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3 years ago

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The mean absolute deviation (MAD) for the first set of data is 1.2 and the MAD for the second set of data is 0.6. Approximately

how many times the variability in the heights of the seventh graders is the variability in the heights of the sixth graders? (Round all values to the tenths place.)

1 answer:

Mekhanik [1.2K]3 years ago

3 0

The answer is 2.0 I promise

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A common blood test performed on pregnant women to screen for chromosome abnormalities in the fetus measures the human chorionic

goldfiish [28.3K]

Answer:

(a) The proportion of women who are tested, get a negative test result is 0.82.

(b) The proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

Step-by-step explanation:

The Bayes' theorem states that the conditional probability of an event <em>E</em> $_{i}$ , of the sample space <em>S,</em> given that another event <em>A</em> has already occurred is:

$P(E_{i}|A)=\frac{P(A|E_{i})P(E_{i})}{\sum\liits^{n}_{i=1}{P(A|E_{i})P(E_{i})}}$

The law of total probability states that, if events <em>E</em>₁, <em>E</em>₂, <em>E</em>₃... are parts of a sample space then for any event <em>A</em>,

$P(A)=\sum\limits^{n}_{i=1}{P(A|B_{i})P(B_{i})}$

Denote the events as follows:

<em>X</em> = fetus have a chromosome abnormality.

<em>Y</em> = the test is positive

The information provided is:

$P(X)=0.04\\P(Y|X)=0.90\\P(Y^{c}|X^{c})=0.85$

Using the above the probabilities compute the remaining values as follows:

$P(X^{c})=1-P(X)=1-0.04=0.96$

$P(Y^{c}|X)=1-P(Y|X)=1-0.90=0.10$

$P(Y|X^{c})=1-P(Y^{c}|X^{c})=1-0.85=0.15$

(a)

Compute the probability of women who are tested negative as follows:

Use the law of total probability:

$P(Y^{c})=P(Y^{c}|X)P(X)+P(Y^{c}|X^{c})P(X^{c})$

$=(0.10\times 0.04)+(0.85\times 0.96)\\=0.004+0.816\\=0.82$

Thus, the proportion of women who are tested, get a negative test result is 0.82.

(b)

Compute the value of P (X|Y) as follows:

Use the Bayes' theorem:

$P(X|Y)=\frac{P(Y|X)P(X)}{P(Y|X)P(X)+P(Y|X^{c})P(X^{c})}$

$=\frac{(0.90\times 0.04)}{(0.90\times 0.04)+(0.15\times 0.96)}$

$=0.20$

Thus, the proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

6 0

3 years ago

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side <em>so proof is complete</em>

8 0

3 years ago

Read 2 more answers

Negative nine is greater than or equal to the sum of two - fifths times a number and 9.

Bumek [7]

Answer:

W is either -22.5, or less than -22.5

Step-by-step explanation:

-9>=2/5 times w (I wasn't sure what you meant by "and" 9. Like 9w?)

W has to be a negative number, or else it would be greater than 9, which would make the equation wrong.

2/5 times -22 1/2=-9. So w can be -22 1/2 or -22.5

W can then be anything "lower" (on the left side) of -22.5.

W<-22.5

5 0

3 years ago

Yara multiplies and divides a certain number by the same

Butoxors [25]

For this case we have to:

x: Represents the unknown number

y: Represents the exponent of base 10

So, considering a system of equations we have:

$x * 10 ^ y = 40,000\\\frac {x} {10 ^ y} = 0.000004$

From the first equation we have:

$10 ^ y = \frac {40,000} {x}$

Substituting in the second equation:

$\frac {x} {\frac {40,000} {x}} = 0.000004\\\frac {x ^ 2} {40,000} = 0.000004\\x ^ 2 = 40,000 * 0.000004\\x ^ 2 = 0.16\\x = \pm \sqrt {0.16}$

We choose the positive value:

$x = 0.4$

So, the number is 0.4.

So:

$10 ^ y = \frac {40,000} {0.4}\\10 ^ y = 100,000\\y = 5$

Thus, the exponent is 5.

Answer:

Yara use 0.4 and $10 ^ 5$

8 0

3 years ago

Given f(x) = (3x – 12) +2, what is the domain in set notation?

sattari [20]

Answer:4

Step-by-step explanation:

6 0

3 years ago