A standard deck has 52 cards.
A standard deck has 4 jacks.
A standard deck has 13 clubs.
From this, we can derive the following:
The probability of drawing a jack is 4/52 or 1/13
The probability of drawing a club is 13/52 or 1/4
But since the problem asks for drawing jack or club, therefore we should add the 2 probabilities, making 17/52. This is not the final answer yet. We know that there is a jack of clubs, therefore we need to subtract 1 from the probabilities since jack of clubs were considered in the 2 categories of probability.
With that being said, the probability of drawing a club or a jack is 16/52 or 4/13
Bonus Question:
The first thing you need to do here is find the probability of each scenario. First let's do what is given, the probability of drawing 2 aces. Since there are 4 aces in a deck of 52, we can easily say that the probability of drawing an ace is 4/52. However for our second draw, the probability of drawing a different ace is 3/51. This is so since we already drew a card that is an ace, hence we need to subtract one from the total aces (4-1) and from the total cards in the deck (52-1). In getting the probability of drawing two aces, we need to multiply the said probabilities: 4/52 and 3/51, resulting to 1/221.
For the second scenario, the drawing of 2 red cards, we just use the same concept but in this, we are already considering the 2 red cards in the first scenario, therefore the chance of drawing a red on our first draw is 24/52. For our second, we just need to subtract one card, therefore 23/51. Multiply these two and we will get 46/221.
Now, the problem asks for the chance of drawing either 2 reds or 2 aces, therefore we add the probabilities of the 2 scenarios:
46/222 + 1/221 = 47/221
Summary:
First Scenario:
4/52 + 3/51 = 1/221
Second Scenario:
24/52 + 23/51 = 46/221
Chances of drawing 2 red cards or 2 aces:
1/221 + 46/221 = 47/221
For every c substitute 4 and for every d substitute -2
c=4
d=-2
6c + 5d - 4c - 3d + 3c - 6d
= 6(4)+ 5(-2)- 4(4)- 3(-2)+ 3(4)- 6(-2)
=24+(-10)-16-(-6)+12-(-12)
=24-10-16+6+12+12
=28
The long side of an isosolies triangle
Answer:
alright all you need to do is Use the slope-intercept form to find the slope and y-intercept
So the Answer to the question is
Slope: 1/5
Y-Intercept: 3 Hope this helps :)
Step-by-step explanation:
Answer:
The ball will be at 100 feet at time x=0.5 sec and at time x=7.5 sec
Step-by-step explanation:
<u><em>The correct question is</em></u>
A ball is launched upward from a height 40 feet above ground level. the ball's height at t seconds is given by h(t)=-16t^2+128t+40 At what time(s) will the ball be at 100 feet?
we have
so
For h(t)=100 ft
substitute in the equation and solve for x
The formula to solve a quadratic equation of the form
is equal to
in this problem we have
so
substitute in the formula
therefore
The ball will be at 100 feet at time x=0.5 sec and at time x=7.5 sec
