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harina [27]
2 years ago
8

I need help plz quickly!!:)

Mathematics
1 answer:
galben [10]2 years ago
6 0

Answer:

the first option (2x+3y=-6)

Step-by-step explanation:

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ASAP!!! ILL GIVE BRAINLIEST
Ostrovityanka [42]

Answer:

The correct options are 2 and 4.

Step-by-step explanation:

From the given box plot it is clear that

\text{Minimum value}=20

Q_1=25

Median=40

Q_3=50

\text{Maximum value}=110

We know that these number divides the data in four equal parts.

Q_1=25\%

Median=50%

Q_3=75\%

25% of the data values lies between 50 and 110. Therefore option 1 is incorrect.

Seventy-five percent of the data values lies between 20 and 50. Therefore option 2 is correct.

It is unlikely that there are any outliers. This statement is not true because the is a huge difference between third quartile and maximum value.

Therefore option 3 is incorrect.

The interquartile range is

IQR=Q_3-Q_1=50-25=25

Therefore option 4 is correct.

The range is

Range = Maximum-Minimum

Range=110-20=90

Therefore option 5 is incorrect.

4 0
3 years ago
When are two distinct lines parallel?
sergey [27]

Answer:

if the line keeps going and theynever touch

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
If a math test has questions on spelling, the test does not<br><br> have this type of validity.
Arada [10]

Answer:

yes

Step-by-step explanation:

because its true

8 0
2 years ago
Read 2 more answers
X to the 2nd power minus 6 equals 1
alexandr402 [8]

Answer:

2.646

Step-by-step explanation:

You have to 6 to the 1 and then take the square root of the number which is seven leaving you with 2.646

3 0
2 years ago
Read 2 more answers
300 ml of pure alcohol is poured from a bottle containing 2 l of pure alcohol. Then, 300 ml of water is added into the bottle. A
Ede4ka [16]

Answer:

The present percentage of pure alcohol in the solution is 72.25% of pure alcohol

Step-by-step explanation:

The volume of pure alcohol poured from the 2 l bottle of pure alcohol = 300 ml of pure alcohol

The volume of water added into the bottle after pouring out the pure alcohol = 300 ml of water

The volume of diluted alcohol poured out of the bottle = 300 ml of diluted alcohol

The volume of water added into the bottle of diluted alcohol after pouring out the 300 ml of diluted alcohol = 300 ml of water

Step 1

After pouring the 300 ml of pure alcohol and adding 300 ml of water to the bottle, the percentage concentration, C%₁ is given as follows;

C%₁ = (Volume of pure alcohol)/(Total volume of the solution) × 100

The volume of pure alcohol in the bottle = 2 l - 300 ml = 1,700 ml

The total volume of the solution = The volume of pure alcohol in the bottle +  The volume of water added = 1,700 ml + 300 ml = 2,000 ml = 2 l

∴ C%₁ = (1,700 ml)/(2,000 ml) × 100 = 85% percent alcohol

Step 2

After pouring out 300 ml diluted alcohol from the 2,000 ml, 85% alcohol and adding 300 ml of water, we have;

Volume of 85% alcohol = 2,000 ml - 300 ml = 1,700 ml

The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol = 85/100 × 1,700 = 1,445 ml

The total volume of the diluted solution = The volume of the 85% alcohol in the solution + The volume of water added

∴ The total volume of the twice diluted solution = 1,700 ml + 300 ml = 2,000 ml

The present percentage of pure alcohol in the solution, C%₂ = (The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol)/(The total volume of the diluted solution) × 100

∴ C%₂ = (1,445 ml)/(2,000 ml) × 100 = 72.25 %

The present percentage of pure alcohol in the solution, C%₂ = 72.25%

3 0
3 years ago
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