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3 years ago

What are some study habits that helped? I have tried loads of ideas but I just can't to seem to find one that works for me.

1 answer:

7 0

For me being 100% honest:

grade 9-11 no homework, pay attention in class and study night before.

Grade 12: pay attention more in class, do some homework when concepts hard, study night before.

Uni: Read notes, do class homework, read textbook, study week before.

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If I read 2/3 of a book that has 222 pages how many pages do I have left to read

joja [24]

Multiply 2/3 by 222 then subtract the number from 222 and you should get your answer

3 0

3 years ago

Read 2 more answers

An engineer wishes to determine the width of a particular electronic component. If she knows that the standard deviation is 1.6

nexus9112 [7]

Answer:

n=1705

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=1.6)$

And the distribution for $\bar X$ is:

$\bar X \sim N(\mu, \frac{1.6}{\sqrt{n}})$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=\pm 2.58$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And we have everything to replace into the formula:

$n=(\frac{2.58(1.6)}{0.1})^2 =1704.03$

And if we round up the answer we see that the value of n to ensure the margin of error required $\pm=0.1$ mm is n=1705.

6 0

3 years ago

Solve simultaneous equations 3x-2y=17 +2x-2y=10

Zarrin [17]

Answer:

(7;2)

Step-by-step explanation:

$3x - 2y = 17 \\ - 2y = 17 - 3x \\ 2x + 17 - 3x = 10 \\ 2x - 3x = 10 - 17 \\ - x = - 7 \\ x = 7 \\ - 2y = 17 - 21 \\ - 2y = - 4 \\ y = 2$

3 0

3 years ago

Which of the numbers below could be terms in the sequence an=4n-7 A)29 B)21 C)20 D)28

natali 33 [55]

Answer:

A)29 and B) 21,

Step-by-step explanation:

First,in the sequence $a_n=4n-7$ , the parameter $n$ must to be a integer.

Second, we need to solve the equations by $n$ .

$n=(a_n+7)/4$

All the option in the problem represent an $a_n$

Then, we need to prove all number in the options, if the result is a integer number, this option can be part of the sequence.

For A)

$n=(29+7)/4=9\\$

For B)

$n=(21+7)/4=7\\$

For C)

$n=(20+7)/4=6.75\\$

For D)

$n=(28+7)/4=8.75\\$

Only A) and B) only A and B meet the requirement

4 0

3 years ago

Find the directional derivative of f(x,y,z)=z3−x2yf(x,y,z)=z3−x2y at the point (−5,5,2)(−5,5,2) in the direction of the vector v

olga_2 [115]

We are given

$f=z^3 -x^2y$

Firstly, we can find gradient

so, we will find partial derivatives

$f_x=0 -2xy$

$f_x=-2xy$

$f_y=0 -x^2$

$f_y=-x^2$

$f_z=3z^2$

now, we can plug point (-5,5,2)

$f_x=-2*-5*5=50$

$f_y=-(-5)^2=-25$

$f_z=3(2)^2=12$

so, gradient will be

$gradf=(50,-25,12)$

now, we are given that

it is in direction of v=⟨−3,2,−4⟩

so, we will find it's unit vector

$|v|=\sqrt{(-3)^2+(2)^2+(-4)^2}$

$|v|=\sqrt{29}$

now, we can find unit vector

$v'=(\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })$

now, we can find dot product to find direction of the vector

$dir=(gradf) \cdot (v')$

now, we can plug values

$dir=(50,-25,12) \cdot (\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })$

$dir=(-\frac{150}{\sqrt{29} } - \frac{50}{\sqrt{29} } - \frac{48}{\sqrt{29} })$

$dir=-\frac{248\sqrt{29}}{29}$ .............Answer

7 0

3 years ago

Read 2 more answers