Ask question

Ask question

All categories

Lady bird [3.3K]

3 years ago

11

What is 200,000(70,981 - 43,005)+22,984 ?

1 answer:

BigorU [14]3 years ago

6 0

Answer:

5,595,222,984 I think this is right it is what my calculator gave me.

You might be interested in

I need some help with this one

k0ka [10]

I believe the answer is 3

8 0

3 years ago

Hi can you help me in mate plis simplify the following algebraic expressions

mars1129 [50]

Answer:

$\frac{3 {x}^{2} }{ {y}^{2} {z}^{6} }$

Step-by-step explanation:

I have attached the explanation above. hopefully this will help

4 0

3 years ago

HELP 7TH GRADE I WILL GIVE BRAINLIEST

Feliz [49]

Answer:

a

Step-by-step explanation:

theres a 90 degree angle and a has a 95 so thats pretty close

5 0

3 years ago

PLEASE help with these math questions

kaheart [24]

1. C
2. A
3. D
4. 62/25
5. 4/33

By the way it is not fair. posting 5 sums for just 7 points. :)

anyway, if it helped u
do click thx.

3 0

4 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

3 years ago