All categories

2 years ago

26. A single card is drawn from a deck. Find the probability that the card is either red or a face card.

Mathematics

1 answer:

Lana71 [14]2 years ago

7 0

Answer:

8/13

Step-by-step explanation:

There are total 26 red cards in the deck of 52 cards.

There are also 12 face cards in the deck ( which is 6 of them are red, while 6 of them are black)

Since we only need either one of them, 26 + 6 = 32

The probability of drawn a card that either red or a face card is 32/52

The simplest form of this fraction is 8/13.

You might be interested in

QUICK HELP!!! PLEASE!! 20 POINTS!

NikAS [45]

Answer:

X/-1 + Y/2 = 1

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Using Cavalieri's Principle, what is the general formula for the volume of a cone?

Anit [1.1K]

Answer:

The volume for a cone and pyramid are the same, V = 1/3 Bh where B is the area of the base.

Step-by-step explanation:

So even though the base is a different shape, as long as the areas and heights are the same, they will have the same volume.

4 0

3 years ago

Which graph represents this system? <br> 2x-5y=-5<br> Y=2/5x+1

zhenek [66]

both equations line up

7 0

3 years ago

Read 2 more answers

Two garden plots are to have the same area. One is square and one is rectangular. The rectangular plot is 4 meters wide and 16 m

AlexFokin [52]

side = 8 m

the area of the rectangle = length × width = 16 × 4 = 64 m²

the area of the square = s² ( where s is the length of a side ), hence

s² = 64 ( since areas are equal )

take the square root of both sides

s = √64 = 8 m

7 0

3 years ago

Evaluate the series

dalvyx [7]

Answer:

the value of the series;

$\sum_{k=1}^{6}(25-k^2) = 59$

C) 59

Step-by-step explanation:

Recall that;

$\sum_{1}^{n}a_n = a_1+a_2+...+a_n\\$

Therefore, we can evaluate the series;

$\sum_{k=1}^{6}(25-k^2)$

by summing the values of the series within that interval.

the values of the series are evaluated by substituting the corresponding values of k into the equation.

$\sum_{k=1}^{6}(25-k^2) =(25-1^2)+(25-2^2)+(25-3^2)+(25-4^2)+(25-5^2)+(25-6^2)\\\sum_{k=1}^{6}(25-k^2) =(25-1)+(25-4)+(25-9)+(25-16)+(25-25)+(25-36)\\\sum_{k=1}^{6}(25-k^2) =24+21+16+9+0+(-11)\\\sum_{k=1}^{6}(25-k^2) = 59\\$

So, the value of the series;

$\sum_{k=1}^{6}(25-k^2) = 59$

6 0

3 years ago