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elena-14-01-66 [18.8K]

3 years ago

8

WILL GIVE BRAINIEST FOR WHO ANSWERS IT CORRECTLY!! Create a list of steps, in order, that will solve the following equation. 1/2

(x+5)^2+2=42.5

1 answer:

Colt1911 [192]3 years ago

5 0

Answer: subtract 2 and then multiply 2 then spuare root both sides then subtract 5

Step-by-step explanation:hope this helps god bless

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Mean birthweight is studied because low birthweight is an indicator of infant mortality. A study of babies in Norway published i

ANEK [815]

Answer: 20

Step-by-step explanation:

Formula to find the minimum sample size(n) when prior population standard deviation $(\sigma)$ is known.

$n=(\dfrac{z^c\times\sigma}{E})^2$ , where E = Margin of error , $z^c$ = Critical z-value for c confidence interval.

Given : E = 225 g , $\sigma=600$ g

Critical z value for 90% confidence = 1.645

Now, $n=(\dfrac{1.645\times600}{225})^2$

$n=(\dfrac{987}{225})^2$

$n=(4.38666666667)^2=19.2428444\approx20$

Hence, the required minimum sample size = 20

7 0

3 years ago

For the function given below, find a formula for the Riemann sum obtained by dividing the interval [0,5] into n equal subinterva

sergij07 [2.7K]

Given

we are given a function

$f(x)=x^2+5$

over the interval [0,5].

Required

we need to find formula for Riemann sum and calculate area under the curve over [0,5].

Explanation

If we divide interval [a,b] into n equal intervals, then each subinterval has width

$\Delta x=\frac{b-a}{n}$

and the endpoints are given by

$a+k.\Delta x,\text{ for }0\leq k\leq n$

For k=0 and k=n, we get

$\begin{gathered} x_0=a+0(\frac{b-a}{n})=a \\ x_n=a+n(\frac{b-a}{n})=b \end{gathered}$

Each rectangle has width and height as

$\Delta x\text{ and }f(x_k)\text{ respectively.}$

we sum the areas of all rectangles then take the limit n tends to infinity to get area under the curve:

$Area=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)$

Here

$f(x)=x^2+5\text{ over the interval \lbrack0,5\rbrack}$ $\Delta x=\frac{5-0}{n}=\frac{5}{n}$ $x_k=0+k.\Delta x=\frac{5k}{n}$ $f(x_k)=f(\frac{5k}{n})=(\frac{5k}{n})^2+5=\frac{25k^2}{n^2}+5$

Now Area=

$\begin{gathered} \lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{5}{n}(\frac{25k^2}{n^2}+5) \\ =\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{125k^2}{n^3}+\frac{25}{n} \\ =\lim_{n\to\infty}(\frac{125}{n^3}\sum_{k\mathop{=}1}^nk^2+\frac{25}{n}\sum_{k\mathop{=}1}^n1) \\ =\lim_{n\to\infty}(\frac{125}{n^3}.\frac{1}{6}n(n+1)(2n+1)+\frac{25}{n}n) \\ =\lim_{n\to\infty}(\frac{125(n+1)(2n+1)}{6n^2}+25) \\ =\lim_{n\to\infty}(\frac{125}{6}(1+\frac{1}{n})(2+\frac{1}{n})+25) \\ =\frac{125}{6}\times2+25=66.6 \end{gathered}$

So the required area is 66.6 sq units.

3 0

1 year ago

(2x^0y^2)^-3 multiply 2yx^3

hodyreva [135]

Answer:

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Step-by-step explanation:

Simplifica la expresión

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telo118 [61]

Answer: 30!

Step-by-step explanation:

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3 years ago

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