A box of chocolates contains five milk chocolates, three dark chocolates, and four white chocolates. You randomly select and eat three chocolates. The first piece is milk
chocolate, the second is white chocolate, and the third is milk chocolate. Find the probability of this occuring.

Answers

Answer 1

Answer:

60/220

Step-by-step explanation:

we use combination,

[tex] (\frac{5}{1} ) \times ( \frac{4}{1} ) \times ( \frac{3}{1} )[/tex]

[tex]5 \times 4 \times 3 = 60[/tex]

then, all divided by,

[tex] (\frac{12}{3}) = 220 [/tex]

[tex]60 \div 220[/tex]

Answer 2

The probability of the first piece being milk chocolate, the second being white chocolate, and the third being milk chocolate is 0.06.

What is Probability?

The probability helps us to know the chances of an event occurring.

[tex]\rm Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]

The sample contains five milk chocolates, three dark chocolates, and four white chocolates. Therefore, the probability that the first piece is milk chocolate is

[tex]\rm Probability=\dfrac{\text{Number of Milk choclates}}{\text{Total number of choclates}}[/tex]

[tex]\rm Probability=\dfrac{5}{12}[/tex]

Now, since the chocolate is been eaten the sample size will reduce from 12 chocolates in total to 11 chocolates in total (four milk chocolates, three dark chocolates, and four white chocolates). Therefore, the probability of the second piece being white chocolate is

[tex]\rm Probability=\dfrac{\text{Number of White choclates}}{\text{Total number of choclates}}[/tex]

[tex]\rm Probability=\dfrac{4}{11}[/tex]

Now, as the chocolate is been eaten the sample size will reduce from 11 chocolates in total to 10 chocolates in total (four milk chocolates, three dark chocolates, and three white chocolates). Therefore, the probability of the third piece being milk chocolate is

[tex]\rm Probability=\dfrac{\text{Number of Milk choclates}}{\text{Total number of choclates}}[/tex]

[tex]\rm Probability=\dfrac{4}{10}[/tex]

Thus, the probability of the first piece being milk chocolate, the second being white chocolate, and the third being milk chocolate is

[tex]\rm Probability=\dfrac{5}{12}\times \dfrac{4}{11} \times \dfrac{4}{10} = \dfrac{80}{1320} = 0.06[/tex]

Hence, the probability of the first piece being milk chocolate, the second being white chocolate, and the third being milk chocolate is 0.06.

Learn more about Probability:

https://brainly.com/question/795909


Related Questions

Kenji earned the test scores below in English class.
79, 91, 93, 85, 86, and 88
What are the mean and median of his test scores?

Answers

Answer:

mean=87

median=87

Step-by-step explanation:

mean=sum of test score/number of subject

mean=79+91+93+85+86+88/6

mean=522/6

mean=87

Literal meaning of median is medium.

To find the number which lies in the medium, we must rearrange the number in ascending.

79, 91, 93, 85, 86, 88

79, 85, 86, 88, 91, 93

86+88/2=87

Hope this helps ;) ❤❤❤

Let me know if there is an error in my answer.

What is the most precise name for quadrilateral ABCD with vertices A(–5,2), B(–3, 5),C(4, 5),and D(2, 2)?

Answers

Answer: ABCD is a parallelogram.

Step-by-step explanation:

First we plot these point on a graph as given in attachment.

From the attachment we can observe that AD || BC || x-axis .

also, AB ||CD, that will make ABCD a parallelogram ,  but to confirm we check the property of parallelogram "diagonals bisect each other" , i.e . "Mid point of both diagonals are equal".

Mid point of AC= [tex](\dfrac{-5+4}{2},\dfrac{2+5}{2})=(\dfrac{-1}{2},\dfrac{7}{2})[/tex]

Mid point of BD= [tex](\dfrac{-3+2}{2},\dfrac{5+2}{2})=(\dfrac{-1}{2},\dfrac{7}{2})[/tex]

Thus, Mid point of AC=Mid point of BD

i.e. diagonals bisect each other.

That means ABCD is a parallelogram.

Answer: ABCD is a parallelogram.

Step-by-step explanation:

First, we plot these points on a graph as given in the attachment. From the attachment, we can observe that AD || BC || x-axis. Also, AB ||CD, which will make ABCD a parallelogram, but to confirm, we check the parallelogram property "diagonals bisect each other," i.e., "Midpoint of both diagonals is equal."

