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2 years ago

What is the fiftieth term in the sequence 5, 7, 9, 11, 13 …?

Mathematics

1 answer:

Komok [63]2 years ago

5 0

The 15th term in the given A.P. sequence is a₁₅ = 33.

According to the statement

we have given that the A.P. Series with the a = 5 and the d is 2.

And we have to find the 15th term of the sequence.

So, for this purpose we know that the

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

And the formula is a

an = a + (n-1)d

After substitute the values in it the equation become

an = 5 + (15-1)2

a₁₅ = 5 + 28

Now the 15th term is a₁₅ = 33.

So, The 15th term in the given A.P. sequence is a₁₅ = 33.

Learn more about arithmetic progression here

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Problem #4:

madreJ [45]

Answer:

Step-by-step explanation:

Prepare closing entireties for the concert hall bond fund

6 0

3 years ago

Read 2 more answers

An n × n matrix B has characteristic polynomial p(λ) = −λ(λ − 3) 3 (λ − 2) 2 (λ + 1). Which of the following statements is false

asambeis [7]

Answer:

Only d) is false.

Step-by-step explanation:

Let $p=p(\lambda)=\lambda(\lambda-3)^3 (\lambda-2)^2 (\lambda+1)$ be the characteristic polynomial of B.

a) We use the rank-nullity theorem. First, note that 0 is an eigenvalue of algebraic multiplicity 1. The null space of B is equal to the eigenspace generated by 0. The dimension of this space is the geometric multiplicity of 0, which can't exceed the algebraic multiplicity. Then Nul(B)≤1. It can't happen that Nul(B)=0, because eigenspaces have positive dimension, therfore Nul(B)=1 and by the rank-nullity theorem, rank(B)=7-nul(B)=6 (B has size 7, see part e)

b) Remember that $p(\lambda)=\det(B-\lambda I)$ . 0 is a root of p, so we have that $p(0)=\det(B-0 I)=\det B=0$ .

c) The matrix T must be a nxn matrix so that the product BTB is well defined. Therefore det(T) is defined and by part c) we have that det(BTB)=det(B)det(T)det(B)=0.

d) det(B)=0 by part c) so B is not invertible.

e) The degree of the characteristic polynomial p is equal to the size of the matrix B. Summing the multiplicities of each root, p has degree 7, therefore the size of B is n=7.

8 0

3 years ago

PLZ HELP!!! Use limits to evaluate the integral.

Marrrta [24]

Split up the interval [0, 2] into n equally spaced subintervals:

$\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]$

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

$r_i=\dfrac{2i}n$

where $1\le i\le n$ . Each interval has length $\Delta x_i=\frac{2-0}n=\frac2n$ .

At these sampling points, the function takes on values of

$f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}$

We approximate the integral with the Riemann sum:

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3$

Recall that

$\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4$

so that the sum reduces to

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}$

Take the limit as n approaches infinity, and the Riemann sum converges to the value of the integral:

$\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}$

Just to check:

$\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28$

4 0

3 years ago

If RSTU is a rhombus, find the measure of angle RUT.

Schach [20]

Finding x:

We know that the diagonals of a rhombus bisect its angles

So, since US is a diagonal of the given rhombus:

∠RUS = ∠TUS

10x - 23 = 3x + 19 [replacing the given values of the angles]

7x - 23 = 19 [subtracting 3x from both sides]

7x = 42 [adding 23 on both sides]

x = 6 [dividing both sides by 7]

Finding ∠RUT:

We can see that:

∠RUT = ∠RUS + ∠TUS

Since we are given the values of ∠RUS and ∠TUS:

∠RUT = (10x - 23) + (3x + 19)

∠RUT = 13x - 4

We know that x = 6:

∠RUT = 13(6)- 4

∠RUT = 74°

8 0

3 years ago

The mean preparation fee H&R Block charged retail customers in 2012 was $183 (The Wall Street Journal, March 7, 2012). Use t

astraxan [27]

Answer:

a)0.6192

b)0.7422

c)0.8904

d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

Step-by-step explanation:

Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Then

z(p)= $\frac{ME*\sqrt{N}}{s }$ where

Me is the margin of error from the mean

s is the standard deviation of the population

N is the sample size

z(p)= $\frac{8*\sqrt{30}}{50 }$ ≈ 0.8764

by looking z-table corresponding p value is 1-0.3808=0.6192

z(p)= $\frac{8*\sqrt{50}}{50 }$ ≈ 1.1314

by looking z-table corresponding p value is 1-0.2578=0.7422

z(p)= $\frac{8*\sqrt{100}}{50 }$ ≈ 1.6

by looking z-table corresponding p value is 1-0.1096=0.8904

Minimum required sample size for 0.95 probability is

N≥ $(\frac{z*s}{ME} )^2$ where

N is the sample size

z is the corresponding z-score in 95% probability (1.96)

s is the standard deviation (50)

ME is the margin of error (8)

then N≥ $(\frac{1.96*50}{8} )^2$ ≈150.6

Thus at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

7 0

3 years ago