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2 years ago

What is the axis of symmetry of h(x) = 6x2 − 60x + 147?

Mathematics

2 answers:

Tema [17]2 years ago

8 0

Thank you for posting your math question here at brainly. I think the choices for the above problem is the below, it can be found elsewhere.

a. x = −5
b. x = −3
c. x = 3
d. x = 5

The answer is d. I hope it helps.

ruslelena [56]2 years ago

8 0

Answer-

The axis of symmetry is $x=5$

Solution-

The parabolic equation given is,

$h(x) = 6x^2-60x+147$

Here,

a = 6, b = -60, c = 147

The equation of the axis of symmetry is,

$x=-\dfrac{b}{2a}$

Putting the values,

$\Rightarrow x=-\dfrac{-60}{2\times 6}\\\\\Rightarrow x=\dfrac{60}{12}\\\\\Rightarrow x=5$

Therefore, the axis of symmetry is $x=5$

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Approximately 80,000 marriages took place in the state of New York last year. Estimate the probability that for at least one of

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Answer:

a)0,45119

b)1

Step-by-step explanation:

For part A of the problem we must first find the probability that both people in the couple have the same birthday (April 30)

$P=\frac{1}{365} *\frac{1}{365}=\frac{1}{133225} \\$

Now the poisson approximation is used

λ=nP=80000*1/133225=0,6

Now, let X be the number of couples that birth April 30

P(X ≥ 1) =

1 − P(X = 0) =

$1-\frac{(e^-0.6)*(-0,6)^{0} }{0!}$

P(X ≥ 1) = 0,45119

B) Now want to find the

probability that both partners celebrated their birthday on th, assuming that the year is 52 weeks and therefore 52 thursday

$P=52*\frac{1}{365} *\frac{1}{365}=\frac{52}{133225} \\$

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λ=nP=80000*52/133225=31.225

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2 years ago

Laplace Transform Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = [infinity] e−stf(t) dt 0 is said to be the

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Answer:

The laplace transform is $\frac{2}{(s+6)^3}$

Step-by-step explanation:

We will solve this problem by applying the laplace transform properties (their proofs are beyond the scope of this explanation).

Consider first the function f(t) = 1. By definition of the laplace transform, we have

$F(s) = \int_{0}^{\infty}f(t)e^{-st}dt$

when f(t) = 1 we get

$F(s) = \int_{0}^{\infty}e^{-st}dt = \left.\frac{-e^{-st}}{s}\right|_{0}^{\infty} = \frac{1}{s}$

We will apply the following properties: Define L(f) as applying the laplace transform

$L(e^{at}f(t)) = F(s-a)$ (this means, multiplying by an exponential corresponds to a shift in the s parameter of the transform of f)

$L(t^nf(t)) = (-1)^n\frac{d^n F}{ds^n}$ (this is, multypling by $t^n$ is equivalent to taking the n-th derivative of the transform.

We are given the function $g(t) = t^2e^{-6t}\cdot 1$ . Since the transform of the constant function 1 is 1/s, by applying the first property we get

$L(e^{-6t}\cdot 1 ) = \frac{1}{s+6}$

By applying the second property we get

$L(g(t)) = L(t^2 e^{-6t} \cdot 1) = (-1)^2\frac{d^2}{ds^2}(\frac{1}{s+6}) = \frac{2}{(s+6)^3}$

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2 years ago

Researchers wondered if there was a difference between males and females in regard to some common annoyances. They asked a rando

inysia [295]

Answer:

The proportions of the females and males who took the survey who are annoyed by the behavior in question are 0.3301 and_0.3791 respectively

D) The data come from a Population that is Normally distributed

F) the samples are independent

c)

Null hypothesis: H₀: $p_{m} ^{-} - p^{-} _{fm} \leq 0$

Alternative Hypothesis H₁: $p_{m} ^{-} - p^{-} _{fm} \geq 0$

Step-by-step explanation:

Given data Among the 530 males surveyed, 175 responded "Yes

The sample proportion of males

$p_{m} ^{-} = \frac{x}{n} = \frac{175}{530} = 0.3301$

The sample proportion of 575 Females surveyed, 218 responded "Yes

$p_{fem} ^{-} = \frac{x}{n} = \frac{218}{575} = 0.3791$

The proportions of the females and males who took the survey who are annoyed by the behavior in question are 0.3301 and_0.3791 respectively

The data come from a Population that is normally distributed.

The two samples are independent

Null hypothesis: H₀: $p_{m} ^{-} - p^{-} _{fm} \geq 0$

Alternative Hypothesis H₁: $p_{m} ^{-} - p^{-} _{fm} \leq 0$

We will use two samples Z - hypothesis test

$Z = \frac{p^{-} _{m} - p^{-} _{fem} }{se(p^{-} _{m}-p^{-} _{fem} ) }$

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$Se(p_{m} -p_{fem}) = \sqrt{\frac{0.3301(1-0.3301) }{530 }+\frac{0.3791(1-0.3791)}{575} }$

$Se(p_{m} ^{-} - p^{-} _{fem} ) = \sqrt{0.000826} = 0.0287$

The test statistic

$Z = \frac{0.3301-0.3791}{0.02875}$

Z = -1.7043

|Z| = |-1.7043|

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The calculated Z= 1.7043 < 1.96 at 0.05 level of significance

The null hypothesis is accepted

Conclusion:-

The evidence suggest not a higher proportion of females are annoyed by this behavior

4 0

2 years ago

22. In the diagram below WXZ is a right angle. If the measure of WXV is 11 more than there times the measure of WXY and the meas

Andreyy89

Given: < WXZ = 90 degrees ( Right angle is an angle of 90 degree measure).

m< WXV = 11 more than 3 times m< WXY = (3 * <WXY + 11)

m< YXV = 139 degree.

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Solution : m< WXZ is equal to 90 degrees and is also the sum of angles m<WXY and m<YXZ i.e.

m< WXZ = 90 degrees = m<WXY + m<YXZ -----(1)

m< WXV = (3 * <WXY + 11) -----(2)

m <YXV is equal to 139 degrees and also equal to the sum of angles m<WXY and m<WXV i.e.

m <YXV = 139 degree = m<WXY + m<WXV -----(3)

Substituting m< WXV = (3 * <WXY + 11) from equation (2) in equation (3)

139 degree = m<WXY + m<WXV

=> 139 = m<WXY + 3 * m<WXY + 11 (Subtracting 11 from both sides)

=> 139 -11 = m<WXY + 3 * m<WXY + 11 -11

=> 128 = 4<WXY (Adding m<WXY + 3 * m<WXY)

=> 32 = < WXY (Dividing both sides by 4)

Substituting < WXY = 32 from above in equation (1)

90 degrees = m<WXY + m<YXZ

90 = 32 + m<YXZ (Subtracting 32 from both sides)

90-32 = 32 + m<YXZ -32

58 = m<YXZ.

Therefore, <YXZ =58 degrees.

5 0

2 years ago

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bazaltina [42]

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2 years ago