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1 year ago

Use part 1 of the fundamental theorem of calculus to find the derivative of the function. y = 5 u3 1 + u2 du 4 − 3x

1 answer:

3 0

Looks like

$y(x)=\displaystyle\int_5^{4-3x}u^3(1+u^2)\,\mathrm du$

in which case the FTC asserts that

$\dfrac{\mathrm dy}{\mathrm dx}=(4-3x)^3(1+(4-3x)^2)\cdot\dfrac{\mathrm d(4-3x)}{\mathrm dx}$

$\dfrac{\mathrm dy}{\mathrm dx}=-3(4-3x)^3(1+(4-3x)^2)$

You might be interested in

An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. In an earlier s

anyanavicka [17]

Answer:

1,496 new car buyers

Step-by-step explanation:

The sample size n in Simple Random Sampling is given by

$\bf n=\frac{z^2p(1-p)}{e^2}$

where

z = 1.645 is the critical value for a 90% confidence level (*)

p= 0.33 is the population proportion.

e = 0.02 is the margin of error

$\bf n=\frac{(1.645)^20.33*0.67}{0.02^2}=1,495.76\approx 1,496$

(*)This is a point z such that the area under the Normal curve N(0,1) inside the interval [-z, z] equals 90% = 0.9

It can be obtained in Excel or OpenOffice Calc with

NORMSINV(0.95)

6 0

2 years ago

A pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome). In the first sta

Akimi4 [234]

Answer:

95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].

Step-by-step explanation:

We are given that a pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome).

In the first stages of a clinical trial, it was successful for 7 out of the 14 women.

Firstly, the pivotal quantity for 95% confidence interval for the true proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of women who find success with this new treatment = $\frac{7}{14}$ = 0.50

n = sample of women = 14

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the true proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.50-1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } }$ , $0.50+1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } }$ ]