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3 years ago

How to write 229,909 227,011

Mathematics

1 answer:

Alex787 [66]3 years ago

8 0

Answer:

two hundred twenty-nine billion nine hundred nine million two hundred twenty- seven thousand and eleven

Step-by-step explanation:

Include all the details i put down

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What is the value of x in the figure below? In this diagram, ABD ~ CAD.

Shalnov [3]

Answer:

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Step-by-step explanation:

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2 years ago

У=-3x+3<br> y = -9x + 15

julia-pushkina [17]

Answer:

(2, - 3 )

Step-by-step explanation:

Given the 2 equations

y = - 3x + 3 → (2)

y = - 9x + 15 → (2)

Substitute y = - 3x + 3 into (2)

- 3x + 3 = - 9x + 15 ( add 9x to both sides )

6x + 3 = 15 ( subtract 3 from both sides )

6x = 12 ( divide both sides by 6 )

x = 2

Substitute x = 2 into either of the 2 equations and solve for y

Substituting into (1)

y = - 3(2) + 3 = - 6 + 3 = - 3

solution is (2, - 3 )

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3 years ago

Find the area of a triangle with a base of 14 ft and a <br> height of 11 ft. A = bh/2)

Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

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3 years ago

Part of the population of 7,750 elk at a wildlife preserve is infected with a parasite. A random sample of 50 elk shows that 11

Sergio039 [100]

Answer:

1,705 elk

Step-by-step explanation:

7,750 / 50 = 155 groups of elk

155 groups x 11 per group = 1,705 elk.

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3 years ago

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2.a. If the distri

zalisa [80]

Answer:

$P(\= X \ge 51 ) =0.0062$

$P(\= X \ge 51 ) = 0$

Step-by-step explanation:

From the question we are told that

The mean value is $\mu = 50$

The standard deviation is $\sigma = 1.2$

Considering question a

The sample size is n = 9

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{9} }$

=> $\sigma_x = 0.4$

Generally the probability that the sample mean hardness for a random sample of 9 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_{x}} \ge \frac{51 - 50 }{0.4 } )$

$\frac{\= X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ \= X )$

$P(\= X \ge 51 ) = P( Z \ge 2.5 )$

=> $P(\= X \ge 51 ) =1- P( Z < 2.5 )$

From the z table the area under the normal curve to the left corresponding to 2.5 is

$P( Z < 2.5 ) = 0.99379$

=> $P(\= X \ge 51 ) =1-0.99379$

=> $P(\= X \ge 51 ) =0.0062$

Considering question b

The sample size is n = 40

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{40} }$

=> $\sigma_x = 0.1897$

Generally the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_x} \ge \frac{51 - 50 }{0.1897 } )$

=> $P(\= X \ge 51 ) = P(Z \ge 5.2715 )$

=> $P(\= X \ge 51 ) = 1- P(Z < 5.2715 )$

From the z table the area under the normal curve to the left corresponding to 5.2715 and

=> $P(Z < 5.2715 ) = 1$

$P(\= X \ge 51 ) = 1- 1$

=> $P(\= X \ge 51 ) = 0$

5 0

3 years ago