0.13 over 0.2 turned into a decimal

1 answer:

4 0

The answer is 0.65
This is the same as 1.3/2 since the decimal must be moved one place to the right for both divisor and dividend.

You might be interested in

The life in hours of a 75-watt light bulb is known to be normally distributed with standard deviation σ = 25 hours. A random sam

irina [24]

Answer: a) (1008.34,1019.658) b) (1009.24,1018.76)

Step-by-step explanation:

Since we have given that

n = 75

mean = 1014 hours

Standard deviation = 25 hours

At 95% two sided , z = 1.96

So, confidence interval would be

$\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=1014\pm 1.96\dfrac{25}{\sqrt{75}}\\\\=1014\pm 5.658\\\\=(1014-5.658,1014+5.658)\\\\=(1008.34,1019.658)$

(b) Construct a 95% lower confidence bound on the mean life.

z = 1.65

So, confidence interval would be

$\bar{x}\pm z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1014\pm 1.65\times \dfrac{25}{\sqrt{75}}\\\\=1014\pm 4.76\\\\=(1014-4.76,1014+4.76)\\\\=(1009.24,1018.76)$

Hence, a) (1008.34,1019.658) b) (1009.24,1018.76)

4 0

4 years ago

Plz..... Solve this questions for me...

faust18 [17]

$6.\\\text{We know}\ 9=3^2.\ \text{Therefore}\\\\9^2=(3^2)^2\\\\\text{use}\ (a^n)^m=a^{nm}\\\\9^2=(3^2)^2=3^{2\cdot2}=3^4\\\\\boxed{3^4=9^2}\\=================$

$7.\\METHOD\ 1:\\\\\sqrt{7^4}=\sqrt{7^{2\cdot2}}\\\\\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt{(7^2)^2}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=\boxed{7^2}\\\\METHOD\ 2:\\\\\text{use}\ a^\frac{m}{n}=\sqrt[n]{a^m}\\\\\sqrt{7^4}=7^\frac{4}{2}=\boxed{7^2}\\=================$

$8.\\5^a\times5^b=5^{11}\\\\a)\\\text{use}\ a^n\times a^m=a^{n+m}\\\\5^a\times5^b=5^{a+b}\\\\5^{a+b}=5^{11}\Rightarrow a+b=11\\\\11\ \text{is odd. The sum of even and odd is odd. Therefore}\ a\ \text{and}\ b\ \text{can't be even.}\\--------------------\\\\b)\\5\ \text{solutions}\\\\11=5+6\\11=4+7\\11=3+8\\11=2+9\\11=1+10$