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2 years ago

How many natural numbers lie between 75² and 76²

Mathematics

1 answer:

vodomira [7]2 years ago

5 0

Answer:

Natural numbers between $75^{2}$ and $76^{2}$ is

(76 * 76) - (75 * 75)

=> 5776 - 5625

=> 151

Therefore, there are 151 natural numbers between $75^{2}$ and $76^{2}$

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Louis started a table showing a multiplication patterns complete the table describe a pattern you see in the products

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One pattern that you can see in a multiplication table is the perfect square numbers. It runs from the top left hand corner directly through the middle to the bottom right hand corner. A perfect square is a number that is multiplied by itself. The perfect square numbers are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144. They keep going forever on but those are the main ones from 1x1 to 12x12.

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2 years ago

Two major automobile manufacturers have produced compact cars with engines of the same size. We are interested in determining wh

snow_tiger [21]

Answer:

Step-by-step explanation:

Corresponding fuel efficiencies of manufacturer A's car and manufacturer B's car form matched pairs.

The data for the test are the differences between the efficiencies of manufacturer A's car and manufacturer B's car

μd = fuel efficiency of manufacturer A's car minus the fuel efficiency of manufacturer B's car.

A B diff

32 28 4

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26 27 - 1

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25 24 1

29 25 4

31 28 3

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Sample mean, xd

= (4 + 5 - 1 + 2 + 1 + 4 + 3 - 2)/8 = 2

xd = 2

Standard deviation = √(summation(x - mean)²/n

n = 8

Summation(x - mean)² = (4 - 2)^2 + (5 - 2)^2 + (- 1 - 2)^2 + (2 - 2)^2 + (1 - 2)^2 + (4 - 2)^2 + (3 - 2)^2 + (- 2 - 2)^2 = 44

Standard deviation = √(44/8

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For the null hypothesis

H0: μd = 0

For the alternative hypothesis

H1: μd ≠ 0

This is a two tailed test and the distribution is a students t. Therefore, degree of freedom, df = n - 1 = 8 - 1 = 7

2) The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (2 - 0)/(2.35/√8)

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We would determine the probability value by using the t test calculator.

p = 0.047

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2 years ago

(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX

pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable X follows a Normal distribution with mean μ = 10 and variance σ² = 36.

(a)

Compute the value of P (X > 5) as follows:

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(b)

Compute the value of P (4 < X < 16) as follows:

$P(4$

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(c)

Compute the value of P (X < 8) as follows:

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Thus, the value of P (X < 8) is 0.3707.

(d)

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Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

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Thus, the value of P (X > 16) is 0.1587.

**Use a z-table for the probabilities.

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2 years ago

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They are both correct in their computation

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2 years ago

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Answer: B. Blind experiment

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2 years ago

How many natural numbers lie between 75² and 76²​

How many natural numbers lie between 75² and 76²