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Setler79 [48]
3 years ago
11

What is the product? 43 .4-3 0) 1) 4) 6)

Mathematics
2 answers:
9966 [12]3 years ago
7 0

Answer:

1

Step-by-step explanation:

So we can actually simplify the exponents here. The power rule for exponents is that when multiplying, you add the exponents. So,

3 + (-3) = 0

so the number is now 4^0.

Everything by the power of 0 is equal to 1, so the answer is 1.

Hope this helps !!

-Ketifa

Tasya [4]3 years ago
3 0
The answer is 1, i hope that helps
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Which expression is equivalent to 6^-3
nikdorinn [45]
You didn'tlist any but anyway

x^{-m}=\frac{1}{x^m}
so
6^{-3}=\frac{1}{6^3}=\frac{1}{216}


so the possible expressions could be \frac{1}{6^3} and/or \frac{1}{216}
4 0
3 years ago
Read 2 more answers
Construc t a 95% confidence interval for the population standard deviation σ of a random sample of 15 men who have a mean weight
GrogVix [38]

The 95% confidence interval for the population standard deviation will be,

$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\

simplifying the equation, we get

$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288

The 95% confidence interval exists (9.887, 21.288).

<h3>What is the  95% of confidence interval?</h3>

A random sample of 15 men exists selected. The mean weight exists at $165.2 pounds and the standard deviation exists at $13.5 pounds. The population exists normally distributed.

So, $n=15, \bar{X}=165.2, s=13.5$

where n exists the sample size, $\bar{X}$ exists the sample size

s exists the sample standard deviation.

The degrees of freedom will be n - 1 i.e.15 - 1 = 14.

For a 95% confidence interval, the level of significance will be $\alpha=0.05$.

The 95% confidence interval for the population standard deviation will be,

$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\

substitute the values in the above equation, we get

$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi^{2}} 0.05} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0}^{2}}} \\

$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.975}^{2}}} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.025}^{2}}} \\

simplifying the above equation, we get

$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288

The 95% confidence interval exists (9.887, 21.288).

To learn more about confidence interval refer to:

brainly.com/question/14825274

#SPJ4

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2 years ago
Marco Invests $760 into a savings account. The account pays 4% simple interest.
Lemur [1.5K]

Answer:

Step-by-step explanation:

The answer is 43,796

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Complete the equation of the line through (1,4) and (2,2).<br> y=
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Answer:

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Step-by-step explanation:

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