All categories

2 years ago

HELPPPPPPPPP Write the quadratic expression to represent the pattern to the right

Mathematics

1 answer:

Lerok [7]2 years ago

4 0

Answer:

n^2 + 2n - 1

Step-by-step explanation:

Figure 1

2*2 - 2

Figure 2

3*3 - 2

Figure 3

4*4 - 2

Equation for nth term

(n + 1)^2 - 2 =
n^2 + 2n + 1 - 2 =
n^2 + 2n - 1

You might be interested in

Working alone, Ryan can dig a 10 ft by 10 ft hole in 5 hours Castel can dig the same hole in 6 hours how long would it take them

Fed [463]

Answer:

The time taken by both boys to complete work = 2.73 hours

Step-by-step explanation:

Time taken by rayan to dig hole $T_{1}$ = 5 hours

Time taken by castel to dig hole $T_{2}$ = 6 hours

Now the time taken by both persons working together

$T = \frac{T_{1} T_{2} }{T_{1} + T_{2}}$

Put all the values in above formula we get

$T = \frac{ 30 }{6 + 5}$

$T = \frac{30}{11}$

T = 2.73 hours

Therefore the time taken by both boys to complete work = 2.73 hours

6 0

2 years ago

Richard was thinking of a number. Richard adds 9.5 to the number, then doubles the result to get an answer of 59.4. Form an equa

Jet001 [13]

The answer is 49.9 because he thought of a number and added 9.5 and the result was 59.4 then you should do 59.4-9.5 and the answer is 49.9

7 0

2 years ago

Read 2 more answers

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

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

7 0

2 years ago

A company tests the life of its rechargeable batteries over six months of use. They test how long a battery will last after a fu

jonny [76]

Answer:

18.6 months

Step-by-step explanation:

Given that :

Best fit line from scatterplot :

y=-12.05x +224.26

x = Number of month

y = charge on battery

Number of months a typical battery uses before being dead completely :

When battery is dead completely ; charge =0, y = 0

y = -12.05x + 224.26

0 = - 12.05x + 224.26

12.05x = 224.26

x = 224.26 / 12.05

x = 18.610788

Hence, 18.6 months before battery is completely dead.

4 0

2 years ago

Solve: 2x + 6 = 10......

Luda [366]

Answer:

x=2

Step-by-step explanation:

2x+6=10

2x=10-6

2x=4

x=4/2

x=2

6 0

3 years ago

Read 2 more answers