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1 year ago

12

Consider a standard deck of 52 cards (13 cards each in 4 suits, clubs, spades, diamonds and hearts). What is the probability of

picking out 5 cards in a row which are all diamonds, without replacing any card to the deck?

1 answer:

Goryan [66]1 year ago

8 0

5/52 as you are taking 5 cards out

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What would be the value of X

Alex_Xolod [135]

To find the value of x, you need to set up a proportion.This is the proportion you would set up:

2.5 x
---- = ------
3.75 60

x=40

7 0

3 years ago

Read 2 more answers

If a point is equidistant from the endpoint of segment, then it is

Mashutka [201]

Hi

.... then it is located on the perpendicular bisector.

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3 years ago

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Delvig [45]

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3 years ago

If the range of f(x)= square root mx and the range of g(x)=m square root x are the same, which statement is true about the value

AveGali [126]

M can be any positive real number.

Explanation:

From f(x) = √(mx) ; if x is posive m has to be positive; if x is negative m has to be negative; if x is cero m can have any value, and the range will always be 0 or positve

From g(x) = m √x; x can only be 0 or positive and the range will have the sign of m.

Given we concluded that the range of f(x) can only be 0 or positive, then me can only be 0 or positive.

3 0

2 years ago

Use the Fundamental Theorem of Calculus to find the area of the region between the graph of the function x5 + 8x4 + 2x2 + 5x + 1

belka [17]

Answer:

The area of the region between the graph of the given function and the x-axis = 25,351 units²

Step-by-step explanation:

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15

If 'f' is a continuous on [a ,b] then the function

$F(x) = \int\limits^a_b {f(x)} \, dx$

By using integration formula

$\int{x^n} \, dx = \frac{x^{n+1} }{n+1} +c$

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]

$\int\limits^6_^-6} (x^{5} + 8 x^{4} + 2 x^{2} + 5 x + 15) )dx$

<em>On integration , we get</em>

= $(\frac{x^{6} }{6} + \frac{8 x^{5} }{5} + 2 \frac{x^{3} }{3} +\frac{5 x^{2} }{2} + 15 x)^{6} _{-6}$

$F(x) = \int\limits^a_b {f(x)} \, dx = F(b) -F(a)$

= $(\frac{6^{6} }{6} + \frac{8 6^{5} }{5} + 2 \frac{6^{3} }{3} +\frac{5 6^{2} }{2} + 15X 6) - ((\frac{(-6)^{6} }{6} + \frac{8 (-6)^{5} }{5} + 2 \frac{(-6)^{3} }{3} +\frac{5 (-6)^{2} }{2} + 15 (-6))$

After simplification and cancellation we get

= $\frac{2 X 8 X (6)^{5} }{5} + \frac{2 X 2 X (6)^3}{3} + 2 X 15 X 6$

on calculation , we get

= $\frac{124,416}{5} + \frac{864}{3} + 180$

On L.C.M 15

= $\frac{124,416 X 3 + 864 X 5 + 180 X 15}{15}$

= 25 351.2 units²

<u><em>Conclusion</em></u>:-

<em>The area of the region between the graph of the given function and the x-axis = 25,351 units²</em>

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2 years ago