The midpoint of AC=. The midpoint of BD=. Thus, the Midpoint of AC=Mid point of BD diagonals bisects each other. That means ABCD is a parallelogram.

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2. You wish to test H0: μ = 100 versus H1: μ ≠ 100 with a sample of n = 9 specimens.
A. If the acceptance region is defined as 98.5 le x- 101.5, find the type I error probability alpha.
B. Find beta for the case where the true mean heat evolved is 103.
C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

Answers

Answer:

A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

[tex]\mathtt{H_o: \mu = 100}[/tex]

[tex]\mathtt{H_1: \mu \neq 100}[/tex]

A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .

Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]

[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]

[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]

when  [tex]\mu = 100[/tex]

[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]

[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]

From the standard normal distribution tables

[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]

[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]

[tex]\mathbf{\alpha = 0.0244 }[/tex]

Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]

Thus;

β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 103[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]

[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 105[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]

[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

logx-log(x-l)^2=2log(x-1)​

Answers

Answer:

  x = 1.00995066776

  x = 2.52925492433

Step-by-step explanation:

This sort of equation is best solved using a graphing calculator. For that purpose, I like to rewrite the equation as a function whose zeros we're seeking. Here, that becomes ...

  [tex]f(x)=\log{(x)}-\log{(x-1)}^2-2\log{(x-1)}[/tex]

The attached graph shows zeros at

  x = 1.00995066776 and 2.52925492433

_____

Comment on the equation

Note that we have taken the middle term to be the square of the log, rather than the log of a square. For the latter interpretation, see mberisso's answer at https://brainly.com/question/17210068

Comment on the answer refinement

We have used Newton's method iteration to refine the solutions to this equation. The solution near 1.00995 requires the initial guess be very close for that method to work properly. Fortunately, the 1.01 value shown on the graph is sufficient for the purpose.

The mean salary of federal government employees on the General Schedule is $59,593. The average salary of 30 state employees who do similar work is $58,800 with \sigmaσσ= $1500. At the 0.01 level of significance, can it be concluded that state employees earn on average less than federal employees? What is the critical value? Round your answer to the nearest hundredths.

Answers

Answer:

Yes it can be concluded that state employees earn on average less than federal employees

  The critical value is  [tex]Z_{\alpha } = - 2.33[/tex]

Step-by-step explanation:

From the question we are told that

   The  population mean is  [tex]\mu = \$ 59593[/tex]

   The sample size is  n =  30

    The  sample mean is [tex]\= x = \$ 58800[/tex]

     The  standard deviation is  [tex]\sigma = \$ 1500[/tex]

     The significance level is  [tex]\alpha = 0.01[/tex]

   

The null hypothesis is  [tex]H_o : \mu = \$ 59593[/tex]

 The  alternative hypothesis is  [tex]H_a : \mu < \$ 59593[/tex]

The critical value of [tex]\alpha[/tex] from the normal distribution table is  [tex]Z_{\alpha } = - 2.33[/tex]

 Generally the test statistics is  mathematically evaluated as

            [tex]t = \frac{\= x - \mu}{ \frac{ \sigma }{ \sqrt{n} } }[/tex]

=>         [tex]t = \frac{ 58800 - 59593 }{ \frac{ 1500 }{ \sqrt{30} } }[/tex]  

=>          [tex]t = -2.896[/tex]

The  p-value is obtained from the z-table

   [tex]p-value = P(t < -2.896) = 0.0018898[/tex]

Since [tex]p-value < \alpha[/tex] , we reject the null hypothesis, hence it can be concluded that state employees earn on average less than federal employees  

   

a
A solid metal cone of base radius a cm and height 2a cm is melted and solid
spheres of radius are made without wastage. How many such spheres can be
made?​

Answers

volume of a cone

.

.

.

volume of sphere

.

.

number of spheres that can be made......

.

.

hence a hemisphere can be formed

A mass of 5 kg stretches a spring 10 cm. The mass is acted on by an external force of 10sin( t ) N(newtons) and moves in a medium that imparts a viscous force of 2 N
when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate the initial value problem describing the motion of the mass.
A)Find the solution of the initial value problem in the above problem.
B)Plot the graph of the steady state solution
C)If the given external force is replaced by a force of 2 cos(ωt) of frequency ω , find the value of ω for which the amplitude of the forced response is maximum.

