All categories

3 years ago

FIND.......16 2/3 % of 96 lamps

1 answer:

4 0

Convert percentage to fraction by dividing by 100
16 2/3 = 16 2/3 ÷ 100 = 1/6

16 2/3 % of 96 = 1/6 x 96 = 16

Ans: 16 2/3 % of 96 lamps = 16

You might be interested in

What is 23x4x67+213 divided by 9 -154 x7

Julli [10]

Answer: -6377/1069 (write as fraction)

Step-by-step explanation: 6377/-1069

If you would like for me to simply it let me know!

7 0

2 years ago

A 2003 survey showed that 14 out of 250 Americans surveyed had suffered some kind of identity theft in the past 12 months. What

ivann1987 [24]

Answer:

The lower confidence limit of the 95% confidence interval for the population proportion of Americans who were victims of identity theft is 0.0275.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

A 2003 survey showed that 14 out of 250 Americans surveyed had suffered some kind of identity theft in the past 12 months.

This means that $n = 250, \pi = \frac{14}{250} = 0.056$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.056 - 1.96\sqrt{\frac{0.056*0.944}{250}} = 0.0275$

The lower confidence limit of the 95% confidence interval for the population proportion of Americans who were victims of identity theft is 0.0275.

5 0

2 years ago

What 2 numbers multiply to -36 and add to 16

KatRina [158]

Answer:

Step-by-step explanation:

I understand

6 0

3 years ago

Find integra of xlnxdx

Mademuasel [1]

Answer:

$\dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

$\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C}$\end{minipage}}$

Increase the power by 1, then divide by the new power.

Given indefinite integral:

$\displaystyle \int x \ln x \:\: \text{d}x$

To integrate the given integral, use Integration by Parts:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

$\text{Let }u=\ln x \implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{x}$

$\text{Let }\dfrac{\text{d}v}{\text{d}x}=x \implies v=\dfrac{1}{2}x^2$

Therefore:

$\begin{aligned}\displaystyle \int u \dfrac{dv}{dx}\:dx & =uv-\int v\: \dfrac{du}{dx}\:dx\\\\\implies \displaystyle \int x \ln x\:\:\text{d}x & = \ln x \cdot \dfrac{1}{2}x^2-\int \dfrac{1}{2}x^2 \cdot \dfrac{1}{x}\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x -\int \dfrac{1}{2}x\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}\end{aligned}$