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3 years ago

Ltiply. 5. 20 x 0.25

Mathematics

2 answers:

AlekseyPX3 years ago

4 0

Answer:

1.30

Step-by-step explanation:

When you multiply 5.20x0.25 you get 1.30

Troyanec [42]3 years ago

4 0

Answer:

1.3

Step-by-step explanation:

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A jeweler has 5 rings, each weighing 12 g, made of an alloy of 20% silver and 80% gold. She decides to melt down the rings and a

Aleksandr [31]

We have been given that a jeweler has 5 rings, each weighing 12 g, made of an alloy of 20% silver and 80% gold.

Upon melting all 5 rings, the weight of the material will be 12*5=60 grams.

We have been given that the alloy contains 20% silver and 80% gold. Further, let us assume that x grams of silver are added to the melted material so that it has 64% gold and 36% silver.

Therefore, we can set up an equation:

$\text{Amount of gold before mixing}=\text{Amount of gold after mixing}$

Therefore, we get an equation:

$80\%\text{ of 60 gram}=64\%\text{ of (x+60) gram}$

$0.80*60=0.64(x+60)\\48=0.64x+38.4\\0.64x=9.6\\x=\frac{9.6}{0.64} \\x=15$

Therefore, 15 grams of silver must be added to the melted material in order to reduce the gold content to 64%.

3 0

3 years ago

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

17.4 x 1.0 Will the product be less than 17.4, equal to 17.4, or greater than 17.4? 17.4 × 0.98 Less than 17.4 Equal to 17.4 Gre

nikdorinn [45]

Answer:

1) Equal to 17.4

2)Less than 17.4

Step-by-step explanation:

One thing we must always remember that when we have number 0 after the decimal point, it means we do not need to consider that as it will have no effect on product and on the other operations.

So,

17.4 x 1.0 = 17.4 x 1 = 17.4 (Product is equal to 17.4)

Suppose if a number 'A' which is less than 1 is multiplied by any other number 'B', than the multiplication always result in the number less than B.

For example:

1x0.5 = 0.5 which is less than 1

Similarly for

17.4 x 0.98 = Result will be less than 17.4 as the second number is less than 1.

Product is less than 17.4

3 0

3 years ago

The paper feed on a copying machine has room for a stack of paper 4.0 cm high. If 10 sheets of paper are 0.08 cm thick, how many

Vladimir [108]