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Alex Ar [27]
2 years ago
14

Zinnia and ruby earn $50 per week for delivering pizzas. Zinnia word for X weeks and earned an additional total bonus of $13. Ru

by work for Y weeks. Which expression shows the total money in dollars that Zinnia and ruby on for delivering pizzas?
Mathematics
1 answer:
Pani-rosa [81]2 years ago
8 0

Answer:

Zinnia = 50[x] +13

Rubu= 50[y]

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Dex estimates that 49,892 ÷ 0.89 is about 5,000. Is his estimate reasonable? Why or why not?
SVEN [57.7K]

Answer:yes

Step-by-step explanation:

because 0.89 is a very small number so itd be reasonable for it to be a large number

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Which angles are linear pairs. check all that apply
snow_tiger [21]

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The answer to the problem
larisa86 [58]

She puts 8% of her salary in, so multiply her pay by 8% to get her yearly amount she saves:

45000 x 0.08 = $3,600 per year.

The company puts 6% of that in the account:

3600 x 0.06 = $216

So per year 3600 + 216 = $3,816 is saved.

3,816 x 2 = $7,632 is the 2 year total.

The answer is A.

6 0
3 years ago
Mathematically verify the outlier(s) in the data set using the 1.5 rule.
statuscvo [17]

Given:

The data values are:

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

To find:

The outliers of the given data set.

Solution:

We have,

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

Divide the data set in two equal parts.

(7, 8, 11, 13, 14, 14), (14, 15, 16, 16, 18, 19)

Divide each parenthesis in two equal parts.

(7, 8, 11), (13, 14, 14), (14, 15, 16), (16, 18, 19)

Now,

Q_1=\dfrac{11+13}{2}

Q_1=\dfrac{24}{2}

Q_1=12

And

Q_3=\dfrac{16+16}{2}

Q_3=\dfrac{32}{2}

Q_3=16

The interquartile range is:

IQR=Q_3-Q_1

IQR=16-12

IQR=4

The data values lies outside the interval [Q_1-1.5IQR,Q_3+1.5IQR] are known as outliers.

[Q_1-1.5IQR,Q_3+1.5IQR]=[12-1.5(4),16+1.5(4)]

[Q_1-1.5IQR,Q_3+1.5IQR]=[12-6,16+6]

[Q_1-1.5IQR,Q_3+1.5IQR]=[6,22]

All the data values lie in the interval [6,22]. So, there are no outliers.

Hence, the correct option is 4.

3 0
3 years ago
In the video, the water level was 15 milliliters before adding the chain, and 20 milliliters after adding the chain. When I weig
mr_godi [17]

Answer:

  • 39.7%

Explanation:

<u />

<u>1. First find the density of your chain</u>

  • Density = mass / volume

  • Volume = displaced water volume

                      = Volume of Final level of water - initial level of water

                     = 20 ml - 15 ml = 5 ml

  • Mass = 66.7 g

  • Density = 66.7g / 5 ml = 13.34 g/ml

<u />

<u>2. Second, write the denisty of the chain as the weighted average of the densities of the other metals:</u>

Mass of gold × density of gold + mass of other metals × density of other metals, all divided by the mass of the chain.

Calling x the amount of gold, then the amount of other metals is 66.7 - x:

        \dfrac{19.3\cdot x+9.7\cdot (66.7-x)}{66.7}=13.34

       19.3x+635.66-9.7x=889.778

      9.6x=254.118\\\\x=26.47

Then, there are 26.47 grams of gold in 66.7 grams of chain, which yields a percentage of:

  • (26.47 / 66.7) × 100 = 39.7%
7 0
3 years ago
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