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3 years ago

Find the 9th term of the sequence which has a first term of 1 and has a common ratio of 1.5

Mathematics

1 answer:

Scrat [10]3 years ago

7 0

A₁ = 1; r=1.5

a₉ = a₁ * r⁸ = 1 * 1.5⁸ ≈ 25.63

Answer: c. 25.63

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Mrs anglin is driving her car at a constant speed of 90 miles in 2 hours select three statements that are true based on her spee

Ira Lisetskai [31]

Answer:

She traveled 45 miles in 1 hour
She traveled 270 miles in 6 hours
She traveled 360 miles in 8 hours

Step-by-step explanation:

If Mrs. Anglin traveled at a speed of 90 miles in 2 hours, it means that her speed per hour is:

= 90/2

= 45 mph

If she is travelling at 45mph then in 1 hoour she would have traveled 45 miles.

In 6 hours she would have traveled:

= 45 * 6

= 270 miles

In 8 hours she would have traveled:

= 45 * 8

= 360 miles

6 0

3 years ago

Ingrid Nilsen purchased a $4,000 hydroponic system for her garden. The down payment is 20%, and the

Cerrena [4.2K]

The purchase of the hydroponic system by Ingrid Nilsen for $4,000 with a 20% down payment at 10% APR for 36 months, and the related calculations, are as follows:

a) down payment = $800.00 ($4,000 x 20%)

b) amount financed = $3,200 ($4,000 - $800)

c) monthly payment = $103.25

d) total amount repaid = $4,517

e) finance charge = $517

<h3>How is periodic payment computed?</h3>

The amount of the periodic payment, finance charge, and total repayment can be computed using an online finance calculator as follows:

<h3>Data and Calculations:</h3>

Auto Price = $4,000

Loan Term = 36 months

Interest Rate = 10%

Down Payment = $800 ($4,000 x 20%)

Monthly Payment = $103.25

Total Loan Amount $3,200.00

Upfront Payment = $800.00 ($4,000 x 20%)

Total of 36 Loan Payments = $3,717. ($103.25 x 36)

Total Loan Interest = $517

Total Cost (price, interest, tax, fees) = $4,517

Thus, Ingrid Nilsen repaid a total of $4,517 with $517 in interest (finance charge).

Learn more about periodic payments at brainly.com/question/13098072

#SPj1

5 0

2 years ago

The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

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3 years ago

Find the length of side a<br>will mark brainly

Mekhanik [1.2K]

Answer:

Step-by-step explanation:use the cube to count(cube is in the corner.)

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The area of a round table in Kim’s living room is 201 square inches. What is the circumference of the round table, to the neares

svetoff [14.1K]

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