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3 years ago

13

Give an example of something appropiatly measured in nano meters

2 answers:

garri49 [273]3 years ago

6 0

The thickness of a piece of paper

o-na [289]3 years ago

4 0

Somethings very small such as Atoms and even molecules

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I need help on this

adoni [48]

I am pretty sure the answer is,

B

4 0

3 years ago

How to solve radical numerator fractions

Nonamiya [84]

Y=radical 3
just cancel out the 2 on both sides

3 0

3 years ago

From an industrial area 70 companies were selected at random and 45 of them were panning for expansion next year. Find 95% confi

Lana71 [14]

Answer:

Confidence limit = [52.8%, 75.2%]

Step-by-step explanation:

$P=\frac{45}{70}= 0.64$

$(1-P)=1-0.64=0.36$

$n= 70$

$P$ ± $z$ $\sqrt{\frac{P(1-P)}{n} }$

where the value $z$ will be taken from the z-table for 95% confidence interval

1-0.95= 0.05/2= 0.025

0.95+0.025= 0.0975

From the z-table the value of $z$ corresponding to 0.0975 is 1.96

$0.64$ ± $1.96$ $\sqrt{\frac{0.64*0.36}{70} }$

$0.64$ ± $1.96$ $(0.057)$

$0.64$ ± $0.112$

$64%$ % ± $11.2$ %

so the confidence interval is

$64+11.2=75.2$ %

$64-11.2=52.8$ %

$[52.8, 75.2]$

8 0

3 years ago

At the start of the year, 15 chameleons were introduced into a zoo. The population of chameleons is expected to grow at a rate o

bearhunter [10]

Answer:

option-B

Step-by-step explanation:

We are given

At the start of the year, 15 chameleons were introduced into a zoo

so, $P_0=15$

The population of chameleons is expected to grow at a rate of 41.42% every year

so, r=0.4142

and x represents the number of years since the chameleons were introduced into the zoo

now, we can set equation to find total population

and we get

$P(x)=P_0(1+r)^x$

now, we can plug values

$P(x)=15(1+0.4142)^x$

$P(x)=15(1.4142)^x$

Average rate of change between 2 years and 4 years:

we can use formula

$A_1=\frac{P(4)-P(2)}{4-2}$

now, we can plug values

$A_1=\frac{15(1.4142)^{4}-15(1.4142)^{2}}{4-2}$

$A_1=14.99914$

Average rate of change between 4 years and 6 years:

we can use formula

$A_2=\frac{P(6)-P(4)}{6-4}$

now, we can plug values

$A_2=\frac{15(1.4142)^{6}-15(1.4142)^{4}}{6-4}$

$A_2=29.99770$

Average rate of change between 6 years and 8 years:

we can use formula

$A_3=\frac{P(8)-P(6)}{8-6}$

now, we can plug values

$A_3=\frac{15(1.4142)^{8}-15(1.4142)^{6}}{8-6}$

$A_3=59.99425$

now, we will check each options

option-A:

we can see that

$A_3-A_2=30$

$A_3-A_2=30$

So, this is FALSE

option-B:

$A_1=\frac{1}{2}A_2$

So, this is TRUE

option-C:

This is FALSE

option-D:

we got

$A_1=\frac{1}{2}A_2$

so, this is FALSE

5 0

3 years ago

What is the area of a parallelogram?

denis23 [38]

D is the correct answer

the area of a parallelogram is equal to: base x height

$18 \times 7 = 126$

good luck

3 0

3 years ago

Read 2 more answers