Ask question

Ask question

All categories

3 years ago

10

to maintain a healthy diet, no more than 30% of calories consumed each day should come from fat. a normal teenage girl needs abo

ut 2,000 calories each day how many calories should come from sources other than fat in a teenage girls diet each day?

1 answer:

Ira Lisetskai [31]3 years ago

5 0

1400 becaude 30% of 2000 is 600 and then you subtract that from 2000 to find 70% which is 1400

You might be interested in

Which set of parametric equations over the interval 0 ≤ t ≤ 1 defines a line segment with initial point (–5, 3) and terminal poi

Strike441 [17]

Answer:

x(t) = –5 + 6t; y(t) = 3 – 9t

Step-by-step explanation:

6 0

3 years ago

I have 5 dogs and 3 puppies what percent are puppies

romanna [79]

5+3 = 8
8 is 100%
3/8 = 37.5%
I feel like I might be wrong haha sorry if I am

6 0

3 years ago

Read 2 more answers

In class of 50 students ,20 play football,25 playvolleyball and 15 play none.How many play both??

____ [38]

Hence, 5+2+3=10 students play exactly two of these sports.

Answer: 10

7 0

3 years ago

Read 2 more answers

Historically, there is a 40% chance of having clear sunny skies in Seattle in July. Let's assume that each day is independent fr

Igoryamba

Answer: a) 0.0058

b) 0.0026

Step-by-step explanation:

Given : The probability of having clear sunny skies in Seattle in July : p= 0.40

The number of days spent in Seattle in July: n= 18

a) Using, Binomial probability formula : $P(x)=^nC_xp^x(1-p)^{n-x}$

The probability of having clear sunny skies on at least 13 of those days:-

$P(x\geq13)=P(13)+P(14)+P(15)+P(16)+P(17)+P(18)\\\\=^{18}C_{13}(0.4)^{13}(0.6)^5+^{18}C_{14}(0.4)^{14}(0.6)^4+^{18}C_{15}(0.4)^{15}(0.6)^3+^{18}C_{16}(0.4)^{16}(0.6)^2+^{18}C_{17}(0.4)^{17}(0.6)^1+^{18}C_{18}(0.4)^{18}(0.6)^0$

$=\dfrac{18!}{13!5!}(0.4)^{13}(0.6)^5+\dfrac{18!}{14!4!}(0.4)^{14}(0.6)^4+\dfrac{18!}{15!3!}(0.4)^{15}(0.6)^3+\dfrac{18!}{16!2!}(0.4)^{16}(0.6)^2+(18)(0.4)^{17}(0.6)^1+(1)(0.4)^{18}$

$=0.00447111249474+0.00106455059399+0.000189253438931+0.0000236566798664+0.00000185542587187+0.000000068719476736$

$=0.00575049735288\approx0.0058$

b) On converting binomial to normal distribution, we have

$\text{Mean=}\mu=np= 18\times0.40=7.2\\\\\text{Standard deviation}=\sigma=\sqrt{np(1-p)}\\\\=\sqrt{18(0.40)(1-0.40)}=2.07846096908\approx2.08$

Let x be the number of days having clear sunny skies in Seattle in July.

Then, using $z=\dfrac{x-\mu}{\sigma}$ we have

$z=\dfrac{13-7.2}{2.08}=2.78846153846\approx2.79$

P-value = $P(x\geq13)=P(z\geq2.79)=1-P(z$

$=1-0.9973645=0.0026355\approx0.0026$

6 0

3 years ago

Consider an object of mass m moving along a frictionless surface and attached to a wall by a spring. If we let y(t) denote the s

ICE Princess25 [194]

Answer:

The dimensions of Ki are N/m while the dimension of K3 is N/m³. The Nondimensional equation is:

$\frac{d^2Y}{d\tau^2} -Y-Y^3=0$

with the substitution:

$y=y_0Y \displaystyle\leftrightarrow y_0=\sqrt{\frac{k_i}{k_3}}$

$t=t_0\tau \leftrightarrow \tau=\displaystyle\sqrt{\frac{m}{k_i} }$

Step-by-step explanation:

To determine the dimensions of the constant we first see the first part of the equation, knowing that each term has to have the same dimension.

The first term is mass multiplied with acceleration, this means that the dimension of each term is the unit of force, Newtons.

The second term is distance multiplied with ki. If the term has to have the force dimension, ki dimensions must be N/m.

The third term is distance to the cube multiplied with k3. If the term has to have the force dimension, k3 dimensions must be N/m³.

For the process of Nondimensionalization, we first make the substitution of the main variables (in this case t of time and y of distance) for nondimensional ones (in this case Y and τ):

$y=y_0Y$

$t=t_0\tau$

Where y₀ and t₀ are the characteristic units used to scale the equation.

Substituted in the original equation:

$m\frac{d^2y}{dt^2} -k_iy-k_3y^3=0\\m\frac{y_0}{t_0^2} \frac{d^2Y}{d\tau^2} -k_iy_0Y-k_3y_0^3Y^3=0$

Now we clear the highest degree term of the differential equation:

$\displaystyle\frac{d^2Y}{d\tau^2} -\frac{k_it_0^2}{m} Y-\frac{k_3t_0^2y_0^2}{m} Y^3=0\\\frac{d^2Y}{d\tau^2} -A Y-B Y^3=0$

Now we need that coefficients A and B to be the unit without any dimension, Therefore:

$\displaystyle A=1=\frac{k_it_0^2}{m} \leftrightarrow t_0=\sqrt{\frac{m}{k_i}} \\B=1=\frac{k_3t_0^2y_0^2}{m}=\frac{k_3y_0^2}{k_i} \leftrightarrow y_0=\sqrt{\frac{k_i}{k_3}}$

With this characteristic units the equation is Nondimensional.

5 0

4 years ago