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3 years ago

Diane's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Diane

Mathematics

1 answer:

Shalnov [3]3 years ago

5 0

4.05x+5.25(149-x)=710.25
Solve for x
X=60 ........4.05

149-60=89.....5.25

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Plssssss help me! What is the area of the shaded portion to the nearest hundredth?

Katarina [22]

Step-by-step explanation:

A1= area of the rectangle

A1= 4yd • 10yd

A1= 40 square yd

A2= area of the part of the circular section

A2= 180°• pi • 4 square yd/360°

A2= 180° • 16yd • pi/ 360°

A2= 16yd•pi/2

A2= 8yd•pi

A2= 25,12yd

A= areo of rhe shaded portion

A= A1-A2

A= 40yd-25,12yd

A= 14,88yd

7 0

3 years ago

⬛️ :blank divided by seven > 800

Nikolay [14]

Answer:

7/8 is nearly 1, and 8 1/10 is obviously 8.1 (1/10), which is rounded to 8. So th expression that best estimates the products of those numbers is D, 1 * 8

7 0

2 years ago

Find the cube roots of the following numbers

Gnoma [55]

a. 5

b. 6

c. 8

d. 10

e. 12

f. 15

5 0

3 years ago

Read 2 more answers

What is 8 21/40 I don't understand how to do it

yaroslaw [1]

Hey, there !

If you want to know it as a "improper fraction" , you can

Multiply the denominator from the front number
Whatever the denominator and front number is , you add it to the numerator

So, for this problem we first have the multiply the front number from the denominator

Like: 8 (40)
That gives us the answer of: 320

Now we have to add 320 to 21

Like: 320 + 21

That gives us the answer of:341

We keep our denominator (which is "40" )

Your improper fraction would be:

341 / 40

Good luck on your assignment and enjoy your day!

~MeIsKaitlyn:)

3 0

3 years ago

Read 2 more answers

If X is a r.v. such that E(X^n)=n! Find the m.g.f. of X,Mx(t). Also find the ch.f. of X,and from this deduce the distribution of

astraxan [27]

$M_X(t)=\mathbb E(e^{Xt})$
$M_X(t)=\mathbb E\left(1+Xt+\dfrac{t^2}{2!}X^2+\dfrac{t^3}{3!}X^3+\cdots\right)$
$M_X(t)=\mathbb E(1)+t\mathbb E(X)+\dfrac{t^2}{2!}\mathbb E(X^2)+\dfrac{t^3}{3!}\mathbb E(X^3)+\cdots$
$M_X(t)=1+t+t^2+t^3+\cdots$
$M_X(t)=\displaystyle\sum_{k\ge0}t^k=\frac1{1-t}$

provided that $|t|$ .

Similarly,

$\varphi_X(t)=\mathbb E(e^{iXt})$
$\varphi_X(t)=1+it+(it)^2+(it)^3+\cdots$
$\varphi_X(t)=(1-t^2+t^4-t^6+\cdots)+it(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=(1+it)(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=\dfrac{1+it}{1+t^2}=\dfrac1{1-it}$

You can find the CDF/PDF using any of the various inversion formulas. One way would be to compute

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{e^{itx}\varphi_X(-t)-e^{-itx}\varphi_X(t)}{it}\,\mathrm dt$

The integral can be rewritten as

$\displaystyle\int_0^\infty\frac{2i\sin(tx)-2it\cos(tx)}{it(1+t^2)}\,\mathrm dt$

so that

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt$

There are lots of ways to compute this integral. For instance, you can take the Laplace transform with respect to $x$ , which gives

$\displaystyle\mathcal L_s\left\{\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt\right\}=\int_0^\infty\frac{1-s}{(1+t^2)(s^2+t^2)}\,\mathrm dt$
$=\displaystyle\frac{\pi(1-s)}{2s(1+s)}$

and taking the inverse transform returns

$F_X(x)=\dfrac12+\dfrac1\pi\left(\dfrac\pi2-\pi e^{-x}\right)=1-e^{-x}$

which describes an exponential distribution with parameter $\lambda=1$ .

6 0

3 years ago