Determine the 20th derivative of y = sin(2x).

Mathematics

1 answer:

snow_lady [41]2 years ago

4 0

1st derivative = 2 cos (2x) 2nd derivative = -4sin(2x) 3rd = -8 cos(2x) . . . 20th derivative= 2^20 sin(2x)

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An element with mass 970 grams decays by 27.7% per minute. How much of the element is remaining after 13 minutes, to the nearest

11Alexandr11 [23.1K]

Every 1 minute, you multiply the mass by 27.7% or 0.277

after 1 minute, you multiply once
after 2 minutes, you multiply twice
3, thrice
etc
so after 13 minutes, you multiply 13 times or 0.277^13

so what is 970g *0.277^13 =?

7 0

2 years ago

Read 2 more answers

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.