Answers

Answer:

A) C1 = 0.00187 m = 0.187 cm,  C2 = 0.0062 m = 0.62 cm

B)  A sample of how the graph looks like is attached below ( periodic sine wave )

C) w = [tex]\sqrt[4]{3}[/tex] is when the amplitude of the forced response is maximum

Step-by-step explanation:

Given data :

mass = 5kg

length of spring = 10 cm = 0.1 m

f(t) = 10sin(t) N

viscous force = 2 N

speed of mass = 4 cm/s = 0.04 m/s

initial velocity = 3 cm/s = 0.03 m/s

Formulating initial value problem

y = viscous force / speed = 2 N / 0.04 m/s = 50 N sec/m

spring constant = mg/ Length of spring = (5 * 9.8) / 0.1 = 490 N/m

f(t) = 10sin(t/2) N

using the initial conditions of u(0) = 0 m and u"(0) = 0.03 m/s to express the equation of motion

the equation of motion = 5u" + 50u' + 490u = 10sin(t/2)

A) finding the solution of the initial value

attached below is the solution and

B) attached is a periodic sine wave replica of how the grapgh of the steady state solution looks like

C attached below

Find a cubic polynomial with integer coefficients that has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.

Answers

Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]

$a^{2}=5+2 \sqrt{6}$

$a^{3}=11 \sqrt{2}+9 \sqrt{3}$

The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.

Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$

so fits with the other answers.

Answer:

[tex]y^3 -6y-6[/tex]

please help me in these question ????

A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.
(a) How many different samples of size 4 pens are possible?
(b) How many samples have 3 red pens and 1 black pen?
(c) How many samples of size 4 contain at least two red pens?
(d) How many samples of size 4 contain


If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal distribution.
1- What percentage of a cucumber give the crop amount between and 834 kg?
2- What the probability of cucumber give the crop exceed 900 kg ?

Answers

Answer:

Step-by-step explanation:

A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.

(a) How many different samples of size 4 pens are possible?

12C4=12!/(4!*8!)=495

(b) How many samples have 3 red pens and 1 black pen?

5C3*7C1

5C3=5!/(3!*2!)=10

7C1=7!/(1!*6!)=7

=>5C3*7C1=10*7=70

(c) How many samples of size 4 contain at least two red pens?

(5C2*7C2)+(5C3*7C1)+(5C4*7C0)

5C2=5!/(2!*3!)=10

7C2=7!/(2!*5!)=21

5C3=5!/(3!*2!)=10

7C1=7!/(1!*6!)=7

5C4=5!/(4!*1!)=5

7C0=7!/(0!*7!)=1

=>(5C2*7C2)+(5C3*7C1)+(5C4*7C0)=285

(d) How many samples of size 4 contain at most one black pen?

(7C1*5C3)+(7C0*5C4)

7C1=7!/(1!*6!)=7

7C0=7!/(0!*7!)=1

5C3=5!/(3!*2!)=10

5C4=5!/(4!*1!)=5

=>(7C1*5C3)+(7C0*5C4)=(7*10)+(1*5)=75

Transform the given parametric equations into rectangular form. Then identify the conic.

Answers

Answer:

Solution : Option B

Step-by-Step Explanation:

We have the following system of equations at hand here.

{ x = 5 cot(t), y = - 3csc(t) + 4 }

Now instead of isolating the t from either equation, let's isolate cot(t) and csc(t) --- Step #1,

x = 5 cot(t) ⇒ x - 5 = cot(t),

y = - 3csc(t) + 4 ⇒ y - 4 = - 3csc(t) ⇒ y - 4 / - 3 = csc(t)

Now let's square these two equations. We know that csc²θ - cot²θ = 1, so let's subtract the equations  as well. --- Step #2

 

( y - 4 / - 3 )² = (csc(t))²

- ( x - 5 / 1 )² = (cot(t))²  

___________________

(y - 4)² / 9 - x² / 25 = 1

And as we are subtracting the two expressions, this is an example of a hyperbola. Therefore your solution is option b.

