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3 years ago

How to do this question by proving the identity?

Mathematics

1 answer:

jeyben [28]3 years ago

6 0

$\sin^3 \theta-\cos^3 \theta=(\sin \theta-\cos \theta)(\sin \theta\cos \theta+1)\\\\
(\sin \theta-\cos \theta)(\sin^2 \theta+\sin \theta\cos \theta +\cos^2 \theta)=(\sin \theta-\cos \theta)(\sin \theta\cos \theta+1)\\\\
(\sin \theta-\cos \theta)(\sin \theta\cos \theta+1)=(\sin \theta-\cos \theta)(\sin \theta\cos \theta+1)\\\\$

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What is the range of the values shown on the table? What does the range represent?

hjlf

Answer:

Option D)R: {0 ≤ y ≤ 360}; The range represents the number of miles the car can travel

Step-by-step explanation:

The table in the attached figure

Let

x -----> the amount of gas used in gallons (independent variable)

y ----> the number of miles the car can travel (dependent variable)

In this problem

The domain is the interval -----> [0,12]

$0\ gal \leq x \leq 12\ gal$

The range is the interval ----> [0,360]

$0\ miles \leq y \leq 360\ miles$

3 0

3 years ago

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago

Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show

Dovator [93]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that there exist 60 or more defected children is $P(x \ge 60)=0.0901$

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is $n= 50,000$

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

$= 0.001$

The probability of not observing the defect is mathematically evaluated as

$q = 1-p$

$= 1-0.001$

$= 0.999$

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

$= 50$

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

$= 7.07$

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

$z = \frac{x-\mu }{\sigma }$

$= \frac{59.5 - 50 }{7.07}$

$=1.34$

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

$P(x \ge 60) \approx P(z > 1.34)$