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Marina86 [1]
3 years ago
14

le="3 \sqrt{2} - 4 \div \sqrt{3} - 2" alt="3 \sqrt{2} - 4 \div \sqrt{3} - 2" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
ivann1987 [24]3 years ago
4 0
Simplify the following:
(3 sqrt(2) - 4)/(sqrt(3) - 2)

Multiply numerator and denominator of (3 sqrt(2) - 4)/(sqrt(3) - 2) by -1:
-(3 sqrt(2) - 4)/(2 - sqrt(3))

-(3 sqrt(2) - 4) = 4 - 3 sqrt(2):
(4 - 3 sqrt(2))/(2 - sqrt(3))

Multiply numerator and denominator of (4 - 3 sqrt(2))/(2 - sqrt(3)) by sqrt(3) + 2:
((4 - 3 sqrt(2)) (sqrt(3) + 2))/((2 - sqrt(3)) (sqrt(3) + 2))

(2 - sqrt(3)) (sqrt(3) + 2) = 2×2 + 2 sqrt(3) - sqrt(3)×2 - sqrt(3) sqrt(3) = 4 + 2 sqrt(3) - 2 sqrt(3) - 3 = 1:
((4 - 3 sqrt(2)) (sqrt(3) + 2))/1

((4 - 3 sqrt(2)) (sqrt(3) + 2))/1 = (4 - 3 sqrt(2)) (sqrt(3) + 2):
Answer: (4 - 3 sqrt(2)) (sqrt(3) + 2)
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A recent survey of
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Answer:

The total number of students in a survey is 300.

Let the number of junior male(JM) be x and the number of senior males(SM) be y.

Let the number of junior female(JF) be p and the number of senior males(SF) be q.

It is given that there are 160 males, 80 junior females, 130 seniors.

Since number of males are 160. So the number of females are,

300-160=140

Since number of junior females is 80.

JF+SF=140\\80+SF=140\\SF=60

Since number of seniors are 130.

SM+SF=130\\SM+60=130\\SM=70

Since number of males is 160.

JM+SM=160\\JM+70=160\\JM=90

Therefore, the table and venn diagram is shown below.

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3 years ago
Identity the range of the function shown in the graph.
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Answer:

C) Y > 0

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Answer: the second one

Step-by-step explanation:

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Wesley is traveling from California to Utah. His trip covers a distance of 673.1 miles. For the first 3 hours, he travels at an
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4 0
3 years ago
Suppose that in a random sample of 300 employed Americans, there are 57 individuals who say that they would fire their boss if t
otez555 [7]

Answer:

The 95% confidence interval for the population proportion is (0.1456, 0.2344). This means that we are 95% sure that the true proportion of employed American who say that they would fire their boss if they could is between 0.1456 and 0.2344.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 300, \pi = \frac{57}{300} = 0.19

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.19 - 1.96\sqrt{\frac{0.19*0.81}{300}} = 0.1456

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.19 + 1.96\sqrt{\frac{0.19*0.81}{300}} = 0.2344

The 95% confidence interval for the population proportion is (0.1456, 0.2344). This means that we are 95% sure that the true proportion of employed American who say that they would fire their boss if they could is between 0.1456 and 0.2344.

4 0
2 years ago
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