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grandymaker [24]
3 years ago
12

A company with a fleet of 150 cars found that the emissions systems of only 5 out of the 22 they tested failed to meet pollution

control guidelines. The company initially believed that 20% of the fleet was out of compliance. Is this strong evidence the percentage of the fleet out of compliance is different from their initial thought? Your Question: State the null hypothesis and the alternative hypotheses they should use for completing a hypothesis test.
Mathematics
1 answer:
Ugo [173]3 years ago
3 0

Answer: No, the percentage of the fleet out of compliance is not different from their initial thought.

Step-by-step explanation:

Since we have given that

n = 22

x = 5

So, \hat{p}=\dfrac{x}{n}=\dfrac{5}{22}=0.23

he company initially believed that 20% of the fleet was out of compliance. Is this strong evidence the percentage of the fleet out of compliance is different from their initial thought.

so, p = 0.2

Hypothesis would be

H_0:p=\hat{p}\\\\H_a:p\neq \hat{p}

So, the t test statistic value would be

t=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\\\t=\dfrac{0.23-0.20}{\sqrt{\dfrac{0.2\times 0.8}{22}}}\\\\\\t=\dfrac{0.03}{0.085}\\\\t=0.353

Degree of freedom = df = n-1 = 22-1 =23

So, t{critical value} = 2.080

So, 2.080>0.353

so, we will accept the null hypothesis.

Hence, No, the percentage of the fleet out of compliance is not different from their initial thought.

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Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y
katrin2010 [14]

Answer:

The vertices are (3 , -5) , (-5 , -5)

The foci are (4 , -5) , (-6 , -5)

Step-by-step explanation:

* Lets study the equation of the hyperbola

- The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the x-axis is

  (x - h)²/a² - (y - k)²/b² = 1

- The length of the transverse axis is 2 a

- The coordinates of the vertices are  (h  ±  a  ,  k)

- The coordinates of the foci are (h ± c , k), where c² = a² + b²

- The distance between the foci is  2c

* Now lets solve the problem

- The equation of the hyperbola is (x + 1)²/16 - (y + 5)²/9 = 1

* From the equation

# a² = 16 ⇒ a = ± 4

# b² = 9 ⇒ b = ± 3

# h = -1

# k = -5

∵ The vertices are (h + a , k) , (h - a , k)

∴ The vertices are (-1 + 4 , -5) , (-1 - 4 , -5)

* The vertices are (3 , -5) , (-5 , -5)

∵ c² = a² + b²

∴ c² = 16 + 9 = 25

∴ c = ± 5

∵ The foci are (h ± c , k)

∴ The foci are (-1 + 5 , -5) , (-1 - 5 , -5)

* The foci are (4 , -5) , (-6 , -5)

4 0
3 years ago
Read 2 more answers
Solve |x| &gt; -9.<br><br> No solution <br><br> All reals<br><br> { x | x &lt; -9 or x &gt; 9}
neonofarm [45]

All reals is the answer.

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6 0
3 years ago
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pshichka [43]
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5 0
3 years ago
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mafiozo [28]

Answer:

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Step-by-step explanation:

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6 0
3 years ago
A baker made 20 pies. A Boy Scout troop buys one–fourth of his pies, a preschool teacher buys one–third of his pies, and a cater
n200080 [17]

Answer:

The number of pies does the baker have left is 15.

Step-by-step explanation:

Given : A baker made 20 pies. A Boy Scout troop buys one–fourth of his pies, a preschool teacher buys one–third of his pies, and a caterer buys one–sixth of his pies.

To find : How many pies does the baker have left?

Solution :

Let x be the number of pies does the baker have left.

According to question,

Number of pies buys,

One-fourth =\frac{1}{4}

One-third =\frac{1}{3}

One-sixth =\frac{1}{6}

i.e. x=20-(20\times \frac{1}{4}+20\times \frac{1}{3}+20\times \frac{1}{6})

x=20-(5+20\times \frac{1}{3}+10\times \frac{1}{3})

x=20-(5+\frac{20+10}{3})

x=20-(5+ \frac{30}{3})

x=20-(5+10)

x=20-(15)

x=5

The number of pies does the baker have left is 15.

7 0
3 years ago
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