Salaries of 42 college graduates who took a statistics course in college have a​ mean, ​, of . Assuming a standard​ deviation, ​, of ​$​, construct a ​% confidence interval for estimating the population mean .

Answers

Answer:

The 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).

Step-by-step explanation:

The complete question is:

Salaries of 42 college graduates who took a statistics course in college have a​ mean, [tex]\bar x[/tex] of, $64, 100. Assuming a standard​ deviation, σ of ​$10​,016 construct a ​99% confidence interval for estimating the population mean μ.

Solution:

The (1 - α)% confidence interval for estimating the population mean μ is:

[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]

The critical value of z for 99% confidence interval is:

[tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.57[/tex]

Compute the 99% confidence interval for estimating the population mean μ as follows:

[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]

     [tex]=64100\pm 2.58\times\frac{10016}{\sqrt{42}}\\\\=64100+3987.3961\\\\=(60112.6039, 68087.3961)\\\\\approx (60112.60, 68087.40)[/tex]

Thus, the 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).

A machine used to fill​ gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of ounces and a standard deviation of ounce. You randomly select cans and carefully measure the contents. The sample mean of the cans is ounces. Does the machine need to be​ reset? Explain your reasoning. ▼ Yes No ​, it is ▼ very unlikely likely that you would have randomly sampled cans with a mean equal to ​ounces, because it ▼ lies does not lie within the range of a usual​ event, namely within ▼ 1 standard deviation 2 standard deviations 3 standard deviations of the mean of the sample means.

Answers

Complete question is;

A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.

(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.

Answer:

Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Step-by-step explanation:

We are given;

Mean: μ = 128

Standard deviation; σ = 0.2

n = 35

Now, formula for standard error of mean is given as;

se = σ/√n

se = 0.2/√35

se = 0.0338

Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;

μ ± 2se = 128 ± 0.0338

This gives; 127.9662, 128.0338

So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

The cost, C, in United States Dollars ($), of cleaning up x percent of an oil spill along the Gulf Coast of the United States increases tremendously as x approaches 100. One equation for determining the cost (in millions $) is:

Answers

Complete Question

On the uploaded image is a similar question that will explain the given question

Answer:

The value of k is  [tex]k = 214285.7[/tex]

The percentage  of the oil that will be cleaned is [tex]x = 80.77\%[/tex]

Step-by-step explanation:

From the question we are told that

   The  cost of cleaning up the spillage is  [tex]C = \frac{ k x }{100 - x }[/tex]  [tex]x \le x \le 100[/tex]

     The  cost of cleaning x =  70% of the oil is  [tex]C = \$500,000[/tex]

   

Now at  [tex]C = \$500,000[/tex] we have  

       [tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]

       [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]k = 214285.7[/tex]

Now  When  [tex]C = \$900,000[/tex]

       [tex]x = 80.77\%[/tex]

       

 

Relating a Polynomial Identity to Pythagorean Triples
In this activity you'll relate polynomial identities with Pythagorean triples. Answer the following questions
based on this triangle with side lengths x^2 – 1, 2x, and x^2 + 1.

Answers

Answer:

Step-by-step explanation:

Hello, please consider the following.

For x > 1, we can apply Pythagoras theorem as below.

[tex]\text{Let's estimate this sum of two squares.} \\\\(2x)^2+(x^2-1)^2=4x^2+x^4-2x^2+1=x^4+2x^2+1\\\\\text{Let's estimate this square, now.} \\\\(x^2+1)^2=x^4+2x^2+1\\\\\text{The two expressions are equal, meaning.} \\\\(2x)^2+(x^2-1)^2=(x^2+1)^2\\\\\text{Using Pythagoras' theorem, we can say that this is a right triangle.}[/tex]

Thank you

Suppose that a sample mean is .29 with a lower bound of a confidence interval of .24. What is the upper bound of the confidence interval?

Answers

Answer:

The upper bound of the confidence interval is 0.34

Step-by-step explanation:

Here in this question, we want to calculate the upper bound of the confidence interval.

We start by calculating the margin of error.

Mathematically, the margin of error = 0.29 -0.24 = 0.05

So to get the upper bound of the confidence interval, we simply add this margin of error to the mean

That would be 0.05 + 0.29 = 0.34

HELP ASAP ROCKY!!! will get branliest.​

Answers

Answer:

work pictured and shown

Answer:

Last one

Step-by-step explanation:

● [ ( 3^2 × 5^0) / 4 ]^2

5^0 is 1 since any number that has a null power is equal to 1.

