Ask question

Ask question

All categories

2 years ago

6

In developing an interval estimate of the population mean, if the population standard deviation is unknown _____.

1 answer:

Liula [17]2 years ago

7 0

Answer:

the sample standard deviation and t distribution can be used

Step-by-step explanation:

You might be interested in

Help word problem ......

Nitella [24]

Her call would have lasted 27 minutes. Hope this helps.

4 0

4 years ago

Read 2 more answers

Please help me with this asap

Oduvanchick [21]

Answer:

The answer is 3

Step-by-step explanation:

5 0

3 years ago

Who can help me d e f thanks

12345 [234]

d)

$y = (2ax^2 + c)^2 (bx^2 - cx)^{-1}$

Product rule:

$y' = \bigg((2ax^2+c)^2\bigg)' (bx^2-cx)^{-1} + (2ax^2+c)^2 \bigg((bx^2-cx)^{-1}\bigg)'$

Chain and power rules:

$y' = 2(2ax^2+c)\bigg(2ax^2+c\bigg)' (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} \bigg(bx^2-cx\bigg)'$

Power rule:

$y' = 2(2ax^2+c)(4ax) (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} (2bx - c)$

Now simplify.

$y' = \dfrac{8ax (2ax^2+c)}{bx^2 - cx} - \dfrac{(2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

$y' = \dfrac{8ax (2ax^2+c) (bx^2 - cx) - (2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

e)

$y = \dfrac{3bx + ac}{\sqrt{ax}}$

Quotient rule:

$y' = \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{\left(\sqrt{ax}\right)^2}$

$y'= \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{ax}$

Power rule:

$y' = \dfrac{3b \sqrt{ax} - (3bx+ac) \left(-\frac12 \sqrt a \, x^{-1/2}\right)}{ax}$

Now simplify.

$y' = \dfrac{3b \sqrt a \, x^{1/2} + \frac{\sqrt a}2 (3bx+ac) x^{-1/2}}{ax}$

$y' = \dfrac{6bx + 3bx+ac}{2\sqrt a\, x^{3/2}}$

$y' = \dfrac{9bx+ac}{2\sqrt a\, x^{3/2}}$

f)

$y = \sin^2(ax+b)$

Chain rule:

$y' = 2 \sin(ax+b) \bigg(\sin(ax+b)\bigg)'$

$y' = 2 \sin(ax+b) \cos(ax+b) \bigg(ax+b\bigg)'$

$y' = 2a \sin(ax+b) \cos(ax+b)$

We can further simplify this to

$y' = a \sin(2(ax+b))$

using the double angle identity for sine.

7 0

2 years ago

When rabbits were introduced to the continent of Australia they quickly multiplied and spread across the continent since there w

Lady bird [3.3K]

Answer:

a. $y=6(1.7472)^x$

b. $y=6e^{0.558t}$

c.13.3 months

Step-by-step explanation:

a.-Given the first term at $t_0$ is 6 and the second term at $t_3$ is 32.

-Let's take rabbit population as a function of time to be

$y=ab^x$

where y is the population at time x and a the initial population at $t_0\\$

#We substitute our values to calculate the value of the constant b:

$y_x=ab^x\\\\y_3=ab^3\\\\32=6b^3\\\\b=1.472$

#Replace b in the population function:

$y=ab^x, b=1.7472,a=6\\\\\therefore y=6(1.7472)^x$

Hence, the regression for the rabbit population as a function of time x is $y=6(1.7472)^x$

b. The exponential function in terms of base $e$ is usually expressed as:

$A=A_0e^{kt}$

Where:

$A_0$ -is the initial population at $t_o$

$A$ -is the population at time t.

$k-$ is the exponential growth constant.

$e-$ the exponent

Our function in terms of base exponent is rewritten as:

$y=A_0e^{kt}$

#Substitute with actual figures to solve for t:

$y=A_0e^{kt}, y=32, xt=3, A_0=6\\\\32=6e^{3k}\\\\3k=In (32/6)\\\\k=0.5580$

Hence, the regression equation in terms of base e is $y=6e^{0.558t}$

c. We substitute y with any number higher than 10,000 to estimate the time for the rabbits to exceed 10,000.

-We know that $y=6e^{0.558t}.$

Therefore we calculate t as(take y=10001):

$y=6e^{0.558t}, y=10001\\\\10001=6e^{0.558t}\\\\1666.8333=e^{0.558t}\\\\0.558t= In 1666.8333\\\\t=13.2951$

Hence, it takes approximately 13.3 months for the population to exceed 10000

3 0

4 years ago

Which three R's are associated with decreasing the world's waste?

Fed [463]

Answer:

this is for

Step-by-step explanation:

my boi auto indo

6 0

3 years ago