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OverLord2011 [107]
3 years ago
13

Kashi draws a random card from a standard 52-card deck. What is the probability he draws either a club OR a King?

Mathematics
1 answer:
Aneli [31]3 years ago
6 0

Answer:

17/52

Step-by-step explanation:

1/4 of 52 cards are Clubs

4 of 52 are Kings (or 1/13)

P(Clubs OR Kings) = 1/4 + 1/13

= 13/52 + 4/52

= 17/52

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sveta [45]

Answer:

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7 0
4 years ago
Read 2 more answers
prove that if f is integrable on [a,b] and c is an element of [a,b], then changing the value of f at c does not change the fact
Neko [114]

Answer with Step-by-step explanation:

We are given that if f is integrable  on [a,b].

c is an element which lie in the interval [a,b]

We have to prove that when we change the value of f at c then the value of f does not change on interval [a,b].

We know that  limit property of an  integral

\int_{a}^{b}f dt=\int_{a}^{c}fdt+\int_{c}^{b} fdt

\int_{a}^{b} fdt=f(b)-f(a)....(Equation I)

Using above property of integral then we get

\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b} fdt......(Equation II)

Substitute equation I and equation II are equal

Then we get

\int_{a}^{b}fdt= f(c)-f(a)+{f(b)-f(c)}

\int_{a}^{b}fdt=f(c)-f(a)+f(b)-f(c)=f(b)-f(a)

\int_{a}^{c}fdt+\int_{c}^{b}fdt=f(b)-f(a)

Therefore, \int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b}fdt.

Hence, the value of function does not change after changing the value of function at c.

6 0
3 years ago
Here is a picture of an older version of the flag of Great Britain. There is a rigid transformation that takes Triangle 1 to Tri
Ainat [17]

Answer:

  The side lengths are the same.

Step-by-step explanation:

A rigid transformation does not change any dimensions. The side lengths are the same.

8 0
4 years ago
Read 2 more answers
What’s the value of X ???!!!!
Alexus [3.1K]

Answer:

Step-by-step explanation:

So since triangle jkl and wyz are similar angles j and w must equal each other, angles k and y must equal each other, and angles l and z are equal. So to find x just set 4x-13 equal to 71

4x-13=71

add 13 to both sides to cancel the -13 from the left

4x=84

and divide both sides by 4 to cancel the 4 on the left

x=21

7 0
3 years ago
What is the area of the shaded region below?<br> 7 m<br> 30 m<br> 7 m<br> 21 m
Rama09 [41]

Answer:

322.28

A

Step-by-step explanation:

Area of the rectangle

L = 30

W = 21

Area = 30 * 21

Area = 630

Area of the circles

Radius = 7 m

Area = pi r^2

Area = 3.14 * 7^2

Area = 153.86 m^2

Area of both circles = 307.72

Area of the Shaded Region

Area of shaded region = area of the rectangle - area of 2 circles.

Area of Shaded Region = 630 - 307.72

Area of Shaded Region = 322.28

5 0
3 years ago
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