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3 years ago

Find the circumference of a circle with a radius of 5 cm. Round your

Mathematics

1 answer:

Verizon [17]3 years ago

7 0

31.4
Hope that helps you!

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The proportion of households in a region that do some or all of their banking on the Internet is 0.31. In a random sample of 100

Alenkasestr [34]

Answer:

Approximate probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is less than 0.0005% .

Step-by-step explanation:

We are given that let X be the number that do some or all of their banking on the Internet.

Also; Mean, $\mu$ = 310/1000 or 0.31 and Standard deviation, $\sigma$ = 14.63/1000 = 0.01463 .

We know that Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

Probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is given by P(X <= 130/1000);

P(X <=0.13) = P( $\frac{X-\mu}{\sigma}$ <= $\frac{0.13-0.31}{0.01463}$ ) = P(Z <= -12.303) = P(Z > 12.303)

Since this value is not represented in the z table as the value is very high and z table is limited to x = 4.4172.

So, after seeing the table we can say that this probability is approximately less than 0.0005% .

4 0

3 years ago

Please help me !! due today:(

raketka [301]

Answer:

$\huge\boxed{\sf x = 65 \textdegree}$

Step-by-step explanation:

Using trigonometric ratio cos.

Here,

Theta = θ = x
Adjacent = 12
Hypotenuse = 28

<h3><u>Finding x:</u></h3>

$\displaystyle cos \theta =\frac{adjacent}{hypotenuse} \\\\cos \ x =\frac{12}{28} \\\\cos \ x = 0.43\\\\x = cos^{-1} 0.43\\\\x = 65 \textdegree\\\\\rule[225]{225}{2}$

7 0

2 years ago

1. The volume of box A is 64 cubic inches. The circumference of a regulation baseball is at least 9 inches but no more than 9. 2

likoan [24]

Answer:

look it up its there

Step-by-step explanation:

look up

4 0

3 years ago

OMG PLEASE HELP! THIS ASSIFNMENT DUE IN LIKE 5 MINUTES!! (Sorry if blurry pic)

Ber [7]

Answer:

I think the 3rd answer

Step-by-step explanation:

3 0

3 years ago

The expressions A, B, C, D, and E are left-hand sides of trigonometric identities. The expressions 1, 2, 3, 4, and 5 are right-h

Semenov [28]

Answer:

$A.\ \tan(x) \to 2.\ \sin(x) \sec(x)$

$B.\ \cos(x) \to 5. \sec(x) - \sec(x)\sin^2(x)$

$C.\ \sec(x)csc(x) \to 3. \tan(x) + \cot(x)$

$D. \frac{1 - (cos(x))^2}{cos(x)} \to 1. \sin(x) \tan(x)$

$E.\ 2\sec(x) \to\ 4.\ \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}$

Step-by-step explanation:

Given

$A.\ \tan(x)$

$B.\ \cos(x)$

$C.\ \sec(x)csc(x)$

$D.\ \frac{1 - (cos(x))^2}{cos(x)}$

$E.\ 2\sec(x)$

Required

Match the above with the appropriate identity from

$1.\ \sin(x) \tan(x)$

$2.\ \sin(x) \sec(x)$

$3.\ \tan(x) + \cot(x)$

$4.\ \frac{cos(x)}{1 - sin(x)} + \frac{1 - \sin(x)}{cos(x)}$

$5.\ \sec(x) - \sec(x)(\sin(x))^2$

Solving (A):

$A.\ \tan(x)$

In trigonometry,

$\frac{sin(x)}{\cos(x)} = \tan(x)$

So, we have:

$\tan(x) = \frac{\sin(x)}{\cos(x)}$

Split

$\tan(x) = \sin(x) * \frac{1}{\cos(x)}$

In trigonometry

$\frac{1}{\cos(x)} =sec(x)$

So, we have:

$\tan(x) = \sin(x) * \sec(x)$

$\tan(x) = \sin(x) \sec(x)$ --- proved

Solving (b):

$B.\ \cos(x)$

Multiply by $\frac{\cos(x)}{\cos(x)}$ --- an equivalent of 1

So, we have:

$\cos(x) = \cos(x) * \frac{\cos(x)}{\cos(x)}$

$\cos(x) = \frac{\cos^2(x)}{\cos(x)}$

In trigonometry:

$\cos^2(x) = 1 - \sin^2(x)$

So, we have:

