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3 years ago

PLEASE HELPPPPPPPPPPPPPPPP

Mathematics

1 answer:

iogann1982 [59]3 years ago

6 0

Answer:

Our radius is already given, 8.

8 x 3.14 x 2 = 50.24cm <-------- perimeter/circumference

8^2 x 3.14

64 x 3.14 = 200.96cm2 <------- area

Formula of a circumference:

2πr

2 x pie x radius (pie can be used as 22/7 or 3/4 most likely 3.14)

Formula of an area of a Circle:

πr^2

pie x radius squared (pie can be used as 22/7 or 3.14)

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If the cos of angle x is 8 over 17 and the triangle was dilated to be two times as big as the original, what would be the value

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Answer:

the answer is 8/17

Step-by-step explanation:

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What is the inverse of the following statement?

Ksju [112]

The inverse of the statement is 'If it is Wednesday, then it is not sunny'

<h3>What is the inverse of a word?</h3>

The inverse of a word is simply the contrary nature or quality and could be termed opposite or reverse of a proposition or theorem that is formed by'

contradicting both the subject and predicate
Contradicting both the hypothesis and conclusion of a given proposition or theorem

For instance, the inverse of "if A then B" is "if not-A then not-B" and 'if not A, then B" is 'If A, not B ' is termed a compare contrapositive

From the information given, we have;

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We can see that the inverse of the proposition is;

'If it is Wednesday, then it is not sunny'

Thus, the inverse of the statement is 'If it is Wednesday, then it is not sunny'

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2 years ago

Let z denote a random variable that has a standard normal distribution. Determine each of the probabilities below. (Round all an

Gelneren [198K]

Answer:

(a) P (Z < 2.36) = 0.9909 (b) P (Z > 2.36) = 0.0091

(e) P (-0.77< Z > -0.55) = 0.0705 (f) P (Z > 3) = 0.0014

(g) P (Z > -3.28) = 0.9995 (h) P (Z < 4.98) = 0.9999.

Step-by-step explanation:

Let us consider a random variable, $X \sim N (\mu, \sigma^{2})$ , then $Z=\frac{X-\mu}{\sigma}$ , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, $Z \sim N (0, 1)$ .

In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean. The z-scores are standardized scores.

The distribution of these z-scores is known as the standard normal distribution.

(a)

Compute the value of P (Z < 2.36) as follows:

P (Z < 2.36) = 0.99086

≈ 0.9909

Thus, the value of P (Z < 2.36) is 0.9909.

(b)

Compute the value of P (Z > 2.36) as follows:

P (Z > 2.36) = 1 - P (Z < 2.36)

= 1 - 0.99086

= 0.00914

≈ 0.0091

Thus, the value of P (Z > 2.36) is 0.0091.

(c)

Compute the value of P (Z < -1.22) as follows:

P (Z < -1.22) = 0.11123

≈ 0.1112

Thus, the value of P (Z < -1.22) is 0.1112.

(d)

Compute the value of P (1.13 < Z > 3.35) as follows:

P (1.13 < Z > 3.35) = P (Z < 3.35) - P (Z < 1.13)

= 0.99960 - 0.87076

= 0.12884

≈ 0.1288

Thus, the value of P (1.13 < Z > 3.35) is 0.1288.

(e)

Compute the value of P (-0.77< Z > -0.55) as follows:

P (-0.77< Z > -0.55) = P (Z < -0.55) - P (Z < -0.77)

= 0.29116 - 0.22065

= 0.07051

≈ 0.0705

Thus, the value of P (-0.77< Z > -0.55) is 0.0705.

(f)

Compute the value of P (Z > 3) as follows:

P (Z > 3) = 1 - P (Z < 3)

= 1 - 0.99865

= 0.00135

≈ 0.0014

Thus, the value of P (Z > 3) is 0.0014.

(g)

Compute the value of P (Z > -3.28) as follows:

P (Z > -3.28) = P (Z < 3.28)

= 0.99948

≈ 0.9995

Thus, the value of P (Z > -3.28) is 0.9995.

(h)

Compute the value of P (Z < 4.98) as follows:

P (Z < 4.98) = 0.99999

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Thus, the value of P (Z < 4.98) is 0.9999.

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