3 years ago

What is the greatest possible product of a 2 digit number and a 1 digit number

Mathematics

1 answer:

Molodets [167]3 years ago

3 0

891
because 99(the greatest 2 digit number)*9(the greatest 1 digit number)=891

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Find the length of AB.

umka2103 [35]

Answer:

6.286

Step-by-step explanation:

♢C=⊙/360x 22/7x D

=45/360x22/7x16

=6.286

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2 years ago

In the diagram, △A’B’C’ is an image of the △ABC. Which rule describes the translation? PLEASE HELP SOMEONE!!

gayaneshka [121]

Answer:

(5,3) had to write 20 characters

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3 years ago

Let x ∼ bin(9, 0.4). find

Kobotan [32]

A.
$\mathbb P(X>6)=\mathbb P(X=7)+\mathbb P(X=8)+\mathbb P(X=9)$
$=\dbinom97(0.4)^7(1-0.4)^{9-7}+\dbinom98(0.4)^8(1-0.4)^{9-8}+\dbinom99(0.4)^9(1-0.4)^{9-9}$
$\approx0.025$

b.
$\mathbb P(X\ge2)=1-\mathbb P(X$
$=1-\dbinom90(0.4)^0(1-0.4)^{9-0}-\dbinom91(0.4)^1(1-0.4)^{9-1}$
$\approx0.929$

c.
$\mathbb P(2\le X$
$=\dbinom92(0.4)^2(1-0.4)^{9-2}+\dbinom93(0.4)^3(1-0.4)^{9-3}+\dbinom94(0.4)^4(1-0.4)^{9-4}$
$\approx0.663$

d.
$\mathbb P(2$
$=\dbinom93(0.4)^3(1-0.4)^{9-3}+\dbinom94(0.4)^4(1-0.4)^{9-4}+\dbinom95(0.4)^5(1-0.4)^{9-5}$
$\approx0.669$

e.
$\mathbb P(X=0)=\dbinom90(0.4)^0(1-0.4)^{9-0}\approx0.01$

f.
$\mathbb P(X=7)=\dbinom97(0.4)^7(1-0.4)^{9-7}\approx0.021$

g, h.
For $X\sim\mathcal B(n,p)$ , recall that $\mathbb 
E[X]=\mu_X=np$ and $\mathbb V[X]={\sigma_X}^2=np(1-p)$ . So

$\mu_X=9(0.4)=3.6$
${\sigma_X}^2=9(0.4)(0.6)=2.16$

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3 years ago

Factor the expression 6x^2-5x. <br><br> Plzzzz helppppp!!

marshall27 [118]

Answer:

x(6x-5)

Step-by-step explanation:

take something from both equally

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3 years ago

What is p if 8/p = 3/14

anzhelika [568]

P/8=14/3
P=14*8/3
P=37.3333

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3 years ago