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4 years ago

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Mathematics

2 answers:

Arte-miy333 [17]4 years ago

8 0

There is an infinite number of values that are in both the domain and range.

arsen [322]4 years ago

7 0

Answer:

100% sure There is an infinite number of values that are in both the domain and range.

Step-by-step explanation:

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Deidre is picking out some movies to rent, and she is primarily interested in documentaries and children's movies. She has narro

exis [7]

Answer: 6630

Step-by-step explanation:

Given , Number of documentaries = 17

Number of children's movies = 20

Total movies = 17+20=37

Number of combinations of r things taken out of things = $^nC_r=\dfrac{n!}{r!(n-r)!}$

Now, the number of different combinations of 3 movies can she rent if she wants at least one documentary

= (1 documentary+2 children's movies , 2 documentary+1 children's movies , 3 documentary+0 children's movies)

$=^{17}C_1\times^{20}C_{2}+^{17}C_2\times^{20}C_{1}+^{17}C_3\times^{20}C_{0}$

$=(17)\times\dfrac{20!}{2!18!}+\dfrac{17!}{2!15!}\times(20)+\dfrac{17!}{3!14!}(1)\\\\=3230+2720+680=6630$

Hence, the number of different combinations of 3 movies can she rent if she wants at least one documentary is 6630 .

3 0

4 years ago

Alice searches for her term paper in her filing cabinet, which has several drawers. She knows thatshe left her term paper in dra

katen-ka-za [31]

You made a mistake with the probability $p_{j}$ , which should be $p_{i}$ in the last expression, so to be clear I will state the expression again.

So we want to solve the following:

Conditioned on this event, show that the probability that her paper is in drawer $j$ , is given by:

(1) $\frac{p_{j} }{1-d_{i}p_{i} } , if j \neq i,$ and

(2) $\frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} } , if j = i.$

so we can say:

$A$ is the event that you search drawer $i$ and find nothing,

$B$ is the event that you search drawer $i$ and find the paper,

$C_{k}$ is the event that the paper is in drawer $k, k = 1, ..., n.$

this gives us:

$P(B) = P(B \cap C_{i} ) = P(C_{i})P(B | C_{i} ) = d_{i} p_{i}$

$P(A) = 1 - P(B) = 1 - d_{i} p_{i}$

Solution to Part (1):

if $j \neq i$ , then $P(A \cap C_{j} ) = P(C_{j} )$ ,

this means that

$P(C_{j} |A) = \frac{P(A \cap C_{j})}{P(A)} = \frac{P(C_{j} )}{P(A)} = \frac{p_{j} }{1-d_{i}p_{i} }$

as needed so part one is solved.

Solution to Part(2):

so we have now that if $j$ = $i$ , we get that:

$P(C_{j}|A ) = \frac{P(A \cap C_{j})}{P(A)}$

remember that:

$P(A|C_{j} ) = \frac{P(A \cap C_{j})}{P(C_{j})}$

this implies that:

$P(A \cap C_{j}) = P(C_{j}) \cdot P(A|C_{j}) = p_{i} (1-d_{i} )$

so we just need to combine the above relations to get:

$P(C_{j}|A) = \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} }$

as needed so part two is solved.

8 0

4 years ago

An SRS of 450 450 high school seniors gained an average of ¯ x = 20 x¯=20 points in their second attempt at the SAT Mathematics

maria [59]

Answer: (15.47, 24.53)

Step-by-step explanation:

We know that the confidence interval for population mean is given by :_

$\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}$

, where n= sample size.

$\sigma$ = standard deviation.

$\overline{x}$ = sample mean.

z*= Critical value.

Given : n= 450

$\overline{x}=20$

$\sigma=49$

Critical value for 95% confidence = z*=1.96 [From z-value table]

Then, the 95% confidence interval will be :-

$20\pm (1.96)\dfrac{49}{\sqrt{450}}$

$\approx 20\pm (1.96)(2.31)$

$\approx 20\pm 4.53$

$=(20-4.53,\ 20+4.53)=(15.47,\ 24.53)$

Hence, the 95% confidence interval for the mean change in score μ μ in the population of all high school seniors. : (15.47, 24.53)

3 0

3 years ago

What is the quotient of the fractions below? 7/2 divided by 1/3

notsponge [240]

21/2 is what you should get if you divide those two

8 0

3 years ago

3p + 4r = q solve for p

Aleksandr-060686 [28]

Answer:

$p = \dfrac{q - 4r}{3}$

Step-by-step explanation:

3p + 4r = q

3p = q - 4r

$p = \dfrac{q - 4r}{3}$

7 0

4 years ago

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