MATHEMATICS HIGH SCHOOL

Sin^2(x)-cos^2(x)=0
Find all the solutions in the interval [0,2pi)
Plz help.

Answers

Answer 1
Answer: Hello,
as sin² x -cos² x=cos 2x
The equation is
cos 2x=0
==>2x=π/2+kπ
==>x=π/4+kπ/2 with k integer


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Write 2/3 and 3/4 as a pair of fractions with a common denominator

Answers

in short simply we "multiply one fraction's top and bottom by the other's denominator", so let's do so,

\bf \cfrac{2\cdot 4}{3\cdot 4}\implies \boxed{\cfrac{8}{12}}\qquad \qquad \qquad \qquad \qquad \cfrac{3\cdot 3}{4\cdot 3}\implies \boxed{\cfrac{9}{12}}

Write the ratio as a fraction in simplest form, with whole numbers in the numerator and denominator.12 kg to 28 kg

Answers

Answer:

28/12

Step-by-step explanation:

(x − 2) is a factor of x4 + 2x3 − 7x2 − 8x + 12.

Answers

(x + 3) • (x + 2) • (x - 1) • (x - 2)

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 247 days and standard deviation sigma equals 16 days. Complete parts​ (a) through​ (f) below.

Answers

Answer:

The answer is given below

Step-by-step explanation:

a) What is the probability that a randomly selected pregnancy lasts less than 242 days

First we have to calculate the z score. The z score is used to determine the measure of standard deviation by which the raw score is above or below the mean. It is given by:

z=(x-\mu)/(\sigma)

Given that Mean (μ) = 247 and standard deviation (σ) = 16 days. For x < 242 days,

z=(x-\mu)/(\sigma)=(242-247)/(16)=-0.31

From the normal distribution table, P(x < 242) = P(z < -0.3125) = 0.3783

(b) Suppose a random sample of 17 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies.

If a sample of 17 pregnancies is obtained, the new mean \mu_x=\mu=247, the new standard deviation: \sigma_x=\sigma/√(n) =16/√(17) =3.88

c) What is the probability that a random sample of 17 pregnancies has a mean gestation period of 242 days or less

z=(x-\mu)/(\sigma/√(n) )=(242-247)/(16/√(17) )=-1.29

From the normal distribution table, P(x < 242) = P(z < -1.29) = 0.0985

d) What is the probability that a random sample of 49 pregnancies has a mean gestation period of 242 days or less?

z=(x-\mu)/(\sigma/√(n) )=(242-247)/(16/√(49) )=-2.19

From the normal distribution table, P(x < 242) = P(z < -2.19) = 0.0143

(e) What might you conclude if a random sample of 49 pregnancies resulted in a mean gestation period of 242 days or less?

It would be unusual if it came from mean of 247 days

f) What is the probability a random sample of size 2020 will have a mean gestation period within 11 days of the mean

For x = 236 days

z=(x-\mu)/(\sigma/√(n) )=(236-247)/(16/√(20) )=-3.07

For x = 258 days

z=(x-\mu)/(\sigma/√(n) )=(258-247)/(16/√(20) )=3.07

From the normal distribution table, P(236 < x < 258) = P(-3.07 < z < 3.07) = P(z < 3.07) - P(z < -3.07) =0.9985 - 0.0011 = 0.9939

If p varies directly with q, and p = 10 when q = 5, what is the value of p when q = 20? A.p = 40

B.p = 1

C.p = 100

D.p = 28

Answers

A. p=40 because if you look at your question carefully q is half of p so if q=20 which is half of 40 then that means p=40

If f(x) = 1/9 x-2 , what is f–1(x)?

Answers

Answer:

f^(-1)(x)=9x+2

Step-by-step explanation:

Given: f(x)=(1)/(9)(x-2)

We need to find the inverse of f(x).

Step 1: Set f(x) = y

y=(1)/(9)(x-2)

Step 2: Switch x and y

x=(1)/(9)(y-2)

Step 3: Solve for y, Isolate y

x=(1)/(9)(y-2)

Multiply by 9 both sides

9x=y-2

Add 2 both sides

9x+2=y

y=f^(-1)(x)

Hence, f^(-1)(x)=9x+2
Hello,

y=1/9 x-2
==>x=1/9 y -2
==>9x=y-18
==>y=9x+18

f^(-1)(x)=9x+18
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