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3 years ago

7

Help on number 3 plz explain to me I don't understand this because problem

1 answer:

lawyer [7]3 years ago

8 0

So first you find the area of thr big square which is 14*20=280. Then you find the area of the small rectangle which is 9*5=45. then you subtract the small rectangles are from the big ones which leaves you with 235yd^2

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I don’t know how to do this I have like 15 minutes till it’s due please

Over [174]

The answer is z=sh−4t3

6 0

3 years ago

Use Net to find the lateral area of the prism. Please show how to work the problem out please

qaws [65]

Think of the three rectangles as a single large rectangle and find its area.

For the large rectangle made up of the three small rectangles:
Length = 8 cm + 15 cm + 17 cm = 40 cm
Width = 11 cm

area = length * width

area = 40 cm * 11 cm

area = 440 cm^2

3 0

4 years ago

Read 2 more answers

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex

olga nikolaevna [1]

$f(x_1,\ldots,x_n)=x_1+\cdots+x_n=\displaystyle\sum_{i=1}^nx_i$

${x_1}^2+\cdots+{x_n}^2=\displaystyle\sum_{i=1}^n{x_i}^2=4$

The Lagrangian is

$L(x_1,\ldots,x_n,\lambda)=\displaystyle\sum_{i=1}^nx_i+\lambda\left(\sum_{i=1}^n{x_i}^2-4\right)$

with partial derivatives (all set equal to 0)

$L_{x_i}=1+2\lambda x_i=0\implies x_i=-\dfrac1{2\lambda}$

for $1\le i\le n$ , and

$L_\lambda=\displaystyle\sum_{i=1}^n{x_i}^2-4=0$

Substituting each $x_i$ into the second sum gives

$\displaystyle\sum_{i=1}^n\left(-\frac1{2\lambda}\right)^2=4\implies\dfrac n{4\lambda^2}=4\implies\lambda=\pm\frac{\sqrt n}4$

Then we get two critical points,

$x_i=-\dfrac1{2\frac{\sqrt n}4}=-\dfrac2{\sqrt n}$

or

$x_i=-\dfrac1{2\left(-\frac{\sqrt n}4\right)}=\dfrac2{\sqrt n}$

At these points we get a value of $f(x_1,\cdots,x_n)=\pm2\sqrt n$ , i.e. a maximum value of $2\sqrt n$ and a minimum value of $-2\sqrt n$ .

6 0

4 years ago

I need help with this PLEASE

Ludmilka [50]

Answer:

1230 I believe I think area is when u add em all up that's what I did good luck

3 0

3 years ago

You need to compute the a 90% confidence interval for the population mean. How large a sample should you draw to ensure that the

RSB [31]

Answer:

The sample size required is 102.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for population mean <em>μ</em> is:

$CI=\bar x\pm z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}$

The margin of error for this interval is:

$MOE= z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}$

Given:

<em>σ</em> = 9.2

(1 - <em>α</em>)% = 90%

MOE = 1.5

The critical value of <em>z</em> for 90% confidence level is:

$z_{0.10/2}=1.645$

*Use a <em>z</em>-table.

Compute the value of <em>n</em> as follows:

$MOE= z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}$

$n=[\frac{z_{\alpha/2}\times \sigma}{MOE}]^{2}$

$=[\frac{1.645\times 9.2}{1.5}]^{2}\\=101.795\\\approx102$

Thus, the sample size required is 102.

5 0

3 years ago