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3 years ago

Noah’s dog weighs 15 1⁄3 pounds (lbs). Aiden’s dog weighs three times as much as Noah’s. How many pounds does Aiden’s dog weigh?

Mathematics

1 answer:

SpyIntel [72]3 years ago

4 0

<h3>Answer: 46 pounds</h3>

===============================================

Work Shown:

N = weight of Noah's dog = 15 & 1/3 pounds

A = weight of Aiden's dog = unknown

A = 3*N

A = 3*(15 & 1/3)

A = 3*(15 + 1/3)

A = 3*15 + 3*(1/3) .... distribute

A = 45 + 1

A = 46

Aiden's dog weighs 46 pounds.

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Solve the equation 4+2|3x+4|=-4

ad-work [718]

Answer:

Value of given expression x is -4

Step-by-step explanation:

Given equation in question;

4 + 2|3x + 4| = -4

Find:

Value of given expression

Computation:

4 + 2|3x + 4| = -4

Using BODMAS rule;

⇒ 4 + 2|3x + 4| = -4

⇒ 2|3x + 4| = - 4 - 4

⇒ 2|3x + 4| = - 8

⇒ |3x + 4| = - 8 / 2

⇒ |3x + 4| = - 4

⇒ 3x + 4 = - 8

⇒ 3x = -8 - 4

⇒ 3x = -12

⇒ x = -12 / 3

⇒ x = -4

Value of given expression x is -4

8 0

3 years ago

There are 7 apples on a plate. When these apples are distributed to 7 children, one for each, an apple snows on the plate. How c

blsea [12.9K]

Answer:

magic

Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

Sam borrows $5700 at 4.5% simple interest for 3 years. Find the interest

kherson [118]

Answer:

The interest is

Step-by-step explanation:

Simple interest is given by

$I = \frac{P \times R \times T}{100}$

where

P is the principal

R is the rate

T is the time given

From the question

The principal / P = $ 5700

The rate / R = 4.5%

The time given / T = 3 years

So the interest is

$I = \frac{5700 \times 4.5 \times 3}{100}$

$I = \frac{76950}{100}$

We have the final answer as

I = $ 769.50

Hope this helps you

3 0

3 years ago

if each side of a square classroom is 25 feet then what is the perimeter of the room? what is the area of the floor of the room?

Svetllana [295]

Answer:

P = 100

Area = 625

Step-by-step explanation:

Area = s^2

sqrt(area) = sqrt(s^2)

area = 25*25

sqrt(25*25) = sqrt(s^2)

s = 25

P = 4*s

P = 4 * 25

P = 100

Area = s^2

s = 25

Area = 25^2

Area = 625

4 0

3 years ago

A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

nlexa [21]

Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

$\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8$

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0

3 years ago