What is the sum of the 11th square number and the 7th cube number

1 answer:

7 0

$\: \: \: \: \: \: \: \: \: \: \: \: \: {11}^{2} + {7}^{3} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = 121 + 343 \\ = 464$

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Peter and Henry went to the movies Peter bought the two Movie tickets for 7 dollars each. How much did he spend for the movie ti

ivann1987 [24]

Answer:

tickets: $14

popcorn: $10

total: $24

Step-by-step explanation:

2 movie tickets at $7 each is 2*7 which is $14

2 buckets of popcorn at $5 each is 2*5 which is $10

adding these together we get a total spending of $24

3 0

2 years ago

[6+4=10 points] Problem 2. Suppose that there are k people in a party with the following PMF: • k = 5 with probability 1 4 • k =

kirza4 [7]

Answer:

1). 0.903547

2). 0.275617

Step-by-step explanation:

It is given :

K people in a party with the following :

i). k = 5 with the probability of $\frac{1}{4}$

ii). k = 10 with the probability of $\frac{1}{4}$

iii). k = 10 with the probability $\frac{1}{2}$

So the probability of at least two person out of the 'n' born people in same month is = 1 - P (none of the n born in the same month)

= 1 - P (choosing the n different months out of 365 days) = $1-\frac{_{n}^{12}\textrm{P}}{12^2}$

1). Hence P(at least 2 born in the same month)=P(k=5 and at least 2 born in the same month)+P(k=10 and at least 2 born in the same month)+P(k=15 and at least 2 born in the same month)

= $\frac{1}{4}\times (1-\frac{_{5}^{12}\textrm{P}}{12^5})+\frac{1}{4}\times (1-\frac{_{10}^{12}\textrm{P}}{12^{10}})+\frac{1}{2}\times (1-\frac{_{15}^{12}\textrm{P}}{12^{15}})$