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Alex73 [517]
3 years ago
7

Find the solution to the system of equations.

Mathematics
1 answer:
frutty [35]3 years ago
8 0
(-1, 1)
Find the coordinates of the intersection between the lines.
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The number of red tiles in a sack is 5 more than 3 times the number of green tiles in the sack. The
lys-0071 [83]

Answer:

B. 10

Step-by-step explanation:

10 x 3 = 30+5 = 35

4 0
3 years ago
Suppose that you begin saving up to buy a car by depositing a certain amount at the end of each month in a savings account which
SpyIntel [72]

Answer:

$2,221.6 monthly

Step-by-step explanation:

A = P(1 + r)^n

A is the total amount I intend to save = $15,000

r is the yearly interest rate = 3.6% = 0.036

n is the duration to achieve my goal = 4 and 1/2 years = 54 months

15,000 = P(1 + 0.036)^54

15,000 = P(1.036)^54

P = 15,000/6.752 = 2,221.6

I need to put $2,221.6 into the savings account monthly

6 0
3 years ago
Read 2 more answers
Product of (2+x) and (2-x) is​
mafiozo [28]

Use FOIL,

(2+x)(2-x)=4-2x+2x-x^2=4-x^2.

8 0
3 years ago
Read 2 more answers
What is adjacents definition
marishachu [46]

Answer:

next to or adjoining something else.

or

GEOMETRY

(of angles) having a common vertex and a common side.

5 0
3 years ago
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The oil prices for 2014, rounded to the nearest dollar, were: 95, 101, 101, 102, 102, 106, 104, 97, 93, 84, 76, 59. what is the
kipiarov [429]
  • Interquartile Range (IQR) = Q_3-Q_1 , with Q_3 as the upper quartile and Q_1 as the lower quartile.

Firstly, rearrange the data so that it's in ascending order: \{59,76,84,93,95,97,101,101,102,102,104,106\}

Next, find the median:

\{59,76,84,93,95,\boxed{97,101,}101,102,102,104,106\}\\\\\frac{97+101}{2}\\\\\frac{198}{2}\\\\99

Now to find the lower quartile, find the "median" of the data set that's to the left of 99:

\{\overbrace{59,76,84,93,95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\{\overbrace{59,76,\boxed{84,93}95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\frac{84+93}{2}\\\\\frac{177}{2}\\\\88.5

Now to find the upper quartile, it's the similar process as finding the lower quartile, except that you are finding the "median" of the data set to the right of 99:

\{59,76,84,93,95,97,\overbrace{101,101,102,102,104,106}^{\textsf{to the right of the median}}\}\\\\\{59,76,84,93,95,97,\overbrace{101,101,\boxed{102,102} 104,106}^{\textsf{to the right of the median}}\}\\\\\frac{102+102}{2}\\\\\frac{204}{2}\\\\102

Now that we have the upper and lower quartile, subtract them:

102-88.5=13.5

<u>In short, the IQR of this data set is 13.5.</u>

8 0
3 years ago
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