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2 years ago

???????????????????????????

Mathematics

1 answer:

Pie2 years ago

8 0

X ≈ √17
The number on the number line is roughly 4.1, and if you square 4.1, then you get 16.81, which 17 is the closest to.

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You are trying to estimate the average amount a family spends on food during a year. In the past the standard deviation of the a

svetoff [14.1K]

Answer:

$n=(\frac{2.58(1000)}{50})^2 =2662.56 \approx 2663$

So the answer for this case would be n=2663 rounded up to the nearest integer

Step-by-step explanation:

We have the following info:

$ME = 50$ margin of error desired

$\sigma = 1000$ the standard deviation for this case

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =50 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

The critical value for 99% of confidence interval now can be founded using the normal distribution. The significance is $\alpha=0.01$ . And for this case would be $z_{\alpha/2}=2.58$ , replacing into formula (b) we got:

$n=(\frac{2.58(1000)}{50})^2 =2662.56 \approx 2663$

So the answer for this case would be n=2663 rounded up to the nearest integer

5 0

2 years ago

Instruction

slavikrds [6]

<h3>Answer: Choice C. (-x, -y)</h3>

Explanation:

Focus on one point, such as A = (1,2). Note how it moves to point A ' = (-1, -2). Both x and y coordinates have been flipped from positive to negative. The rule therefore is $(x,y) \to (-x,-y)$ . This describes a 180 degree rotation (either clockwise or counterclockwise, it doesn't matter). Points B and C follow the same idea.

Side note: Lines AA', BB' and CC' all go through the origin (0,0).

8 0

2 years ago

'Polonium-210 is a radioactive substance with a half-life of 138 days. If a nuclear

Tatiana [17]

Answer:

<u>59.0891 g (rounded to 4 decimal places)</u>

Explanation:

<em>Half-life time</em> of a radioactive substance is the time for half of the substance to decay.

Thus, the amount of the radioactive substance that remains after a number n of half-lives is given by:

$A=A_0\cdot (1/2)^n$

Where:

A is the amount that remains of the substance after n half-lives have elapses, and

A₀ is the starting amount of the substance.

In this problem, you have that the half-live for your sample (polonium-210) is 138 days and the number of days elapsed is 330 days. Thus, the number of half-lives elapsed is: