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Julli [10]
3 years ago
8

How do you write 23.6 in scientific notation?

Mathematics
2 answers:
crimeas [40]3 years ago
8 0
The first factor must be at least 1 and less than 10.
The second factor must be a power of 10.

Move the decimal point to the left until the number is between 1 and 10. Count how many places you move the decimal point.23.6 → 2.36

You moved the decimal point 1 place to the left. The power of 10 is 10 to the power of 1

23.6 = 2.36 × 10 to the power of 1
9966 [12]3 years ago
6 0
In scientific notation you have to have the decimal after the first digit so you would write 2.36*10^#.
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Step-by-step explanation:

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In the expression 852.763 x 10?, what is the missing exponent that would result in a number that is 1,000 times the value of 852
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Answer:

The missing exponent = power of 3

852.763 x 10³ = 852763

Step-by-step explanation:

The expression 852.763 x 10?

A number that is 1,000 times the value of 852.763 is calculated as:

852.763 × 1000 = 852763

Converting this to exponent

852.763 × 10³ = 852763

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Which expression is equivalent to 24x + 12?
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Test the claim that the mean GPA of night students is larger than 3.1 at the .10 significance level. The null and alternative hy
Alborosie

Answer:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

df = n-1=75-1 =74

The p value is:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is: 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

Step-by-step explanation:

For this case we want to test the claim that mean GPA of night students is larger than 3.1 at the .10 significance level. The claim needs to be on the alternative hypothesis so then we have the following system of hypothesis:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

We have the following info given:

\bar X = 3.13 , s =0.03 , n =75

The statistic to check the hypothesis is given by:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

Replacing the info given we got:

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

The degrees of freedom are given by:

df = n-1=75-1 =74

The p value since is a right tailed ted is given by:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

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