What is the GCF of 36 and 84

A SINGLE CARD IS DRAWN AT RANDOM FROM A STANDARD DECK OF 52 CARDS. FIND THE PROBABILITY OF DRAWING THE FOLLOWING CARDS. PLEASE R

erma4kov [3.2K]

Answer:

A. $\frac{17}{52}$

B. $\frac{17}{52}$

C. $\frac{2}{13}$

Step-by-step explanation:

A.

There are 52/4 diamonds in the deck and 4 '5's in the dech of cards

52/4 = 13 + 4 = 17

Therefore, you have a $\frac{17}{52}$ chance of drawing one of those cards.

B.

There are 13 hearts in the deck and 4 jacks. Therefore, your odds are the same : $\frac{17}{52}$

C.

There are 4 jacks in a deck of cards and 4 '8's in a deck of cards

Therefore your probability is $\frac{8}{52}$ which simplifies to $=\frac{2}{13}$

As per brainly guidelines I can only answer 3 questions in one answer

7 0

2 years ago

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If the outlier is removed, which measure will not change? mean median mode range

Elena-2011 [213]

Answer:

Out of mean, median, mode, and range, the answer to this question is mode. If an outlier is removed, mode will not change. Hope this helps!

5 0

3 years ago

What is 124/87 SOMEONE?

Rufina [12.5K]

Answer:

1.53

Step-by-step explanation:

5 0

3 years ago

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A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. (If an answer does

NNADVOKAT [17]

The question is not clear, but it is possible to obtain distance, s, from the given function. This, I would show.

Answer:

s = 17 units

Step-by-step explanation:

Given f(t) = t³ - 8t² + 27t

Differentiating f(t), we have

f'(t) = 3t² - 16 t + 27

At t = 0

f'(t) = 27

This is the required obtainaible distance, s.

6 0

3 years ago

Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their p

GuDViN [60]

Answer:

A) points at which paths intersect : (1,1,1) ; (2,4,8)

B) DNE

Step-by-step explanation:

A) To find the points in which the particle paths intersect, it is necessary to find the values of t for which the three components of both vectors are equal:

$t_1=1+2t_2\\\\t_1^2=1+6t_2\\\\t_1^3=1+14t_2$

you replace t1 from the first equation in the second equation:

$(1+2t_2)^2=1+6t_2\\\\1+4t_2+4t_2^2=1+6t_2\\\\4t_2^2-2t_2=0\\\\t_2(2t_2-1)=0\\\\t_2=0\\\\t_2=\frac{1}{2}$

Then, for t2 = 0 and t2=1/2 you obtain for t1:

$t_1=1+2(0)=1\\\\t_1=1+2(\frac{1}{2})=2$

Hence, for t1=1 and t2=0 the paths intersect. Furthermore, for t1=2 and t2=1/2 the paths also intersect.

The points at which the paths intersect are:

$r_1(1)=(1,1,1)=r_2(0)=(1,1,1)\\\\r_1(2)=(2,4,8)=r_2(\frac{1}{2})=(2,4,8)$

B) You have the following two trajectories of two independent particles:

$r_1(t)=(t,t^2,t^3)\\\\r_2(t)=(1+2t,1+6t,1+14t)$

To find the time in which the particles collide, it is necessary that both particles are in the same position on the same time. That is, each component of the vectors must coincide:

$t=1+2t\\\\t^2=1+6t\\\\t^3=1+14t$

From the first equation you have:

$t=1+2t\\\\t=-1$

This values does not have a physical meaning, then, the particle do not collide

answer: DNE

5 0

3 years ago