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inessss [21]
3 years ago
8

Find the values for x, y, and z.

Mathematics
2 answers:
PSYCHO15rus [73]3 years ago
5 0
Ok The volume of x is 2x+=3 hope this helps
miv72 [106K]3 years ago
3 0

Answer:

X = 65

Y = 55

Z = 25

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Complete the two-column proof by providing the missing statements or reasons.
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Step-by-step explanation:

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Tim went shopping and spent $31.79 at each 3 stores
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What I'm getting out of it is that you want to know how much he spent in all and if you mean that he spent $95.37 in all.
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6 0
3 years ago
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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
4 years ago
Absolute value of [4-9]​
Firdavs [7]

Answer:

-5

Step-by-step explanation:

==> [4-9]

==>[-5]

==> -5

6 0
2 years ago
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