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UNO [17]
3 years ago
9

Red queens and black jacks are removed from a pack of 52 cards. A card is drawn at random from the remaining cards, after reshuf

fling them. Find the probability that the drawn card is (i) king (ii) of red colour (iii) a face card (iv) a queen.
Mathematics
1 answer:
FinnZ [79.3K]3 years ago
3 0

Answer:

option 2

Step-by-step explanation:

just did it

mark brainlyist

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The following distribution shows the frequency of daily revenue for an Italian restaurant in Wilmington, Delaware. Revenue Frequ
sergejj [24]

Answer:

The probability that a randomly selected day will generate more than $6,000 in revenue is 0.325.

Step-by-step explanation:

The frequency distribution of daily revenue for an Italian restaurant in Wilmington, Delaware is:

     <u>Revenue</u>      <u>Frequency</u>

   Under $2,000        18

$2,000 -  $4,000      10

$4,000 -  $6,000     26

$6,000 -  $8,000      14

$8,000 - $10,000       8

   Over $10,000           4

          Total                 80

The probability of an event <em>E</em> is defined as:

P(E)=\frac{n(E)}{N}

Here,

n (E) = favorable outcome

N = total outcomes

The favorable outcomes of the event <em>X</em> : "more than $6,000 in revenue" is defined as:

n (X) = 14 + 8 + 4 = 26

Total number of outcomes is, <em>N</em> = 80.

Compute the probability of the event <em>X</em> is:

P(X)=\frac{n(X)}{N}=\frac{26}{80}=0.325

Thus, the probability that a randomly selected day will generate more than $6,000 in revenue is 0.325.

6 0
3 years ago
A high percentage of people who fracture or dislocate a bone see a doctor for that condition. Suppose the percentage is 99%. Con
marishachu [46]

Answer:

(i) 0.15708

(ii) 0.432488

(iii) 3

Step-by-step explanation:

Given that, 99% of people who fracture or dislocate a bone see a doctor for that condition.

There is only two chance either the person having fracture or dislocation of bone will either see the doctor or not.

As per previous data, if one person got a fracture or dislocation of bone, the chance of seeing the doctor is 0.99. Assuming this chance is the same for every individual, so the total number of people having fractured or dislocated a bone can be considered as Bernoulli's population.

Let p be the probability of success represented by the chances of not seeing a doctor by any one individual having fractured or dislocated a bone.

So, p=1-0.99=0.01

According to Bernoulli's theorem, the probability of exactly r success among the total of n randomly selected from Bernoulli's population is

P(r)=\binom{n}{r}p^r(1-p)^{n-r}\cdots(i)

(i) The total number of persons randomly selected, n=400.

The probability that exactly 5 of them did not see a doctor

So, r=5 , p=0.01

Using equation (i),

P(r=5)=\binom{400}{5}(0.01)^5(1-0.01)^{400-5}

=\frac{400!}{(400-5)!\times 5!}(0.01)^5(0.99)^{395}

=0.15708

(ii) The probability that fewer than four of them did not see a doctor

=P(r

=P(r=0)+P(r=1)+P(r=2)+P(r=3)

=\binom{400}{0}(0.01)^0(0.99)^{400}+\binom{400}{1}(0.01)^1(0.99)^{399}+\binom{400}{2}(0.01)^2(0.99)^{398}+\binom{400}{3}(0.01)^3(0.99)^{397}

=0.017951+0.072527+0.146154+0.195856

=0.432488

(iii) The expected number of people who would not see a doctor

=np

=300\times 0.01

=3

7 0
3 years ago
Marco weighs 80 pounds. His bones make up 1/5 of his body weight. How much do his bones weight?
Afina-wow [57]
80 /5 = 16

answer:

His bones weight 16 lbs
6 0
3 years ago
Read 2 more answers
Match the products of rational expressions with their simplest forms.
Alborosie
Lets simplify each one of the rational expressions using the product rule of exponents: a^n*a^m=a^{n+m} and the quotient rule of exponents: \frac{a^n}{a^m} =a^{n-m}.

1.  \frac{-3x^2}{2y^2} * \frac{y^2}{9x} =- \frac{x}{6} = \frac{-1}{6} x

2. \frac{5y^2}{10x^2} * \frac{4x^2}{y} =2y

3. \frac{-9y^2}{5x} * \frac{-10x^2}{3y} =6xy

4. \frac{14x^2}{5y} * \frac{-10y}{7x} =-4x

We can conclude that you should pair the rational expressions with their simplest form as follows:
6xy-----\ \textgreater \ \frac{-9y^2}{5x} * \frac{-10x^2}{3y}
\frac{-1}{6} x-----\ \textgreater \  \frac{-3x^2}{2y^2} * \frac{y^2}{9x}
-4x-----\ \textgreater \ \frac{14x^2}{5y} * \frac{-10y}{7x}
2y-----\ \textgreater \ \frac{5y^2}{10x^2} * \frac{4x^2}{y}
7 0
3 years ago
Find all solutions to the differential equation dy dt = 1 − y 2
klemol [59]
Try this:
\frac{dy}{dt}=1-y^2; \ =\ \textgreater \  \   \frac{dy}{y^2-1}=-dt; \ =\ \textgreater \  \ ln| \frac{y-1}{y+1}|=-2t+C.
If it is possible, modify result to form more compact.
7 0
3 years ago
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