●[ (3^2 ×1 ) / 4 ]^2

● (9/4)^2

● 81 / 16

one third multiplied by the sum of a and b

Answers

Answer:

1/3(a+b)

hope it helps :>

a+b/3
This is the answer of ur question

Foram prescritos 500mg de dipirona para uma criança com febre.Na unidade tem disponivel ampola de 1g/2ml.Quantos g vão ser administrados no paciente

Answers

De acordo com a disponibilidade da unidade, há apenas a seguinte dosagem: 1g/2mL - ou seja, uma grama de dipirona a cada 2mL

O enunciado está meio mal formulado, pois é dito que foram prescritos 500mg de dipirona e é essa quantidade de farmaco que a criança tem que tomar. Deseja-se saber quantos mL deverao ser administrados.

Fazendo a classica regra de 3, podemos chegar no volume desejado:

(atentar que 500mg = 0,5g)

     g               mL

     1    ---------   2

    0,5  ---------  X    

1 . X = 0,5 . 2

X = 1mL

The quotient of 3 and the cube
of y+2

Answers

Answer:

  [tex]\dfrac{3}{(y+2)^3}[/tex]

Step-by-step explanation:

Maybe you want this written using math symbols. It will be ...

  [tex]\boxed{\dfrac{3}{(y+2)^3}}[/tex]

5x+4(-x-2)=-5x+2(x-1)+12

Answers

Answer:

x=9/2

Step-by-step explanation:

Let's solve your equation step-by-step.

5x+4(−x−2)=−5x+2(x−1)+12

Step 1: Simplify both sides of the equation.

5x+4(−x−2)=−5x+2(x−1)+12

5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)

5x+−4x+−8=−5x+2x+−2+12

(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)

x+−8=−3x+10

x−8=−3x+10

Step 2: Add 3x to both sides.

x−8+3x=−3x+10+3x

4x−8=10

Step 3: Add 8 to both sides.

4x−8+8=10+8

4x=18

Step 4: Divide both sides by 4.

4x/4=18/4

x=9/2

The cost of a daily rental car is as follows: The initial fee is $39.99 for the car, and it costs $0.20 per mile. If Julie's final bill was $100.00 before taxes, how many miles did she drive?

Answers

Answer:

300.05 miles

Step-by-step explanation:

initial fee= $39.99

final bill = $ 100

cost =$ 0.20 per mile

remaining amount = $ 60.01

solution,

she drive = remaining amount / cost

=60.01/0.20

=300.05 miles

Answer:

500 miles

Step-by-step explanation:

Let us use cross multiplication to find the unknown amount.

Given:

1) Cost for 1 mile=$0.20

2)Cost for x miles=$100

Solution:

No of miles                             Cost

1) 1                                             $0.20

2)x                                             $100

By cross multiplying,

100 x 1= 0.20x

x=100/0.20

x=500 miles

Thank you!

BRAINLIST AND A THANK YOU AND 5 stars WILL BE REWARDED PLS ANSER

Answers

Answer:

The first picture's answer would be (6, 21)

Step-by-step explanation:

You have to find the points on the 8th and the 9th day, and then you would add them together, and then divide by two finding the average, which would be 24 and 18, so when added, you get 42, divided by 2 you get 21. You look on the graph for the point with 21, and you find it is on 6.

Claire has to go to the movie theater the movie starts at 4:15 pm it is a 25min walk to the theater from her home what time dose the have to leave the house to get there on time

Answers

Answer:

claire has to leave at 3:50 from her house.

Answer:

She needs to leave by 3:50 to get there on time.

Step-by-step explanation:

4:15 - 0:25 = 3:50.

Time

(minutes)

Water

(gallons)

1

16.50

1.5

24.75

2

33

find the constant of proportionality for the second and third row

Answers

Answer:

16.50

Step-by-step explanation:

Constant of proportionality = no of gallons of water per 1 minute.

In the first row, we have 16.50 gallons of water per 1 minute.

