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3 years ago

13

Steve is choosing between two jobs at Smithsonian museums. One job is funded through the federal government and offers a pension

plan. The other job is privately funded and offers a 403(b) plan similar to a 401(k) plan, in which the Smithsonian would contribute 7% of Steve's salary. What factor would make Steve prefer the federal job and pension?

2 answers:

Lyrx [107]3 years ago

7 0

Answer:

The government career are likely to be more stable than the private sector

Step-by-step explanation:

The federal job will likely to be more stable. This is because of the decided benefits that are offered by the employer after finishing the career. This is unlike the privately funded job that offers the decided contribution, but can become easily less stable and affected by the changes in the economy. Hence, as a result of the factors, Steve will prefer a federal job.

lora16 [44]3 years ago

4 0

Private funding can run out. federal funding will likely not run out

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Brilliant_brown [7]

you shouldnt answer if you really dont know you noodle head , the answer is 2.54

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3 years ago

I will mark either the best or the first answer brainliest.

posledela

Answer:

11.x=16 12.x=37

Step-by-step explanation:

11.

20^2-12^2=x^2

256=x^2

x=√256

x=16

12.

13^2-5^2=y^2

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y=12

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5 0

3 years ago

Read 2 more answers

Need help don’t understand

sineoko [7]

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3 years ago

519 divided by 6 and 915 divided by 7 and 439 divided by 7 and 812 divided by 9 which one does not have a two digit quotient.

Setler [38]

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5 0

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:

a

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

b

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

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

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

So

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

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

5 0

2 years ago