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2 years ago

I also need help with this.

Mathematics

1 answer:

Brrunno [24]2 years ago

3 0

The answer is 1.

If there is no number in front of a multiplicity, the number will always be a 1.
6 x 1(9 - 4) = 30
(1 x 9) - (1 x 4) = 5
6 x 5 = 30

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Consider this sequence

pochemuha

Hello,

$u_{1} =1*(8-7)=1*(2*4-(6+1))\\
 u_{2} =2*(10-7)=2*(2*5-(6+2))=2*(2*(4+1)-(6+2))\\
 u_{3} =3*(12-9)=3*(2*6-(6+3))=3*(2*(4+2)-(6+3))\\
 u_{4} =4*(14-10)=4*(2*7-(6+4))=4*(2*(4+3)-(6+4))\\

$

$u_{10}=10*(2*(4+10)-(6+10))\\
=10*(28-16)
$

$...\\\\
u_{n} =n*(2*(n+n-1)-(6+n))\\
=n*(4n-2-6-n)=n(3n-8)\\

$

5 0

3 years ago

Use induction to prove: For every integer n > 1, the number n5 - n is a multiple of 5.

nignag [31]

Answer:

we need to prove : for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

1) check divisibility for n=1, $f(1)=(1)^{5}-1=0$ (divisible)

2) Assume that $f(k)$ is divisible by 5, $f(k)=(k)^{5}-k$

3) Induction,

$f(k+1)=(k+1)^{5}-(k+1)$

$=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1$

$=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k$

Now, $f(k+1)-f(k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k$

$f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k$

Take out the common factor,

$f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)$ (divisible by 5)

add both the sides by f(k)

$f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)$

We have proved that difference between $f(k+1)$ and $f(k)$ is divisible by 5.

so, our assumption in step 2 is correct.

Since $f(k)$ is divisible by 5, then $f(k+1)$ must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

3 0

3 years ago

The number of undergraduates at Johns Hopkins University is approximately 2000, while the number at Ohio State University is app

Likurg_2 [28]

The remaining part of Question:

4) what can we conclude about sampling variability in the sample proportion calculated in the sample at John Hopkins as compared to that calculated in the sample at Ohio State.

5) The number of undergraduates at Johns Hopkins is approximately 2000 while the number at Ohio State is approximately 40000, suppose instead that at both schools, a simple random sample of about 3% of the undergraduates Will be taken.

Answer:

4) The sample proportion from Johns Hopkins will have about the same sampling variability as that from Ohio State

5) The sample proportion from John Hopkins will have more sampling variability than that from Ohio State

Step-by-step explanation:

Note: The sampling variability in the sample proportion decreases with increase in the sample size.

4) since the sample size at both Johns Hopkins and Ohio State is the same (i.e. n = 50), the sample variability of the sample proportion will be the same for both cases.

5) 3% of the population are selected for the observation in both cases.

At Johns Hopkins, sample size, n = 3% * 2000

n = 60

At Ohio State, sample size, n = 3% * 40000

n = 1200

Since sampling variability in the sample proportion decreases with increase in the sample size, the sampling variability in sample proportion will be higher at Johns Hopkins than at Ohio State.

5 0

3 years ago

What is 124 decreased by 67%

inysia [295]

Answer:

40.92

Step-by-step explanation:

hope this helps

5 0

3 years ago

Read 2 more answers

Which student evaluated the power correctly?

Colt1911 [192]

Answer:jerry

Step-by-step explanation:

Just plug in the calculator and see which

6 0

3 years ago