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3 years ago

You are dealth one card from a 52-card deck. find the probability that you are dealt a two or a black card

1 answer:

6 0

The probability of getting a 2 or a getting a black card, find individual probabilities;

A standard deck has 52 cards.

There are 4 2's in a normal deck; probability of getting it is 4/25

The probability of getting a black card is; 26/52 since half the deck is red and black.

Now add up the probabilities since it says "or"

(4/52)+(26/52)=30/52 probability of the card that you were dealt being a two or a black card.

Hope I helped :)

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Ten thousand raffle tickets are sold for $1 each. One first prize of $2000, 4 second prizes of $700 each, and 8 third prizes of

Nina [5.8K]

Answer:

72 cents.

Step-by-step explanation:

The expected winnings is the amount times the probability that you will get that amount.

2,000 * (1/10,000) = 2,000 / 10,000 = 2 / 10 = 0.2.

700 * (4 / 10,000) = 2,800 / 10,000 = 28 / 100 = 0.28.

300 * (8 / 10,000) = 2,400 / 10,000 = 24 / 100 = 0.24.

0.2 + 0.28 + 0.24 = 0.72.

Hope this helps!

4 0

2 years ago

15c² - 5c. Factorise completely

KATRIN_1 [288]

Looking at the problem, what we must do to complete this question is to completely factor the expression that was provided. The expression that was provided is $15c^2 - 5c$ .

The first step that we must do is to take a look at the expression and see what the two pieces of the expression have in common. We can see that both $15c^2$ and $-5c$ have the number 5 and the variable c associated with them so we can factor out those two.

<u>Factor out 5c</u>

$15c^2 - 5c$

$5c(\frac{15c^2}{5c}) - 5c(\frac{5c}{5c})$

$5c(3c) - 5c(1)$

$5c(3c - 1)$

Now we have completely factored out the expression that was provided in the problem statement and we came to final answer of $5c(3c - 1)$ .

7 0

1 year ago

Read 2 more answers

Estimate the sum of 487 and 310

Ronch [10]

487 + 310 ≈ 500 + 300 ≈ 800

3 0

2 years ago

Read 2 more answers

When an electric current passes through two resistors with resistance r1 and r2, connected in parallel, the combined resistance,

kondaur [170]

Answer:

The combined resistance of a circuit consisting of two resistors in parallel is given by:

$\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}$

where

R is the combined resistance

$r_1, r_2$ are the two resistors

We can re-write the expression as follows:

$\frac{1}{R}=\frac{r_1+r_2}{r_1r_2}$

$R=\frac{r_1 r_2}{r_1+r_2}$

In order to see if the function is increasing in r1, we calculate the derivative with respect to r1: if the derivative if > 0, then the function is increasing.

The derivative of R with respect to r1 is:

$\frac{dR}{dr_1}=\frac{r_2(r_1+r_2)-1(r_1r_2)}{(r_1+r_2)^2}=\frac{r_2^2}{(r_1+r_2)^2}$

We notice that the derivative is a fraction of two squared terms: therefore, both factors are positive, so the derivative is always positive, and this means that R is an increasing function of r1.

To solve this part, we use again the expression for R written in part a:

$R=\frac{r_1 r_2}{r_1+r_2}$

We start by noticing that there is a limit on the allowed values for r1: in fact, r1 must be strictly positive,

$r_1>0$