + 18 = 50 (Subtract 18 on both sides to get x)

x = 32

Prove: 32 + 6(3) = 50 ; 32+18 = 50 ; 50 = 50 True

7 0

3 years ago

Tickets for a dance recital cost $15 for adults and $7 for children. The dance company sold 253 tickets, and the total receipts

elixir [45]

Answer:

Adult tickets =125

Child tickets = 128

Step-by-step explanation:

Let the number of adult and children tickets sold be x and y respectively

So that

x+y= 253--------1

Since total sales/receipt is $2,771

And given that tickets for a dance recital cost $15 for adults and $7 for children hence

15x+7y= 2771-------2

Solving equation 1 and 2 simultaneously we have

x+y= 253---------------1

15x+7y= 2771----------2

Let us multiply equation 1 by 15 to get equation 3 to eliminate x and subtract equation 2 from 3

15x+15y=3795---------3

-{15x+7y= 2771----------2

0+8y=1024

8y= 1024

y= 1024/8

y= 128 tickets

So solve for x let us put y= 128 in equation 1

x+ 128=253

x=253-128

x= 125 tickets

8 0

3 years ago

Given 2 events a and b, and that p(a)=0.30, p(b)=0.45, p(a u b)=0.60. find the probability p(a∩b).

pickupchik [31]

P(a U b) = p(a) + p(b) - p(a ∩ b)

<=>

p(a ∩ b) = p(a) + p(b) - p(a U b)

I let you do the maths :)

3 0

3 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

3 0

2 years ago