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2 years ago

A 14B 8C 11D 20I'm pose to simplify it

Mathematics

2 answers:

Tomtit [17]2 years ago

5 0

Try photo math it shows you the answer

UkoKoshka [18]2 years ago

5 0

Hopes this helps:

Answer: 11

1. Simplify 3^2 to 9.
9 + 1 • 2

2. Simplify 1 • 2 to 2.
9 + 2

3. Simplify.
11

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What is 4x squared minus x squared

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3x^2 or 3x squared

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4x^2-x^2=3x^2

4x squared-x squared equals x squared

when subtracting squared it's like subtracting normal numbers

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2 years ago

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2 years ago

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

3 years ago

A 14B 8C 11D 20I'm pose to simplify it ​

A 14B 8C 11D 20I'm pose to simplify it