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2 years ago

The nth term of a sequence is 20 - n2

Mathematics

1 answer:

maks197457 [2]2 years ago

5 0

Answer:

Fifth term

Step-by-step explanation:

The third term is:

20 − (3)² = 11

If the term is negative:

20 − n² < 0

n² > 20

n > √20

n > 4.47

The first term to have a negative value is the fifth term.

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Coordinates, gradient and tangent work (see image).

Ivenika [448]

Answer:

a) $P(x,2x^2-5),\ Q(x+h,2(x+h)^2-5)$

b) $4x+2h$

c) $4x$

Step-by-step explanation:

Given the curve

$y=2x^2-5$

a) If the x-coordinate of P is $x$ , then the y-coordinate is $2x^2-5,$ so point P has coordinates $(x,2x^2-5)$

If the x-coordinate of Q is $x+h$ , then the y-coordinate is $2(x+h)^2-5$ so point Q has coordinates $(x+h,2(x+h)^2-5)$

b) The gradient of the secant RQ is

$\dfrac{y_Q-y_P}{x_Q-x_P}\\ \\=\dfrac{(2(x+h)^2-5)-(2x^2-5)}{(x+h)-x}\\ \\=\dfrac{2(x+h)^2-5-x^2+5}{x+h-x}\\ \\=\dfrac{2(x+h)^2-2x^2}{h}\\ \\=\dfrac{2x^2+4xh+2h^2-2x^2}{h}\\ \\=\dfrac{4xh+2h^2}{h}\\ \\=4x+2h$

c) If $h\rightarrow 0,$ then the gradient $4x+2h\rightarrow 4x$

5 0

2 years ago

Of the 500 sample households in the previous exercise, 7 had three or more large-screen TVs. (a) The percentage of households in

likoan [24]

Answer:

a) The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

$\hat p=\frac{7}{500}=0.014$

And that represent the 1.4%.

b) And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=7 represent the households with three or more large-screen TVs

$\hat p=\frac{7}{500}=0.014$ estimated proportion of households with three or more large-screen TVs

$\alpha=0.05$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

p= population proportion of households with three or more large-screen TVs

Part a

The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

$\hat p=\frac{7}{500}=0.014$

And that represent the 1.4%.

Part b

Yes is possible. We hav that $np>10$ and $n(1-p)>10$ so we have the assumption of normality to find the interval.

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.