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3 years ago

6

A 5-digit combination lock with digits 0-9 can be opened only if a correct combination of digits is chosen. Find the probability

of guessing the correct combination if:

2 answers:

Schach [20]3 years ago

5 0

My best guess is a 1/225 chance in guess the right combination

NemiM [27]3 years ago

5 0

You would have a 5/9 chance

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bagirrra123 [75]

Answer:

= -14y

Step-by-step explanation:

3 0

3 years ago

Consider two functions f and g on [1, 8] such that integral^8_1 f(x) dx = 9, integral^8_1 g(x) dx = 5, integral^8_5 f(x) dx = 4,

Gelneren [198K]

I'll abbreviate the definite integral with the notation,

$I(f(x),a,b)=\int_a^bf(x)\,\mathrm dx$

We're given

$I(f,1,8)=9$
$I(g,1,8)=5$
$I(f,5,8)=4$
$I(g,1,5)=3$

Recall that the definite integral is additive on the interval $[a,b]$ , meaning for some $c\in[a,b]$ we have

$I(f,a,b)=I(f,a,c)+I(f,c,b)$

The definite integral is also linear in the sense that

$I(kf+\ell g,a,b)=kIf(a,b)+\ell I(g,a,b)$

for some constant scalars $k,\ell$ .

Also, if $a\ge b$ , then

$I(f,a,b)=-I(f,b,a)$

a. $I(2f,1,5)=2I(f,1,5)=2(I(f,1,8)-I(f,5,8))=2(9-4)=\boxed{10}$

b. $I(f-g,1,8)=I(f,1,8)-I(g,1,8)=9-5=\boxed{4}$

c. $I(f-g,1,5)=I(f,1,5)-I(g,1,5)=\dfrac{I(2f,1,5)}2-I(g,1,5)=10-3=\boxed{7}$

d. $I(g-f,5,8)=I(g,5,8)-I(f,5,8)=(I(g,1,8)-I(g,1,5))-I(f,5,8)=(5-3)-4=\boxed{-2}$

e. $I(7g,5,8)=7I(g,5,8)=7(5-3)=\boxed{14}$

f. $I(3f,5,1)=3I(f,5,1)=-3I(f,1,5)=-\dfrac32I(2f,1,5)=-\dfrac32(10)=\boxed{-15}$

4 0

3 years ago

Hey.... what's <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B343%7D%20" id="TexFormula1" title=" \sqrt[3]{343} " alt

kap26 [50]

Answer:

³√343 = 7 .....ans .....

5 0

2 years ago

Read 2 more answers

7) 15:5 +6.5<br> I do not no how to do this

deff fn [24]

Answer:

15 divided by 5 =3

3 +6=9

9 - 5 = 4

so 4 is the answer

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

La fábrica Font Magdalena elabora 6 845 magdalenas diarias y, al empaquetarlas por docenas, utiliza 570 bolsas y le sobran 5 mag

AleksandrR [38]

Answer:

5 magdalenas

Step-by-step explanation:

Ya que vamos a estimar la cantidad de magdalenas que sobraron en la fábrica de María Magdalena. Debemos razonar como sigue.

Dado que hay el doble de magdalenas producidos en Maria Magdelena en comparación con el número producido en Font Magdelena y los magdalenas de la primera se empaquetan en dos docenas, se deduce que el número de magdalenas que quedan en Maria Magdelena sigue siendo 5 porque el número de magdalenas y el número envasado aumentó exactamente en la misma cantidad.

7 0

3 years ago