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3 years ago

PLEASE I NEED HELP AGAIN ATTACHMENT BELOW

Mathematics

1 answer:

Anastaziya [24]3 years ago

5 0

Answer:

A.) 1/81

Step-by-step explanation:

This is your equation:

f(x) = 9^x

They want you to solve for x = -2, so substitute -2 in for x.

f(-2) = 9^-2

When you solve this, you get, 1/81.

So, the answer is A.)

I hope this helps! :)

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A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of

notsponge [240]

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.01 = 0.99$ , so Z = 2.327.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031$

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.0001 = 2.327\frac{0.0003}{\sqrt{n}}$

$0.0001\sqrt{n} = 2.327*0.0003$

$\sqrt{n} = \frac{2.327*0.0003}{0.0001}$

$(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2$

$n = 48.73$

Rounding up

49 measurements are needed.

7 0

3 years ago

A chain lying on the ground is 10 m long and its mass is 70 kg. How much work (in J) is required to raise one end of the chain t

DerKrebs [107]

Answer:

$W= 34.3 \frac{kg}{s^2} (4^2-0^2)m^2 =548.8 \frac{kg m^2}{s^2} =548.8 J$

Step-by-step explanation:

Data Given: m = 70 kg , g = 9.8 ms^-2, h =10m.

For this case we can use the following formula:

$W = \int_{x_i}^{x_f} F(x) dx$

For this case we need to find an expression for the force in terms of the distance. And since on this case the total distance is 10 m long we can write the expression like this:

$F(x) = \frac{ma}{10m}= \frac{mg}{10m} x$

The only acceleration on this case is the gravity and if we replace the values given we got:

$\frac{70 kg *9.8 m/s^2}{10m} x=68.6 x\frac{kg}{s^2}$

Now we can find the required work with the following integral:

$W= 68.6 \frac{kg}{s^2} \int_{0}^4 x dx$

$W= 34.3 \frac{kg}{s^2} x^2 \Big|_0^4$

$W= 34.3 \frac{kg}{s^2} (4^2-0^2)m^2 =548.8 \frac{kg m^2}{s^2} =548.8 J$

7 0

4 years ago

Write a number that is less than - 10

ra1l [238]

Any number from -10 to -∞ would be correct. One example is -79.

8 0

3 years ago

What is another name for line m?

Ganezh [65]

I’m pretty sure it’s the slope

5 0

3 years ago

I need help on this question!!!!

blsea [12.9K]

D
just subtract 14 from 26

6 0

3 years ago

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