In the 2nd row, we have 24.75 gallons of water in 1.5 minutes. In 1 minute, we will have 24.75 ÷ 1.5 = 16.50 gallons

In the 3rd row, we have 33 gallons in 2 minutes. In 1 minute, we will have 33 ÷ 2 = 16.50 gallons.

We can see that there seems to be the same constant of proportionality for the 2nd and 3rd row, which is 16.50.

Thus, a relationship between gallons of water (w) and time (t), considering the constant, 16.50, can be written as: [tex] w = 16.50t [/tex]

This means the constant of proportionality, 16.50, is same for all rows.

For the following graph, state the polar coordinate with a positive r and positive q (in radians). Explain your steps as to how you determined the coordinate (in your own words). I'm looking for answers that involve π, not degrees for your angles. State the polar coordinate with (r, -q). Explain how you found the new angle. State the polar coordinate with (-r, q). Explain how you found the new angle. State the polar coordinate with (-r, -q). Explain how you found the new angle.

Answers

the graph has 12 segments so angle enclosed by each segment is [tex] {2\pi\over 12}=\frac{\pi}6[/tex]

anti-clockwise is taken as positive, so if you want positive q, you need to rotate 8 segments [tex] q=8\frac,{\pi}6=\frac{4\pi}3 [/tex] , and and 8 circles or units so r=8

and for a negative angle, you need to rotate clockwise

Which is 4 segments from the horizontal line. so [tex]q=-\frac{2\pi}3[/tex] and r will be same, 8 units.

[not sure about -r so I won't include it in answer]

Answer:

Points : ( 8, - 2π/3 ), ( - 8, π/3 ), ( - 8, - 5π/3 )

Step-by-step explanation:

For the first two cases, ( r, θ ) r would be > 0, where r is the directed distance from the pole, and theta is the directed angle from the positive x - axis.

So when r is positive, we can tell that this point is 8 units from the pole, so r is going to be 8 in either case,

( 8, 240° ) - because r is positive, theta would have to be an angle with which it's terminal side passes through this point. As you can see that would be 2 / 3rd of 90 degrees more than a 180 degree angle,or 60 + 180 = 240 degrees.

( 8, - 120° ) - now theta will be the negative side of 360 - 240, or in other words - 120

Now let's consider the second two cases, where r is < 0. Of course the point will still be 8 units from the pole. Again for r < 0 the point will lay on the ray pointing in the opposite direction of the terminal side of theta.

( - 8, 60° ) - theta will now be 2 / 3rd of 90 degrees, or 60 degrees, for - r. Respectively the remaining degrees will be negative, 360 - 60 = 300, - 300. Thus our second point for - r will be ( - 8, - 300° )

_________________________________

So we have the points ( 8, 240° ), ( 8, - 120° ), ( - 8, 60° ), and ( - 8, - 300° ). However we only want 3 cases, so we have points ( 8, - 120° ), ( - 8, 60° ), and ( - 8, - 300° ). Let's convert the degrees into radians,

Points : ( 8, - 2π/3 ), ( - 8, π/3 ), ( - 8, - 5π/3 )

Which of the following represents "next integer after the integer n"? n + 1 n 2n

Answers

Answer:

n + 1

Step-by-step explanation:

Starting with the integer 'n,' we represent the "next integer" by n + 1.

Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6

Answers

Answer:

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Step-by-step explanation:

Given that:

[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]

recall that:

cos (A-B) = cos AcosB + sin A sin B

[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]

[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]

[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]

[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]

[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Find usubscript10 in the sequence -23, -18, -13, -8, -3, ...

Answers

Step-by-step explanation:

utilise the formula a+(n-1)d

a is the first number while d is common difference

Answer:

22

Step-by-step explanation:

Using the formular, Un = a + (n - 1)d

Where n = 10; a = -23; d = 5

U10 = -23 + (9)* 5

U10 = -23 + 45 = 22

The following shape is based only on squares, semicircles, and quarter circles. Find the area of the shaded part.

Answers

Answer:

this? hope it helps ........

Answer:

The answer is area=32pi-64 and the perimeter is 8pi

Step-by-step explanation:

602/100 into a decimal describe plz

Answers

Answer:

6.02

six point zero two

Step-by-step explanation:

Answer:

602 / 100= 6,02

Step-by-step explanation:

602 to divide 100 = 6,02

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