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guajiro [1.7K]
3 years ago
10

A store sells a package of comic books with a poster. A poster and three comic books cost $10.50. A poster and 12 comics cost $1

9.50. Write a linear function rule that models the cost Y of a package containing any number of X comics. Suppose another store sells a similar package, modeled by a linear function rule with initial value $8.75. Use y = mx + b
Mathematics
1 answer:
NikAS [45]3 years ago
4 0

answer:

there you goooo

Explanation

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Simplify the expression: z2 – z(z + 3) + 3z a. 2z2 + 6z b. z + 3 c. 6z d. 0
Mama L [17]
This stuff is way to hard
5 0
3 years ago
How to solve this ?<br>​
beks73 [17]

Answer:

1350

Step-by-step explanation:

use binomial general term expression for (r+1)th term

get the power of x in terms of r.

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5 0
1 year ago
One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst
grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...

(a)

Compute the probability of <em>X</em> = 0 as follows:

P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498

Thus, the  probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of <em>X</em> > 5  as follows:

P (X > 5) = 1 - P (X ≤ 5)

              =1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of <em>X</em> ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

             =\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0
3 years ago
Someone please help me I need help
pochemuha
Find how many mm are in 25m (by multiplying 25 by 1000) and convert that answer to km. 
3 0
3 years ago
Will give brainliest answer
makvit [3.9K]

Answer:

28 . 26cm

Step-by-step explanation:

First find the radius

d = 2r \\ 9 = 2r \\  \frac{9}{2}  =  \frac{2r}{2}  \\ r = 4.5

Now find the CIRCUMFERENCE

c = 2\pi \: r \\  = 2 \times \pi \times 4.5 \\  = 9 \times \pi \\  = 28.26

5 0
3 years ago
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