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shusha [124]
3 years ago
9

you draw two cards from a well-shuffled deck of 52 cards. After you draw the first card, you do not replace it in the deck. What

is the probability of P(get ace of spades on first draw and get queen of hearts second draw?
Mathematics
1 answer:
Nuetrik [128]3 years ago
6 0
52 total cards...
P(ace of spades on first draw) = 1/52
without replacing
P(queen of hearts) = 1/51

P(both events happening) = 1/52 * 1/51 = 1/2652 <==
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Write an equation for the line parallel to y= -2x+1 that contains (-2,5)
grin007 [14]

Answer:

y = - 2x + 1

Step-by-step explanation:

Parallels line have the same slope, when an equation is in the form y= mx + b, m is the slope. In this problem slope = - 2

Now with the slope what is missing is the y-intercept, the problem says that the line contains the point (-2, 5), replacing that point in the equation you can solve it to find the y-intercept

y = mx + b

5 = (-2)(-2) + b

5 = 4 + b

1 + b

y = - 2x +1

5 0
3 years ago
Help please!!! i’ve been stuck on this for days!
Snezhnost [94]

Answer:

Step-by-step explanation:

What is the full question?

3 0
3 years ago
Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x =
lord [1]
Using washers, the volume is given by

\displaystyle\pi\int_{x=1}^{x=e}((1+3)^2-(\ln x+3)^2)\,\mathrm dx
=\displaystyle\pi\int_{x=1}^{x=e}(7-6\ln x-(\ln x)^2)\,\mathrm dx
5 0
2 years ago
A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient
ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

y=(x-a) \sqrt{(x-b)}  --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

0=(b+1-a) \sqrt{(b+1-b)}

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

y^{\prime}=u^{\prime} y+y^{\prime} u

so, we get

{u}^{\prime}=1

v^{\prime}=\frac{1}{2} \sqrt{(x-b)}

y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}

By putting value of point A and putting value of eq 2 we get

y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}

y^{\prime}=\frac{d y}{d x}=1

Hence proved that the gradient of the curve at A is 1.

7 0
2 years ago
2(x−4)+3x=7 solve for x
sesenic [268]

Answer:

x = 3

Step-by-step explanation:

5x - 8 + 8 = 7 + 8

5x = 15

5x/5 = 15/5

15 divided by 5 equals 3

Hope this helped you out :)

7 0
3 years ago
Read 2 more answers
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