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qwelly [4]
3 years ago
13

Locate the foci of ellipse. Show your work. X^2/36 +y^2/11 =1

Mathematics
1 answer:
Eva8 [605]3 years ago
6 0
<span>In this equation, 36>11 and so the ellipse has major axis along x axis
standard equation of ellipse with major axis is along x axis
x^2/a^2+y^2/b^2=1,
the foci are at (ae,0) and (–ae,0)
eccentricity e is given by
ae=√(a^2-b^2)

In this case ae=√(36-11)= √25=+/- 5

Focii are(-5,0) and(5,0) </span>
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Can someone pleaseeee help if you’re correct I’ll give u brainlist
nadezda [96]

Answer:

4

Step-by-step explanation:

this is an equilateral triangle, meaning all sides and angles are the same

5 0
2 years ago
3. IF ON = 5x - 7, LM = 4x +4, NM = x - 7, and OL = 3y – 4, find the values of x and y for which LMNO must
kkurt [141]

Answer:

x = 11 and y = \frac{8}{3}

Step-by-step explanation:

If LMNO is a parallelogram then it's opposite sides will be equal.

So, LM = ON

⇒ 4x + 4 = 5x - 7

⇒ x = 11

Now, MN = OL  

⇒ x - 7 = 3y - 4

⇒ 11 - 7 = 3y - 4 {Since we already have, x = 11}

⇒ y = \frac{8}{3} (Answer)

3 0
3 years ago
A recent estimate by a large distributor of gasoline claims that 60% of all cars stopping at their service stations chose unlead
Anton [14]

Answer:

We reject the null hypothesis and conclude that at least 2 proportions differ from the stated value.

Step-by-step explanation:

There are 3 types of gas listed in the question.

Thus;

n = 3

DF = n - 1

DF = 3 - 1

DF = 2

Let's state the hypotheses;

Null hypothesis; H0: P_regular = P_super unleaded = 20%; P_i leaded = 60%

Alternative hypothesis; Ha: At least 2 proportions differ from the stated value.

Observed values are;

Regular gas; O = 51

Unleaded gas; O = 261

Super Unleaded; O = 88

Total observed values = 51 + 261 + 88 = 400

We are told that super unleaded and regular were each selected 20% of the time and that unleaded gas was chosen 60% of the time.

Thus, expected values are;

Regular gas; E = 20% × 400 = 80

Unleaded gas; E = 60% × 400 = 240

Super Unleaded; E = 20% × 400 = 80

Formula for chi Square goodness of fit is;

X² = Σ[(O - E)²/E]

X² = (51 - 80)²/80) + (261 - 240)²/240) + (88 - 80)²/80)

X² = 13.15

From the chi Square distribution table attached and using; DF = 2 and X² = 13.15, we can trace the p-value to be approximately 0.001

Also from online p-value from chi Square calculator attached, we have p to be approximately 0.001 which is similar to what we got from the table.

Now, if we take the significance level to be 0.05, it means the p-value is less than it and thus we reject the null hypothesis and conclude that at least 2 proportions differ from the stated value.

4 0
3 years ago
Use the distributive property to expand the expression. −(3a3 − 4b2) (12x − 9y − 5)
aalyn [17]
The answer to that is -3a+4b+12x-9y-4
6 0
3 years ago
Why are 9th Graders still working on Fractions/Decimals/Percentages?
Georgia [21]

Answer:

Rigged

Step-by-step explanation:

The 9th grade system is rigged.

6 0
3 years ago
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