Find numerator that makes statement true? 2/4 is less than /6

Find the 96th term of the arithmetic sequence 1, -12, -25, ...

goldenfox [79]

Answer:

The 96th term of the arithmetic sequence is -1234.

Step-by-step explanation:

first term (a)=1

second term (t2)=-12

common difference (d)= t2-a

d=-12-1

d=-13

96th term (t96)=?

We know that,

t96=a+(n-1)d

t96=1+(96-1)(-13)

t96=1+95(-13)

t96=1-1235

t96=-1234

6 0

2 years ago

A sports shop has 60 employees. 20% of the employees work part-time. How many part-time employees does the sports shop have?

torisob [31]

Answer:

12

Step-by-step explanation:

60÷100=0.6 0.6x20=12

Therefore the sports shop has 12 part-time employees

8 0

2 years ago

Find m(angle) ABC 21 132 48 19

Brut [27]

I don't like these algebra problems which pretend to be geometry.

These are supplementary angles adding to 180 degrees; geometry done.

7x - 1 + 3x - 9 = 180

10 x = 190

x = 19

ABC=3x - 9 = 3(19) - 9 = 48

Answer 48°

5 0

2 years ago

Read 2 more answers

A GMC dealer believes that demand for 2005 Envoys will be normally distributed with a mean of 200 and standard deviation of 30.

irina1246 [14]

Answer:

The correct answer to the following question will be "280 units".

Step-by-step explanation:

The given values are:

Mean = 200

Standard deviation = 30

Let's ask that the supplier needs to reach 99.9% of the standard of operation. To reach this quality of operation, she has to preserve an inventory of

Now,

⇒ $Mean + 3\times sigma$

On putting the values, we get

⇒ $200+3\times 30$

⇒ $200+90$

⇒ $290 \ units$

290 units would then accomplish a level of service of 99.9%.

Half unpurchased units can indeed be purchased at $30000 i.e. no failure. If he requests 300, there will be losses on 5 products that are unsold.

Hence, ordering 280 units seems to be preferable

3 0

3 years ago

A population has a standard deviation of 5.5. What is the standard error of the sampling distribution if the sample size is 81?

VladimirAG [237]

Answer:

$\sigma = 5.5$

And we have a sample size of n =81. We want to estimate the standard error of the sampling distribution $\bar X$ and for this case we know that the distribution is given by:

$\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$\sigma_{\bar x}= \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$\sigma_{\bar x}=\frac{5.5}{\sqrt{81}}= 0.611$

Step-by-step explanation:

For this case we know the population deviation given by:

$\sigma = 5.5$

And we have a sample size of n =81. We want to estimate the standard error of the sampling distribution $\bar X$ and for this case we know that the distribution is given by:

$\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$\sigma_{\bar x}= \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$\sigma_{\bar x}=\frac{5.5}{\sqrt{81}}= 0.611$

4 0

2 years ago