Ask question

Ask question

All categories

2 years ago

14

The mass of a an atom of helium is 6.65 x 10−24. Use scientific notation to express the mass of 250 helium atoms. A) 1.6625 x 10

−21 B) 1.6625 x 10−22 C) 1.6625 x 10−24 D) 1.6625 x 10−26

1 answer:

lidiya [134]2 years ago

5 0

Answer: its A

Step-by-step explanation:

i know this is it

You might be interested in

PLEASE HELP ASAP I WILL MARK BRILLIANT WHOEVER ANSWERS THANKS (:

Mila [183]

The answer is 38.5 cm^2

5 0

3 years ago

Read 2 more answers

Triangle JKL is similar to triangle MNO. Find the measure of side OM. Round your answer to the nearest tenth if necessary.

Sonbull [250]

Answer:43.1

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

1.62 as a percentage

goldenfox [79]

1.62 as a percentage would be 162%. Hope this is helpful!

5 0

4 years ago

Read 2 more answers

Need help pls<br>thank you in advance

ira [324]

Answer:

<h3><u>Mean</u></h3>

<u />

$\textsf{Mean}\:\overline{X}=\sf \dfrac{\textsf{sum of all the data values}}{\textsf{total number of data values}}$

$\implies \sf Mean\:(Nilo)=\dfrac{5+6+14+15}{4}=\dfrac{40}{4}=10$

$\implies \sf Mean\:(Lisa)=\dfrac{8+9+11+12}{4}=\dfrac{40}{4}=10$

<h3><u>Standard Deviation</u></h3>

$\displaystyle \textsf{Standard Deviation }s=\sqrt{\dfrac{\sum X^2-\dfrac{(\sum X)^2}{n}}{n-1}}$

$\begin{aligned}\displaystyle \textsf{Standard Deviation (Nilo)} & =\sqrt{\dfrac{(5^2+6^2+14^2+15^2)-\dfrac{(5+6+14+15)^2}{4}}{4-1}}\\\\& = \sqrt{\dfrac{482-\dfrac{40^2}{4}}{3}}\\\\& = \sqrt{\dfrac{82}{3}}\\\\& = 5.23\end{aligned}$

$\begin{aligned}\displaystyle \textsf{Standard Deviation (Lisa)} & =\sqrt{\dfrac{(8^2+9^2+11^2+12^2)-\dfrac{(8+9+11+12)^2}{4}}{4-1}}\\\\& = \sqrt{\dfrac{410-\dfrac{40^2}{4}}{3}}\\\\& = \sqrt{\dfrac{10}{3}}\\\\& = 1.83\end{aligned}$

<h3><u>Summary</u></h3>

Nilo has a mean score of 10 and a standard deviation of 5.23.

Lisa has a mean score of 10 and a standard deviation of 1.83.

The <u>mean</u> scores are the <u>same</u>.

Nilo's standard deviation is higher than Lisa's. Therefore, Nilo's test scores are more <u>spread out</u> that Lisa's, which means Lisa's test scores are more <u>consistent</u>.

5 0

2 years ago

National data indicates that 35% of households own a desktop computer. In a random sample of 570 households, 40% owned a deskt

Elan Coil [88]

Answer:

Yes, this provide enough evidence to show a difference in the proportion of households that own a desktop.

Step-by-step explanation:

We are given that National data indicates that 35% of households own a desktop computer.

In a random sample of 570 households, 40% owned a desktop computer.

<em><u>Let p = population proportion of households who own a desktop computer</u></em>

SO, Null Hypothesis, $H_0$ : p = 25% {means that 35% of households own a desktop computer}

Alternate Hypothesis, $H_A$ : p $\neq$ 25% {means that % of households who own a desktop computer is different from 35%}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of 570 households who owned a desktop computer = 40%

n = sample of households = 570

So, <u><em>test statistics</em></u> = $\frac{0.40-0.35}{{\sqrt{\frac{0.40(1-0.40)}{570} } } } }$

= 2.437

<em>Since, in the question we are not given with the level of significance at which to test out hypothesis so we assume it to be 5%. Now at 5% significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics doesn't lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.</em>

Therefore, we conclude that % of households who own a desktop computer is different from 35%.

3 0

3 years ago