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3 years ago

7

Suppose the professor decides to grade on a curve. If the professor wants 16% of the students to get an A, what is the minimum s

core for an A?

1 answer:

vladimir1956 [14]3 years ago

3 0

There isn't enough info to determine that, I believe. You would need an equation that would allow me to determine the minimum output for an A.

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If ΔABC ≅ ΔEDF where the coordinates of A(0, 2), B(2, 4), and C(2, −1), what is the measure of DF?

allsm [11]

Answer:

C. 5

Step-by-step explanation:

Since Triangle ABC is congruent to EDF it mean that the sides are the same so the length of BC is congruent to the length of DF:

distance formula:

d = √(x2 - x1)^2 + (y2 - y1)^2

d = √(2 - 2)^2 + (-1 - 4)^2

d = √(0)^2 + (-5)^2

d = √0 + 25

d = √25

d = 5

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3 years ago

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A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

3 years ago

8(r+9) What is the answer?

Andreyy89

Answer:

8r+72

Step-by-step explanation:

4 0

2 years ago

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Passes through 5,4 slope -3 ASAP

aksik [14]

The answer should be
Y=-3x+19

5 0

3 years ago

10. If 1000 cm = 1 l, find the capacity of the following water tank in litre

Lelu [443]

<h2>Given :</h2>

length = 200 cm

breadth = 150 cm

height = 100 cm

<h2>Solution : </h2>

$\large \boxed{volume = l \times b \times h}$

$200 \times 150 \times 100$

$3000000 \: \: cm {}^{3}$

Through the statement above we have,

$\small\mathrm{1000 \: cm³ = 1 \: \: }l$

$\small\mathrm{1 \: cm {}^{3} = \dfrac{1}{1000} \: }l$

$\small\mathrm{3000000 \: cm {}^{3} = 3000000 \times \dfrac{1}{1000} \: }l$

$\small\mathrm{3000000 \: cm {}^{3} = 3000000 \times \dfrac{1}{1000} \: }l$

$\small\mathrm{ 3000 \: \: litres}$

_____________________________

$\mathrm{ \#TeeNForeveR}$

3 0

3 years ago