$\cos(x) = \frac{1 - \sin^2(x)}{\cos(x)}$

Split

$\cos(x) = \frac{1}{\cos(x)} - \frac{\sin^2(x)}{\cos(x)}$

Rewrite as:

$\cos(x) = \frac{1}{\cos(x)} - \frac{1}{\cos(x)}*\sin^2(x)$

Express $\frac{1}{\cos(x)}\ as\ \sec(x)$

$\cos(x) = \sec(x) - \sec(x) * \sin^2(x)$

$\cos(x) = \sec(x) - \sec(x)\sin^2(x)$ --- proved

Solving (C):

$C.\ \sec(x)csc(x)$

In trigonometry

$\sec(x)= \frac{1}{\cos(x)}$

and

$\csc(x)= \frac{1}{\sin(x)}$

So, we have:

$\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)}$

Multiply by $\frac{\cos(x)}{\cos(x)}$ --- an equivalent of 1

$\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)} * \frac{\cos(x)}{\cos(x)}$

$\sec(x)csc(x) = \frac{1}{\cos^2(x)}*\frac{\cos(x)}{\sin(x)}$

Express $\frac{1}{\cos^2(x)}\ as\ \sec^2(x)$ and $\frac{\cos(x)}{\sin(x)}\ as\ \frac{1}{\tan(x)}$

$\sec(x)csc(x) = \sec^2(x)*\frac{1}{\tan(x)}$

$\sec(x)csc(x) = \frac{\sec^2(x)}{\tan(x)}$

In trigonometry:

$tan^2(x) + 1 =\sec^2(x)$

So, we have:

$\sec(x)csc(x) = \frac{\tan^2(x) + 1}{\tan(x)}$

Split

$\sec(x)csc(x) = \frac{\tan^2(x)}{\tan(x)} + \frac{1}{\tan(x)}$

Simplify

$\sec(x)csc(x) = \tan(x) + \cot(x)$ proved

Solving (D)

$D.\ \frac{1 - (cos(x))^2}{cos(x)}$

Open bracket

$\frac{1 - (cos(x))^2}{cos(x)} = \frac{1 - cos^2(x)}{cos(x)}$

$1 - \cos^2(x) = \sin^2(x)$

So, we have:

$\frac{1 - (cos(x))^2}{cos(x)} = \frac{sin^2(x)}{cos(x)}$

Split

$\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \frac{sin(x)}{cos(x)}$

$\frac{sin(x)}{\cos(x)} = \tan(x)$

So, we have:

$\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \tan(x)$

$\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) \tan(x)$ --- proved

Solving (E):

$E.\ 2\sec(x)$

In trigonometry

$\sec(x)= \frac{1}{\cos(x)}$

So, we have:

$2\sec(x) = 2 * \frac{1}{\cos(x)}$

$2\sec(x) = \frac{2}{\cos(x)}$

Multiply by $\frac{1 - \sin(x)}{1 - \sin(x)}$ --- an equivalent of 1

$2\sec(x) = \frac{2}{\cos(x)} * \frac{1 - \sin(x)}{1 - \sin(x)}$

$2\sec(x) = \frac{2(1 - \sin(x))}{(1 - \sin(x))\cos(x)}$

Open bracket

$2\sec(x) = \frac{2 - 2\sin(x)}{(1 - \sin(x))\cos(x)}$

Express 2 as 1 + 1

$2\sec(x) = \frac{1+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}$

Express 1 as $\sin^2(x) + \cos^2(x)$

$2\sec(x) = \frac{\sin^2(x) + \cos^2(x)+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}$

Rewrite as:

$2\sec(x) = \frac{\cos^2(x)+1 - 2\sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}$

Expand

$2\sec(x) = \frac{\cos^2(x)+1 - \sin(x)- \sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}$

Factorize

$2\sec(x) = \frac{\cos^2(x)+1(1 - \sin(x))- \sin(x)(1-\sin(x))}{(1 - \sin(x))\cos(x)}$

Factor out 1 - sin(x)

$2\sec(x) = \frac{\cos^2(x)+(1- \sin(x))(1-\sin(x))}{(1 - \sin(x))\cos(x)}$

Express as squares

$2\sec(x) = \frac{\cos^2(x)+(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}$

Split

$2\sec(x) = \frac{\cos^2(x)}{(1 - \sin(x))\cos(x)} +\frac{(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}$

Cancel out like factors

$2\sec(x) = \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}$ --- proved

3 0

3 years ago