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Gala2k [10]
2 years ago
13

Can someone please help me to find the remainder when f(x) is divided by (x^2-4)?

Mathematics
1 answer:
yuradex [85]2 years ago
8 0

Answer:

Step-by-step explanation: Given : f(x)=x

4

−3x

2

+4

f(2)=(2)

4

−3(2)

2

+4

=16−12+4=8

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How many three-digit positive integers have three different digits and at least one prime digit?
Blababa [14]

The number of three-digit positive integers that have three different digits and at least one prime digit are 7960.

The only two components in prime numbers are 1 and the number itself.

Any whole number greater than one is a prime number.

It has exactly two factors—1 and the actual number.

There is just one 2-digit even prime number.

Every pair of prime numbers is always a co-prime.

The product of prime numbers can be used to represent any number.

Three-digit positive integers that have three different digits and at least one prime digit = 3!*4!*10*9 = 7960

Learn more about prime numbers here:

brainly.com/question/145452

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6 0
2 years ago
A family purchased tickets to a museum and spent a total of $45.00. The family purchased 5 tickets. There was $1.50 processing f
Arisa [49]
7.50$ for each ticket without the processing fee
4 0
2 years ago
An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed
masya89 [10]

Answer:

P(X\leq 1) = 0.331

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

Determine the probability of renting no more than 1 apartment

A Poisson distribution is given as;

P(X = x) = \frac{y^xe^{-y}}{x!}

Where y represents λ (average)

y = 2.3

<em>Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment</em>

<em />

Using probability notations;

P(X\leq 1) = P(X=0) + P(X =1)

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}

P(X = 0) = \frac{1 * e^{-2.3}}{1}

P(X = 0) = e^{-2.3}

P(X = 0) = 0.10025884372

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}

P(X = 1) = \frac{2.3 * e^{-2.3}}{1}

P(X = 1) =2.3 * e^{-2.3}

P(X = 1) = 2.3 * 0.10025884372

P(X = 1) = 0.23059534055

P(X\leq 1) = P(X=0) + P(X =1)

P(X\leq 1) = 0.10025884372 + 0.23059534055

P(X\leq 1) = 0.33085418427

P(X\leq 1) = 0.331

Hence, the required probability is 0.331

6 0
2 years ago
Landon scored an 85% on his test. If he got 34 problems correct, how many were on the test?
Sergeu [11.5K]

Answer:

There are 40 problems on the test

Step-by-step explanation:

Let p = the number of problems on the test

The number of problems on the test times the percent correct is the number correct

p * 85% = 34

p * .85 = 34

Divide each side by .85

p * .85/.85 = 34/.85

p = 40

There are 40 problems on the test

6 0
2 years ago
Why can you multiply or divide both terms of a ratio by the same number without changing the value of the ratio
o-na [289]
The ratio is rational and by multiplying or dividing the numbers included in the ratio by the same number, it's just like taking a fraction and taking it out of simplest form, it doesn't change the value of the ratio or the fraction in simplest form, it just changes the numbers

6 0
3 years ago
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