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goldfiish [28.3K]
3 years ago
11

A is a common factor of 72 and 108

Mathematics
2 answers:
elena-14-01-66 [18.8K]3 years ago
7 0

Answer:

36/20 = 9/5

Step-by-step explanation:

Factors of 72: 2×2×2×3×3

Factors of 108: 2×2×3×3×3

A: 2, ..., 36

Multiples of 4: 4, 8, 12, 16, 20..

Multiples of 10: 10, 20, 30..

B: 20, 40, ..

Highest fraction value when:

Denominator least

Numerator max

36/20

Olin [163]3 years ago
4 0

Answer:

36/20=9/5

Step-by-step explanation:

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Lucy went to the pet store to buy 17 fish for her aquarium. She bought Fancy Goldfish for $13.49 each and Plecos for $3.25 each.
tatuchka [14]

Step-by-step explanation:

let gold fish= X

let Plecos fish= Y

X +Y = 17................(I)

13.49X + 3.25Y =198.61........(II)

solve simuteniously to get

X = 14

Y = 3.....therefore

she bought 14 gold fish and 3 plecos fish

8 0
3 years ago
How would you solve this? Are there any steps to take?
Genrish500 [490]
You just have to find what x is. Pretend there isn't the signs and think of it as = signs. Then whatever u do on one side do it to all of them
7 0
4 years ago
The time it takes for a planet to complete its orbit around a particular star is called the? planet's sidereal year. The siderea
BartSMP [9]

Answer:

(a) See below

(b) r = 0.9879  

(c) y = -12.629 + 0.0654x

(d) See below

(e) No.

Step-by-step explanation:

(a) Plot the data

I used Excel to plot your data and got the graph in Fig 1 below.

(b) Correlation coefficient

One formula for the correlation coefficient is  

r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}

The calculation is not difficult, but it is tedious.

(i) Calculate the intermediate numbers

We can display them in a table.

<u>    x   </u>    <u>      y     </u>   <u>       xy     </u>    <u>              x²    </u>   <u>       y²    </u>

   36       0.22              7.92               1296           0.05

   67        0.62            42.21              4489           0.40

   93         1.00            93.00           20164           3.46

 433        11.8          5699.4          233289        139.24

 887      29.3         25989.1          786769       858.49

1785      82.0        146370          3186225      6724

2797     163.0         455911         7823209    26569

<u>3675 </u>  <u> 248.0  </u>    <u>   911400      </u>  <u>13505625</u>   <u> 61504        </u>

9965   537.81     1545776.75  25569715   95799.63

(ii) Calculate the correlation coefficient

r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}\\\\= \dfrac{9\times 1545776.75 - 9965\times 537.81}{\sqrt{[9\times 25569715 -9965^{2}][9\times 95799.63 - 537.81^{2}]}} \approx \mathbf{0.9879}

(c) Regression line

The equation for the regression line is

y = a + bx where

a = \dfrac{\sum y \sum x^{2} - \sum x \sum xy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\= \dfrac{537.81\times 25569715 - 9965 \times 1545776.75}{9\times 25569715 - 9965^{2}} \approx \mathbf{-12.629}\\\\b = \dfrac{n \sum xy  - \sum x \sum y}{n\sum x^{2}- \left (\sum x\right )^{2}} -  \dfrac{9\times 1545776.75  - 9965 \times 537.81}{9\times 25569715 - 9965^{2}} \approx\mathbf{0.0654}\\\\\\\text{The equation for the regression line is $\large \boxed{\mathbf{y = -12.629 + 0.0654x}}$}

(d) Residuals

Insert the values of x into the regression equation to get the estimated values of y.

Then take the difference between the actual and estimated values to get the residuals.

<u>    x    </u>   <u>      y     </u>   <u>Estimated</u>   <u>Residual </u>

    36        0.22        -10                 10

    67        0.62          -8                  9

    93        1.00           -7                  8

   142        1.86           -3                  5

  433       11.8             19               -  7

  887     29.3             45               -16  

 1785     82.0            104              -22

2797    163.0            170               -  7

3675   248.0            228               20

(e) Suitability of regression line

A linear model would have the residuals scattered randomly above and below a horizontal line.

Instead, they appear to lie along a parabola (Fig. 2).

This suggests that linear regression is not a good model for the data.

4 0
3 years ago
Can someone please help me<br> Please answer it right
Likurg_2 [28]

Answer:

25 more inches

Step-by-step explanation:

3 0
3 years ago
The high school band is selling wrapping paper. It costs $4 each week to advertise the wrapping paper. The band buys each roll o
Setler79 [48]
For this case, the first thing we must do is define variables:
 x: rolls of wrapping paper in one week
 We now write the equation that models the profit of the band.
 We have then:
 P (x) = (3.50 - 2) x - 4
 P (x) = 1.50x - 4
 Answer:
 
An equation to model the amount of profit, P, the band makes from x rolls of wrapping paper in one week is:
 
P (x) = 1.50x - 4
8 0
3 years ago
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