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3 years ago

10

Someone draws a card at random from a deck, and then, without replacing the first card, draws a second card. What is the probabi

lity that both cards will be aces?

1 answer:

Flura [38]3 years ago

6 0

2/52 would be the answer i believe

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Please help pls pls<br><br><br>

katrin2010 [14]

Answer:

$g=\frac{\sqrt{(6+T)(6-T)} }{T}$

Step-by-step explanation:

I hope this helps!! <3

5 0

3 years ago

If correct I’ll mark brainlist

tekilochka [14]

Answer:

(0,1)

Step-by-step explanation:

The line meets the y-axis at the point (0,1)

6 0

3 years ago

What is the value of x?<br>what is the value of y?<br>type in an integer or decimal

Vilka [71]

9514 1404 393

Answer:

x = 5.6
y = 65

Step-by-step explanation:

There are a couple of relations that are applicable to these questions.

the product of segment lengths of crossed chords is the same for both chords
the angle formed at crossed chords is the average of the intercepted arc measures

__

The segment lengths relation tells us ...

10x = 8×7 . . . . . . products of segment lengths are equal

x = 56/10 = 5.6 . . . . divide by 10

__

The value of y° is the average of the intercepted arcs:

y° = (85° +45°)/2 = 65°

_____

<em>Additional comment</em>

This diagram does not have enough information to allow computation of z. We would need to know the intercepted arc, or the length of the secant that meets tangent z.

5 0

3 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

For a scavenger hunt, Jim's mom distributed a bag of 460 jelly beans evenly into 20 plastic containers and hid them around the y

Veseljchak [2.6K]

I think he found nine bags

7 0

4 years ago