What is the 20% of 140.79

Can a linear function be continuous but not have a domain and range of all real numbers

ivolga24 [154]

"Mathematically yes - but from an engineering point no. And micro black holes are very unstable. They will evaporate extremely violently" hope this helps

4 0

2 years ago

You may remember that the perimeter of a rectangle is P=2(W+L) where W is the width and L is the length. Suppose that the perime

Anna71 [15]

Answer:

5 feet

Step-by-step explanation:

P = 2(w + l)

P = 34

l = w + 7

34 = 2 (w + w + 7)

34 = 2w + 2w + 14

34 - 14 = 4w

20 = 4w

w = 5 feet

Since l = w + 7

l = 5 + 7

l = 12 feet

5 0

2 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

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

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

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

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

3 years ago

Lilian says that 25% of a number will always be less than 75% of any other number. Complete one inequality to support Lillian's

rjkz [21]

She is incorrect cause if we take the number 20 and take 25% of it it equals 5 so now 20 is only 15. They if we take 75% away from 20 is equals 15 so then we only have 5 from 20 left. Therefore she’s incorrect because 75% leaves 5/20 and 25% leaves 15/20. Hope that makes sense.

6 0

3 years ago

HELP WILL GIVE YOU Brainiest!!!

Phoenix [80]

Answer:

Step-by-step explanation:

So there we have it, <A=135° and <B=45°. This should check when you add them together ("<A and <B are supplementary which means that together their angles equal 180°"). Does 135°+45=180°? YES!

I Really hope it helps

8 0

2 years ago