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Veseljchak [2.6K]
3 years ago
13

Glen, Andre, and Ramón share the cost of renting a vacation cottage. The total cost is $960. Glen pays 30% of the total cost. An

dre pays 0.45 of the total cost. Ramón pays the rest. How much does Ramón pay?
Mathematics
1 answer:
SashulF [63]3 years ago
8 0

Answer:

Ramon pays $240.

Step-by-step explanation:

30% of $960 is $288, so Glen pays $288

0.45 = 45%

45% of $960 is $432, so Andre pays $432

30% + 45% = 75% meaning Ramon pays 100% - 75% which is 25%.

25% of $960 is $240. So, Ramon pays $240.

You could also add up the payments of Andre and Glen and subtract that sum from $960 to get the same answer.

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Step-by-step explanation:

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find the mean and standard deviation of difference.

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Answer:

The sequence of transformations that maps ΔABC to ΔA'B'C' is the reflection across the <u>line y = x</u> and a translation <u>10 units right and 4 units up</u>, equivalent to T₍₁₀, ₄₎

Step-by-step explanation:

For a reflection across the line y = -x, we have, (x, y) → (y, x)

Therefore, the point of the preimage A(-6, 2) before the reflection, becomes the point A''(2, -6) after the reflection across the line y = -x

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5 0
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
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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)

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the your answer is 1000

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