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2 years ago

12

Rewrite the equation by completing the square x^2-4x=0

1 answer:

Dmitriy789 [7]2 years ago

7 0

Answer:

(x+2)(x-2) = 0

Step-by-step explanation:

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2 years ago

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2 years ago

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Which expression is represented by the number line?

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2 years ago

Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that

aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

$A\cap B=$ {HH}

$B\cap C$ ={HH}

$A\cap C$ ={HH}

P(E)= $\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}$

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

$P(A)=\frac{2}{4}=\frac{1}{2}$

Number of favorable cases to event B=2

Number of favorable cases to event C=2

$P(B)=\frac{2}{4}=\frac{1}{2}$

$P(C)=\frac{2}{4}=\frac{1}{2}$

If the two events A and B are independent then

$P(A)\cdot P(B)=P(A\cap B)$

$P(A\cap)=\frac{1}{4}$

$P(B\cap C)=\frac{1}{4}$

$P(A\cap C)=\frac{1}{4}$

$P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

$P(B)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(B)=P(A\cap B)$

Therefore, A and B are independent

$P(B)\cdot P(C)=P(B\cap C)$

Therefore, B and C are independent

$P(A\cap C)=P(A)\cdot P(C)$

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

6 0

2 years ago

The daily average temperature in Santiago, Chile, varies over time in a periodic way that can be modeled approximately by a trig

natali 33 [55]

Answer:

a) the trigonometric function is;

$y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5$

b) $y = 28.36^0 \ C$ ( to two decimal places)

Step-by-step explanation:

This data can be represented by the sinusoidal function of the form :

$\mathbf{y = A sin (Bt -C)+D}$

where A = amplitude and which can be determined via the formula:

$A = \dfrac{largest \ temperature - lowest \ temperature}{2}$

$A = \dfrac{29-14}{2}$

$A = \dfrac{15}{2}$

A = 7.5° C

where B = the frequency;

Since the data covers a period of 3 days ; then $\dfrac{2 \pi}{B } =365$

$B = \dfrac{2 \pi}{365}$ ( where 365 is the time period )

The vertical shift is found by the equation D;

D = $\frac{largest \ temperature + lowest \ temperature}{2}$

D = $\frac{29+14}{2}$

D = 21.5

Replacing the values of A ; B and D into the above sinusoidal function; we have :

$y = 7.5 sin (\frac{2 \pi}{365}t -C) + 21.5$

From the question; when it is 7th of the year ( i.e January 7);

t = 7 and the temperature (y) = 29° C

replacing that too into the above equation; we have:

$29= 7.5 sin (\frac{2 \pi}{365}*7 -C) + 21.5$

$29= 7.5 sin (\frac{14 \pi}{365} -C) + 21.5$

$\frac{29-21.5}{7.5}= sin (\frac{14 \pi}{365} -C)$

$1= sin (\frac{14 \pi}{365} -C)$

$sin^{-1}(1)= (\frac{14 \pi}{365} -C)$

$\frac{\pi}{2}= (\frac{14 \pi}{365} -C)$

$C= (\frac{14 \pi}{365} -\frac{\pi}{2})$

$C= (\frac{28 \pi- 365 \pi}{730} )$

$C= \frac{-337 \pi}{730}$

Thus; the trigonometric function is;

$y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5$

Similarly; to determine the temperature o Jan 31; i.e when t= 31 ; we have :

$y = 7.5 sin ( \frac{2 \pi}{365}*31+ \frac{337 \pi}{730})+ 21.5$

$y = 7.5 sin ( \frac{62 \pi}{365}+ \frac{337 \pi}{730})+ 21.5$

$y = 7.5 sin ( \frac{124 \pi+ 337 \pi }{730})+ 21.5$

$y = 7.5 sin ( \frac{461 \pi }{730})+ 21.5$

$y = 7.5 *( 0.915)+ 21.5$

$y = 6.8689+ 21.5$

$y = 28.36^0 \ C$ ( to two decimal places)

7 0

2 years ago

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