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Dafna11 [192]
2 years ago
11

5: will give brainliest

Mathematics
1 answer:
Fiesta28 [93]2 years ago
6 0

The focal length of the given ellipse is given as (±6, 0)

<h3>Equation of an ellipse</h3>

An ellipse is defined as a regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant or when a cone is cut by an oblique plane which does not intersect the base.

The standard equation of an ellipse is expressed as;

x^2/a^2 + y^2/b^2 = 1

The formula for calculating the focus of the ellipse is given as:

c^2 = b^2 - a^2

Given the equation of an ellipse

(x-7)^2/64 + (y-5)^2/100 = 1

This can also be expressed as:

(x-7)^2/8^2 + (y-5)^2/10^2 = 1

Comparing with the general equation

a = 8 and b = 10

Substitute

c^2 = 10^2 - 8^2

c^2 = 100 - 64

c^2 = 36

c = 6

Hence the focal length of the given ellipse is given as (±6, 0)

Learn more on focus of ellipse here; brainly.com/question/4429071

#SPJ1

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(8000 - 5000) / (200 - 100) = $30

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A company with a fleet of 150 cars found that the emissions systems of only 4 out of the 25 they tested failed to meet pollution
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Answer:

No, there is no strong evidence that the percentage of the fleet out of compliance is different from their initial thought.

Step-by-step explanation:

We are given that a company with a fleet of 150 cars found that the emissions systems of only 4 out of the 25 they tested failed to meet pollution control guidelines.

The company initially believed that 30% of the fleet was out of compliance.

<u><em>Let p = percentage of the fleet that was out of compliance.</em></u>

SO, Null Hypothesis, H_0 : p = 30%   {means that the percentage of the fleet out of compliance is same as their initial thought}

Alternate Hypothesis, H_A : p \neq 30%   {means that the percentage of the fleet out of compliance is different from their initial thought}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

                                 T.S.  = \frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }  ~ N(0,1)

where, \hat p = percentage of the fleet out of compliance = \frac{4}{25} = 16%

           n = sample of systems tested = 25

So, <u><em>test statistics</em></u>  =  \frac{0.16-0.30}{{\sqrt{\frac{0.16(1-0.16)}{25} } } } }

                              =  -1.909

The value of the test statistics is -1.909.

<em>Since in the question we are not given with the level of significance at which hypothesis can be tested, so we assume it to be 5%. Now at 5% significance level, </em><u><em>the z table gives critical values between -1.96 and 1.96 for two-tailed test.</em></u><em> </em>

<em>Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.</em>

Therefore, we conclude that the percentage of the fleet out of compliance is same as their initial thought.

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Initially a pool contains 350 gallons of water. A hose is placed in the pool and the water is turned on. The hose adds 5.2 gallo
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Answer: 350 + 5.2x = V

Step-by-step explanation:

350 + 5.2x = V

5.2x represents adding 5.2 gallons per minuet

350 means adding the initial 350 gallons

= V represents the total amount of water

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Essentially, the $35 is the fixed cost associated with the rental and the $15 is the variable cost to do so. If this were in a linear equation format it would take the form: y = 15x + 35.

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