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rjkz [21]
3 years ago
6

What is 0.0059 in scientific annotation?

Mathematics
1 answer:
Alik [6]3 years ago
5 0
5.9*10^{-3}

To write 0.0059 in scientific notation we need to move the decimal point so there is one number to the left of the decimal point. We need to move decimal point 3 places to the left, meaning that the power of 10 will be negative 3.

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Please help me again!!!!!!!!
Kazeer [188]

Answer:

Huan receives 3.33 pounds (£) every week

Step-by-step explanation:

We can calculate this by dividing the amount for 5 weeks and dividing by 5 to find the amount of pocket money for each week. 16.65/5 = 3.33, which means that he receives 3.33 pounds (£) every week for pocket money.

8 0
3 years ago
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in 1996,Donovan Bailey of Canada ran the 100 meter dash in 9.84 seconds.How fastis that to the nearst second?to the nearst tenth
Fofino [41]
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3 years ago
Fractions closer to 0 than 1
vagabundo [1.1K]

Answer:

Any legal fraction (denominator not equal to zero) with a numerator equal to zero has an overall value of zero. all have a fraction value of zero because the numerators are equal to zero

Step-by-step explanation:

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3 years ago
Hi, can you please help with with this?​
BARSIC [14]

Answer:

Step-by-step explanation:

the ratio is 5:3:2

therefore, Ali receives 5x, Ben receives 3x and Carl receives 2x

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so,

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x = 60/3 = 20

Ali's share = 5x = 5*20

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3 years ago
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Construct a 99% confidthence interval for the population mean .Assume the population has a normal distribution. A group of 19 ra
Kobotan [32]
<h2>Answer with explanation:</h2>

Confidence interval for mean, when population standard deviation is unknown:

\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}

, where \overline{x} = sample mean

n= sample size

s= sample standard deviation

t_{\alpha/2} = Critical t-value for n-1 degrees of freedom

We assume the population has a normal distribution.

Given, n= 19 , s= 3.8 , \overline{x}=22.4

\alpha=1-0.99=0.01

A) Critical t value for \alpha/2=0.005 and degree of 18 freedom

t_{\alpha/2} = 2.8784

B) Required confidence interval:

22.4\pm ( 2.8744)\dfrac{3.8}{\sqrt{19}}\\\\=22.4\pm2.5058\\\\=(22.4-2.5058,\ 22.4+2.5058)=(19.8942,\ 24.9058)\approx(19.9,\ 24.9)

Lower bound = 19.9 years

Uppen bound = 24.9 years

C) Interpretation: We are 99% confident that the true population mean of lies in (19.9, 24.9) .

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