Question 11

Newborn males have weights with a mean of 3272.8 g and a standard deviation of 660.2 g. Newborn females have weights with a mean

Maksim231197 [3]

The correct option is: a female who weighs 1500 g

Explanation

Formula for finding the z-score is: $z= \frac{X-\mu}{\sigma}$

Newborn males have weights with a mean $(\mu)$ of 3272.8 g and a standard deviation $(\sigma)$ of 660.2 g.

So, the z-score for the newborn male who weighs 1500 g will be.......

$z(X=1500)=\frac{1500-3272.8}{660.2}=-2.685... \approx -2.69$

According to the normal distribution table, $P(z=-2.69)=0.0036 = 0.36\%$

Now, newborn females have weights with a mean $(\mu)$ of 3037.1 g and a standard deviation $(\sigma)$ of 706.3 g.

So, the z-score for the newborn female who weighs 1500 g will be.......

$z(X=1500)=\frac{1500-3037.1}{706.3}=-2.176... \approx -2.18$

According to the normal distribution table, $P(z=-2.18)=0.0146 = 1.46\%$

As we can see that the probability that a newborn female has weight of 1500 g is greater than newborn male, so a newborn female has the weight of 1500 g that is more extreme relative to the group from which he came.

5 0

3 years ago

F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0

3 years ago

What is the value of StartFraction 6 Over x EndFraction + 2x2, when x = 3?

vivado [14]

Answer:

hello there,

the answer for this equation is:

The value of Start Fraction 6 Over x End Fraction + 2x2, when x = 3 is 20.

Step-by-step explanation:

: $\frac{6}{x} + 2x^{2} \\ \\\\ \frac{6}{3} + 2(3)^{2}\\\\\\2+2*9 =\\\\\\\\2+18=20$

5 0

3 years ago

Read 2 more answers

For each pair of functions F and G below, find f(g(x)) and g(f(x)).

8090 [49]

Answer:

FG

Step-by-step explanation:

COOL AIDS BRO FAX AND MORE STUFF

8 0

3 years ago

Can someone help me with this it’s for science and it’s due tomorrow

8_murik_8 [283]

Answer:

Hoping someone else can answer with more confidence, but here's what I have:

on the left hand side:

upper is partial/penumbral lunar eclipse

lower is total solar eclipse

on the right hand side:

on the top: it's impossible for the moon to be behind the sun, so I'm not sure what they want there

middle is total solar eclipse

bottom is partial/penumbral lunar eclipse

Hope I helped at least a little

6 0

3 years ago