I need help on 4, 5 and 6

Assume that military aircraft use ejection seats designed for men weighing between 141.8 lb and 218 lb. If women's weights are

Mariulka [41]

Answer:

$P(141.8$

And we can find this probability with this difference and using the normal standard table:

$P(-0.639$

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

Step-by-step explanation:

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(173.6,49.8)$

Where $\mu=173.6$ and $\sigma=49.8$

We are interested on this probability

$P(141.8$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(141.8$

And we can find this probability with this difference and using the normal standard table:

$P(-0.639$

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

8 0

3 years ago

A map that was created using a scale of 1 inch : 3 miles shows a lake with an area of 18 square inches. what is the actual area

Dmitrij [34]

Answer: The area of the lake is actually 182 square miles 1 inch represents one side 3 miles long the other side is also 3 miles. 3×3=9 square miles. 1 Square Inch on the map represents 9 Square Miles.

Step-by-step explanation: If the area on the map is 18 square inches, and each square inch represents 9 square miles, multiply 18×9=182 to get the area of that big lake.

4 0

3 years ago

Read 2 more answers

Pls help help me out help help pls help quick

kotegsom [21]

The answer is b I hope this help

7 0

2 years ago

Read 2 more answers

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

$\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]$

The left and right endpoints of the $i$ -th subinterval, respectively, are

$\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8$

$r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8$

Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

$T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}$

Midpoint rule

We approximate the area for each subinterval by

$M_i=f(m_i)(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}$

Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0

3 years ago

How do you solve 5 3/7 ÷ 5/11

Artyom0805 [142]

Answer:

11 33/35

Step-by-step explanation:

5 3/7 = 5 × 7/7 + 3/7 = 35 + 37 = 38/7

38/7 ÷ 5/11 = 38/7 × 11/5 = 38 × 11/7 × 5 = 418/35

418/35 = 11 33/35

Hope this helps!

6 0

3 years ago