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1 year ago

There are 3 types of gummy bears:

1 answer:

5 0

Using proportions, it is found that it takes 886 more mini-bears than regular-bears to have the same weight as one super-bear.

<h3>What is a proportion?</h3>

A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct(when both increase or both decrease) or inverse proportional(when one increases and the other decreases, or vice versa), can be built to find the desired measures in the problem, or equations to find these measures.

10 mini-bears weights to 12.1 grams, hence the weight of a mini-bear is of:

12.1/10 = 1.21 grams.

10 regular bears weights to 23.1 grams, hence the weight of a regular bear is of:

23.1/10 = 2.31 grams.

1 super bear weights to 2250 grams, hence the proportion between the weight of a super bear and the weight of a mini-bear is:

2250/1.21 = 1860.

The proportion between the weight of a super bear and the weight of a regular bear is:

2250/2.31 = 974.

The difference of proportions is given by:

1860 - 974 = 886.

It takes 886 more mini-bears than regular-bears to have the same weight as one super-bear.

More can be learned about proportions at brainly.com/question/24372153

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Question 11

Black_prince [1.1K]

Answer:

Step-by-step explanation:

Because the cost per pie is less.

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3 years ago

A student's cost for last semester at her community college was 2800. She spent 448 of that on books. what percent of semester's

Diano4ka-milaya [45]

16%

Step-by-step explanation:

Percent can by calculated by part divided by whole multiplied by one hundred so (448/2800)*100=16

4 0

2 years ago

Puzzle in the picture. the answer should be 23, explain how you did it

Delvig [45]

Step-by-step explanation:

try this option, all the details are in the attachment.

Note, the nine numbers. used in the solution are: 1, 2, 3, 4, 5, 6, 7, 8 and 9.

8 0

2 years ago

Solve y" + y = tet, y(0) = 0, y'(0) = 0 using Laplace transforms.

irina1246 [14]

Answer:

The solution of the diferential equation is:

$y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}$

Step-by-step explanation:

Given $y" + y = te^{t}$ ; y(0) = 0 ; y'(0) = 0

We need to use the Laplace transform to solve it.

ℒ[y" + y]=ℒ[ $te^{t}$ ]

ℒ[y"]+ℒ[y]=ℒ[ $te^{t}$ ]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]+s·y(0)-y'(0)=s²·Y(s)

ℒ[y]=Y(s)

ℒ[ $te^{t}$ ]= $\frac{1}{(s-1)^{2}}$

So, the transformation is equal to:

s²·Y(s)+Y(s)= $\frac{1}{(s-1)^{2}}$

(s²+1)·Y(s)= $\frac{1}{(s-1)^{2}}$

Y(s)= $\frac{1}{(s^{2}+1)(s-1)^{2}}$

To be able to separate in terms, we use the partial fraction method:

$\frac{1}{(s^{2}+1)(s-1)^{2}}=\frac{As+B}{s^{2}+1} +\frac{C}{s-1}+\frac{D}{(s-1)^2}$

1=(As+B)(s-1)² + C(s-1)(s²+1)+ D(s²+1)

The equation is reduced to:

1=s³(A+C)+s²(B-2A-C+D)+s(A-2B+C)+(B+D-C)

With the previous equation we can make an equation system of 4 variables.

The system is given by:

A+C=0

B-2A-C+D=0

A-2B+C=0

B+D-C=1

The solution of the system is:

A=1/2 ; B=0 ; C=-1/2 ; D=1/2

Therefore, Y(s) is equal to:

Y(s)= $\frac{s}{2(s^{2} +1)} -\frac{1}{2(s-1)} +\frac{1}{2(s-1)^{2}}$

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[ $\frac{s}{2(s^{2} +1)}$ ]-ℒ⁻¹[ $\frac{1}{2(s-1)}$ ]+ℒ⁻¹[ $\frac{1}{2(s-1)^{2}}$ ]

$y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}$

8 0

3 years ago

If a data set has ssr = 400 and sse = 100, then the coefficient of determination is

BARSIC [14]

To solve this problem you must apply the proccedure shown below:

1. You have that the data set has SSR=400 and SSE=100

2. Therefore you have the coefficient of determination is:

r²=SSR/SSTO

SSTO=SSR+SSE

3. Then, when you substitute the values, you obtain:

SSTO=400+100
SSTO=500

r²=400/500

4. So, you have that the result is:

r²=0.8

Therefore, as you can see, the answer for the exercise shown above is: the coefficient of determination is 0.8

3 0

3 years ago

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