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3 years ago

The box plots below show student grades on the most recent exam compared to overall grades in the class:

Mathematics

1 answer:

ANTONII [103]3 years ago

3 0

Answer:

1) The class and exam medians are almost the same.

Step-by-step explanation:

1) According to the given data (box-plot) the median of Class is 84 and median of Exam is 85.

Thus, the medians of both class and exam are almost same.

Hence, this is the correct option.

2) Since the median of Class is 84 and median of Exam is 85, Thus, we can't say median of exam is much high than median of class.

Hence, this option is not correct.

3) Since Q3 of Class is 89 and Q3 of exam is 94, which is not equal.

Hence, this option is also not correct.

4) Since the median of Class is 84 and median of Exam is 85, Thus, median of exam is not lower than median of Class.

Hence, this option is also not correct.

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Ginny is studying a population of frogs. She determines that the population is decreasing at an average rate of 3% per year. Whe

Rama09 [41]

You can see how this works by thinking through what's going on.

In the first year the population declines by 3%. So the population at the end of the first year is the starting population (1200) minus the decline: 1200 minus 3% of 1200. 3% of 1200 is the same as .03 * 1200. So the population at the end of the first year is 1200 - .03 * 1200. That can be written as 1200 * (1 - .03), or 1200 * 0.97

What about the second year? The population starts at 1200 * 0.97. It declines by 3% again. But 3% of what??? The decline is based on the population at the beginning of the year, NOT based no the original population. So the decline in the second year is 0.03 * (1200 * 0.97). And just as in the first year, the population at the end of the second year is the population at the beginning of the second year minus the decline in the second year. So that's 1200 * 0.97 - 0.03 * (1200 * 0.97), which is equal to 1200 * 0.97 (1 - 0.03) = 1200 * 0.97 * 0.97 = 1200 * 0.972.

So there's a pattern. If you worked out the third year, you'd see that the population ends up as 1200 * 0.973, and it would keep going like that.

So the population after x years is 1200 * 0.97x

5 0

3 years ago

WILL MARK BRAINLIEST!! Given ΔABC : ΔDEF, find x. 10 14 27 12

lilavasa [31]

Answer:

x=12

Step-by-step explanation:

If the triangles are similar:

x/8=18/12

x/8=1.5

x=12

7 0

3 years ago

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Solve for x. 3/x-1 - 1/x^-1=5/x-1 Please show all work

navik [9.2K]

3/(x-1) - 1/(x^2-1) = 5/(x-1)

Subtract 3/(x-1) from both sides

-1/(x^2-1) = 2/(x-1)

Factor x^2 - 1

-1/[(x-1)(x+1] = 2/(x-1)

Multiply by (x-1)(x+1) on both sides

-1 = 2 (x+1)

-1 = 2x + 2

Subtract 2 from both sides

-3 = 2x

Divide by 2 on both sides

-3/2 = x

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2 years ago

Read 2 more answers

The first rule is subtract 6 starting from 48. The second rule is subtract 3 starting from 30. What is the second ordered pair u

Natasha2012 [34]

Answer:

(42,48)?

Step-by-step explanation:

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2 years ago

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(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX

pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable X follows a Normal distribution with mean μ = 10 and variance σ² = 36.

(a)

Compute the value of P (X > 5) as follows:

$P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z$

Thus, the value of P (X > 5) is 0.7967.

(b)

Compute the value of P (4 < X < 16) as follows:

$P(4$

Thus, the value of P (4 < X < 16) is 0.6826.

(c)

Compute the value of P (X < 8) as follows:

$P(X$

Thus, the value of P (X < 8) is 0.3707.

(d)

Compute the value of P (X < 20) as follows:

$P(X$

Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

$P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z$

Thus, the value of P (X > 16) is 0.1587.

**Use a z-table for the probabilities.

8 0

3 years ago