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3 years ago

Help me please i really need it so i can relax this weekend

Mathematics

1 answer:

andreev551 [17]3 years ago

5 0

1. Recrystallization
2. Transpiration
3. Atmosphere
4. Evaporation
5. Gravity
6. Precipitation
7. Change of state

Hope you have a good weekend

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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.

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3 years ago

Two kg of tea and 3 kg of sugar cost rupees 39 in january 1997, However in march 1997 the price of the tea increased by 25% and

Irina18 [472]

Answer: tea = 15 rupees per kg

sugar= 3 rupees per kg

Step-by-step explanation:

Hi, to answer this question we have to write a system of equations with the information given:

"Two kg of tea and 3 kg of sugar cost rupees 39 in january 1997":

2 t + 3 s =39 (a)

Where:

t= price of 1 kg of tea
s = price of 1 kg of sugar

"in march 1997 the price of the tea increased by 25% (1.25)and the price of the sugar increased by 20%(1.20) and the same quantity of tea and sugar cost rupees 48.30. "

2(t1.25)+3(s1.2) = 48.30 (b)

Solving for t in (b)

2t =39-3s

t = (39 -3s)/2

t = 19.5-1.5s

Replacing the value of t in (b)

2 x ((19.5-1.5s)1.25)+ 3 ( 1.2s) =48.30

2x ( 24.375 -1.875s) +3.6s =48.30

48.75 -3.75s+3.6s= 48.30

48.75-48.30 = 3.75s-3.6s

0.45= 0.15s

0.45/0.15 =s

3 =s

Replacing the value of s in (a)

2 t + 3 (3) =39

2 t + 9 =39

2 t =39 -9

2 t =30

t = 30/2

t= 15

Prices in january:

tea = 15 rupees per kg

sugar= 3 rupees per kg

Feel free to ask for more if needed or if you did not understand something.

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