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3 years ago

How many different ways can you add two odd numbers to get a sum of 20 and add examples

1 answer:

7 0

We can add only odd numbers. Let's assume that we're dealing ONLY with positive numbers. Which two odd numbers add to 20? 1+19 qualifies.
So do 3+17, 5+15, 7+13, 9+11.

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Find all integers $n$ for which $\frac{n^2+n+1}{n-1}$ is an integer.

Grace [21]

$\dfrac{n^2+n+1}{n-1}=\dfrac{(n-1)^2+3(n-1)+3}{n-1}=n-2+\dfrac3{n-1}$

The remainder term $\dfrac3{n-1}$ will never vanish, and will always be rational as long as the denominator is not 1, 3, or -3.

$n-1=1\implies n=2$
$n-1=3\implies n=4$
$n-1=-3\implies n=-2$

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3 years ago

Chickens and pigs, 100 heads and 270 legs. How many pigs are there?

Annette [7]

Oop-
I guess 370 if not then

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3 years ago

Define what a prime factor decomposition

nadya68 [22]

It means to factorise a number, and divide it by the first possible prime number.

5 0

3 years ago

Assume that random guesses are made for seven multiple choice questions on an SAT test, so that there are n=7 trials, each with

Vedmedyk [2.9K]

Answer:

P(X < 4) = 0.5

Step-by-step explanation:

For each question, there are only two possible outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent of any other question. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this question, we have that:

$n = 7, p = 0.5$

Find the probability that the number of correct answers is fewer than 4:

This is

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{7,0}.(0.5)^{0}.(0.5)^{7} = 0.0078$

$P(X = 1) = C_{7,1}.(0.5)^{1}.(0.5)^{6} = 0.0547$

$P(X = 2) = C_{7,2}.(0.5)^{2}.(0.5)^{5} = 0.1641$

$P(X = 3) = C_{7,3}.(0.5)^{3}.(0.5)^{4} = 0.2734$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0078 + 0.0547 + 0.1641 + 0.2734 = 0.5$

P(X < 4) = 0.5

6 0

3 years ago

The following list shows the number of video games sold per day at a video store.

TEA [102]

Here is what your frequency/cumulative frequency table would look like. The cumulative frequency is where you add all the frequencies from before it to get a sum after each interval.

<u> Frequency </u> <u>Cumulative Frequency</u>
0-25 8 8
26-50 0 8
51-75 4 12
76-100 2 14

4 0

3 years ago