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ivanzaharov [21]
2 years ago
11

The owner of Animal Lovers Pet Shop has a total of 153 parakeets in 3 cages. Could each cage hold the same number of parakeets?

Explain how you found your answer.
Mathematics
2 answers:
ollegr [7]2 years ago
6 0
If you do 153 divided by 3 you would get the answer 51. So yes you could have the same number of parakeets in the 3 cages. Hope this helps!
Alekssandra [29.7K]2 years ago
6 0

yes because 153 divides equally into 3

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Stolb23 [73]
2x^2 - 32
2(x^2 - 16)
2(x + 4)(x - 4) <==
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2 years ago
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The combined average weight of an okapi and a llama is 450450450 kilograms. The average weight of 333 llamas is 190190190 kilogr
Kisachek [45]
I'm going to assume that you meant 450kg for the combined weight, 190kg more and 3 Llamas. I'm pretty sure Llamas and Okapis don't weigh 450450450kg (that's 993,073,252 pounds). :)

x= Okapi weight
y= Llama weight

EQUATIONS:
There are 2 equations to be written:

1) 450kg is equal to the weight of one Okapi and one Llama

450kg= x + y

2) The weight of 3 llamas is equal to the weight of one Okapi plus 190kg.

3y=190kg + x


STEP 1:
Solve for one variable in one equation and substitute the answer in the other equation.

450kg= x + y
Subtract y from both sides
450-y =x


STEP 2:
Substitute (450-y) in second equation in place of x to solve for y.

3y=190kg + x
3y=190 + (450-y)
3y=640 -y
add y to both sides

4y=640
divide both sides by 4

y=160kg Llama weight


STEP 3:
Substitute 160kg in either equation to solve for x.

3y=190kg + x
3(160)=190 + x
480=190 + x
Subtract both sides by 190

290= x
x= 290kg Okapi weight


CHECK:
3y=190kg + x
3(160)=190 + 290
480=480

Hope this helps! :)
8 0
2 years ago
According to the journal Chemical Engineering, an important property of a fiber is its water absorbency. A random sample of 20 p
Feliz [49]

Answer:

Step-by-step explanation:

a )

sample mean = sum total of given data / no of data

= 415.35 / 20 = 20.76

To calculate the median we arrange the data in ascending order and take the average of 10 th and 11 th term .

= 20.50 + 20.72 / 2

= 20.61

b ) To calculate the 10% trimmed mean , we neglect the largest 10% and smallest 10 % data and then calculate the mean . Here we neglect the first two smallest and last two greatest

(18.92 + 19.25 ..... + 22.43 + 22.85) / 16

= 20.74

c )

We can easily plot the data on number line from 17 to 24

d )

Maximum value of data set = 23.71 and minimum value is 18.04

mean is 20.76 , median is 20.61 and trimmed mean is 20.74

They are between maximum and minimum values of given data . Hence there is no outliers .

4 0
2 years ago
During the summer, Gavin tutors and Eva washes cars.To show their earnings, Gavin makes a graph and Eva makes a table. Who earns
sladkih [1.3K]
To find Eva's slope divide 23 by 2.

23/2 = 11.5

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Gavin earns $8.50 more than Eva.

Hope this helps :)
6 0
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In △ABC, m∠A=15 °, a=10 , and b=11 . Find c to the nearest tenth.
pshichka [43]

Answer:

The answer is:

\bold{c\approx 20.2\ units}

Step-by-step explanation:

Given:

In △ABC:

m∠A=15°

a=10 and

b=11

To find:

c = ?

Solution:

We can use cosine rule here to find the value of third side c.

Formula for cosine rule:

cos A = \dfrac{b^{2}+c^{2}-a^{2}}{2bc}

Where  

a is the side opposite to \angle A

b is the side opposite to \angle B

c is the side opposite to \angle C

Putting all the values.

cos 15^\circ = \dfrac{11^{2}+c^{2}-10^{2}}{2\times 11 \times c}\\\Rightarrow 0.96 = \dfrac{121+c^{2}-100}{22c}\\\Rightarrow 0.96 \times 22c= 121+c^{2}-100\\\Rightarrow 21.25 c= 21+c^{2}\\\Rightarrow c^{2}-21.25c+21=0\\\\\text{solving the quadratic equation:}\\\\c = \dfrac{21.25+\sqrt{21.25^2-4 \times 1 \times 21}}{2}\\c = \dfrac{21.25+\sqrt{367.56}}{2}\\c = \dfrac{21.25+19.17}{2}\\c \approx 20.2\ units

The answer is:

\bold{c\approx 20.2\ units}

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