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mario62 [17]
2 years ago
6

The center of a hyperbola is located at (0, 0). One focus is located at (0, 5) and its associated directrix is represented by th

e line y = 9/5 . Given the standard form of the equation of a hyperbola, what are the values of a and B
y^2/a^2 + x^2/b^2 = 1
A=
B=
Mathematics
2 answers:
Lady bird [3.3K]2 years ago
8 0
For a hyperbola   \dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1
where   a^{2}+b^{2}=c^{2}
the directrix is the line   y=\dfrac{a^{2}}{c}
and the focus is at (0, c).

Here, we have c = 5, a² = 9, so b² = 5² - 9 = 16.
  a = √9 = 3
  b = √16 = 4

Your hyperbola's constants are ...
  a = 3
  b = 4


______
Please note that the equation of a hyperbola has a negative sign for one of the terms. The equation given in your problem statement is that of an ellipse.
tresset_1 [31]2 years ago
4 0

Answer:

a=3, b=4

Step-by-step explanation:

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Marin writes the functions m(x) = StartFraction 3 x Over x + 7 EndFraction and n(x) = StartFraction 7 x Over 3 minus x EndFracti
Svetlanka [38]

Answer:

B.

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Here are the ingredients to make 12 cupcakes.
scoundrel [369]

Answer:

Mark needs 2 kg butter, 2 kg caster sugar, 14 eggs, and 1.5 kg flour.

Step-by-step explanation:

Alright, first you double the number of kids, as Mark wants each kid to have two cupcakes, not one. (Then add 80+80 to get 160)

Now add the adults to children: 160+152=312

now divide by 12 to see how many batches of cupcakes he needs to make.

312÷12=26 So he needs to make 26 batches

Now here are the ingredients needed to make 12 batches of cupcakes (I multiplied the amount of each ingredient by 26)

5200 g (5.2 kg) butter, 5200 g (5.2 kg) caster sugar, 104 eggs, and 6500 g (6.5 kg) flour.

now make the measurements the same as Mark's ingredients (or the other way around)

(to save time I am not showing each one, just the normal ingredients turned into kg and compared)

5 kg butter < 5.2 kg butter

5 kg caster sugar < 5.2 kg butter

90 eggs < 104 eggs

5 kg flour < 6.5 kg flour

Now do a little subtraction.....

5.2 kg-5 kg = 2 kg butter is needed....

5.2 kg - 5 kg = 2 kg caster sugar is needed....

104 - 90 = 14 eggs needed....

6.5 kg - 5 kg = 1.5 kg flour needed....

And there you go!!! Hope this helped!!! :D

4 0
2 years ago
Estimates show that there are 1.4x10^8 pet fish and 9.4x10^6 pet reptiles in the United States. How many pet fish and reptile pe
azamat

Answer:

There are total number of pet fish and reptile pets are 1.49\times 10^8.

Step-by-step explanation:

Number of pet fish are 1.4\times 10^8 and the number of pet reptiles are 9.4\times 10^6.

It is required to find the total number of pet fish and reptile pets in the United States. It can be done by simply adding the above numbers such that total numbers are given by :

T=1.4\times10^{8}+9.4\times10^{6}\\\\T=149400000\\\\T=1.49\times 10^8

So, there are total number of pet fish and reptile pets are 1.49\times 10^8.

8 0
2 years ago
"An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particul
Galina-37 [17]

Answer:

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15) = 0.0173

b) Probability of not rejecting the claim when p = 0.7, P(X > 15) = 0.8106

when p = 0.6, P(X > 15) = 0.4246

c) Check Explanation

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Step-by-step explanation:

p is the true proportion of houses with smoke detectors and p = 0.80

The claim that 80% of houses have smoke detectors is rejected if in a sample of 25 houses, not more than 15 houses have smoke detectors.

If X is the number of homes with detectors among the 25 sampled

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15)

This is a binomial distribution problem

A binomial experiment is one in which the probability of success doesn't change with every run or number of trials (probability that each house has a detector is 0.80)

It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure (we are sampling 25 houses with each of them either having or not having a detector)

The outcome of each trial/run of a binomial experiment is independent of one another.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = less than or equal to 15

p = probability of success = probability that a house has smoke detectors = 0.80

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.80 = 0.20

P(X ≤ 15) = Sum of probabilities from P(X = 0) to P(X = 15) = 0.01733186954 = 0.01733

b) Probability of not rejecting the claim when p= 0.7 when p= 0.6

For us not to reject the claim, we need more than 15 houses with detectors, hence, th is probability = P(X > 15), but p = 0.7 and 0.6 respectively for this question.

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = more than 15

p = probability that a house has smoke detectors = 0.70, then 0.60

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.70 = 0.30

And 1 - 0.60 = 0.40

P(X > 15) = sum of probabilities from P(X = 15) to P(X = 25)

When p = 0.70, P(X > 15) = 0.8105639765 = 0.8106

When p = 0.60, P(X > 15) = 0.42461701767 = 0.4246

c) How do the "error probabilities" of parts (a) and (b) change if the value 15 in the decision rule is replaced by 14.

The error probabilities include the probability of the claim being false.

When X = 15

(Error probability when p = 0.80) = 0.0173

when p = 0.70, error probability = P(X ≤ 15) = 1 - P(X > 15) = 1 - 0.8106 = 0.1894

when p = 0.60, error probability = 1 - 0.4246 = 0.5754

When X = 14

(Error probability when p = 0.80) = P(X ≤ 14) = 0.00555

when p = 0.70, error probability = P(X ≤ 14) = 0.0978

when p = 0.60, error probability = P(X ≤ 14) = 0.4142

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Hope this Helps!!!

6 0
2 years ago
Without properly working spark plugs, a vehicle will not run. For a specific vehicle, the spark plugs are supposed to have a gap
Salsk061 [2.6K]

Answer:

<em>The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.</em>

<em>P(3.9≤X≤4.3)  = 0.9922</em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the mean of the Normal distribution = 4.1mm</em>

<em>Given that the standard deviation of the Normal distribution = 0.0075mm</em>

<em>Let 'X' be the random variable in a normal distribution</em>

Given that X₁ = 3.9mm

Z_{1} = \frac{X_{1} -mean}{S.D}

Z_{1} = \frac{3.9 -4.1}{0.075} =  -2.666

Given that X₂ = 4.3mm

Z_{2} = \frac{X_{2} -mean}{S.D}

Z_{1} = \frac{4.3 -4.1}{0.075} =  2.666

<u><em>Step(ii):-</em></u>

<em>The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.</em>

<em>P(3.9≤X≤4.3) = P(-2.666≤Z≤2.666)</em>

<em>                      = P(Z≤2.666)-P(Z≤-2.666)</em>

<em>                     = 0.5 +A(2.666) - (0.5-A(2.666)</em>

<em>                    = 2 × A(2.666)</em>

<em>                  = 2×0.4961</em>

<em>                  = 0.9922</em>

<u><em>Final answer:-</em></u>

<em>The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.</em>

<em>P(3.9≤X≤4.3)  = 0.9922</em>

<em></em>

3 0
1 year ago
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