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3 years ago

What is 0.02 to the fifth power equal?

Mathematics

2 answers:

Lyrx [107]3 years ago

5 0

3.2E-9

A bit complicated, so PLEASE don't ask for an explanation

Sonbull [250]3 years ago

5 0

I plugged it into desmos. it's 3.2*10^{-9}
or
-0.30149814893105

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Solve for x 27^(2x+4)=1/9^(x-46)

lys-0071 [83]

$27^{2x+4}=(\frac{1}{9})^{x-46} \\
(3^3)^{2x+4}=(3^{-2})^{x-46} \\
3^{3(2x+4)}=3^{-2(x-46)} \\
3^{6x+12}=3^{-2x+92} \\
6x+12=-2x+92 \\
6x+2x=92-12 \\
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8 0

3 years ago

PLEASE HELP ME 50 POINTS!!!! Identify an expression that is equivalent to 0.3(15y).

IgorC [24]

Answer: 4.5y

Step-by-step explanation: You have To Multiply 15 times 0.3 and you get 4.5y

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3 years ago

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mezya [45]

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Step-by-step explanation:

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3 years ago

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Olenka [21]

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7 0

3 years ago

Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

3 years ago