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3 years ago

Select the correct text in the passage.

Mathematics

2 answers:

lukranit [14]3 years ago

3 0

Answer:

His wet feet froze the faster, and his exposed fingers numbed the faster, though they had not yet begun to freeze.

Step-by-step explanation:

Did this for edmentum and got it correct.

Assoli18 [71]3 years ago

3 0

Answer:

His wet feet froze the faster, and his exposed fingers numbed the faster, though they had not yet begun to freeze.

Step-by-step explanation:

I did the test on edmentum and got it right !!!

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Which pairs of numbers have a greatest common factor of 9

liubo4ka [24]

Answer:

81 and 72

Step-by-step explanation:

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3 years ago

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The proportion of high school seniors who are married is 0.02. Suppose we take a random sample of 300 high school seniors; a.) F

cricket20 [7]

Answer:

a) Mean 6, standard deviation 2.42

b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c) 14.85% probability that we find less than 4 of the seniors are married.

d) 99.77% probability that we find at least 1 of the seniors are married

Step-by-step explanation:

For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In this problem, we have that:

$n = 300, p = 0.02$

a.) Find the mean and standard deviation of the sample count X who are married.

Mean

$E(X) = np = 300*0.02 = 6$

Standard deviation

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42$

b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?

This is P(X = 8).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040$

10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c.) What is the probability that we find less than 4 of the seniors are married?

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023$

$P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143$

$P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436$

$P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485$

14.85% probability that we find less than 4 of the seniors are married.

d.) What is the probability that we find at least 1 of the seniors are married?

Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

From c), we have that $P(X = 0) = 0.0023$ . So

$0.0023 + P(X \geq 1) = 1$

$P(X \geq 1) = 0.9977$

99.77% probability that we find at least 1 of the seniors are married

3 0

3 years ago

Help with this please.

Irina-Kira [14]

Step-by-step explanation:

90 degree - 22 degree = 68 degree

use sin rule

sin90/x = sin68/56

x =60.4 degree

3 0

3 years ago

Can someone please help me I got this question wrong

lana66690 [7]

Answer: Answer is C, Associative and Distributive

Step-by-step explanation:

On step one to step two it shows that llz is added to 6z (Both have Z) and from step 2 to 3, 5 is distributed (Multiplied) by 17z which equals 85z and also Distributed (Multiplied) by 29 which gives you 145. Which makes final equation... 85z + 145

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4 years ago

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Use lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y = xyz; x^

Snezhnost [94]

I'm assuming the constraint involves some plus signs that aren't appearing for some reason, so that you're finding the extrema subject to $x^2+2y^2+3z^2=96$ .

Set $f(x,y,z)=xyz$ and $g(x,y,z)=x^2+2y^2+3z^2-96$ , so that the Lagrangian is

$L(x,y,z,\lambda)=xyz+\lambda(x^2+2y^2+3z^2-96)$

Take the partial derivatives and set them equal to zero.

$\begin{cases}L_x=yz+2\lambda x=0\\L_y=xz+4\lambda y=0\\L_z=xy+6\lambda z=0\\L_\lambda=x^2+2y^2+3z^2-96=0\end{cases}$

One way to find the possible critical points is to multiply the first three equations by the variable that is missing in the first term and dividing by 2. This gives

$\begin{cases}\dfrac{xyz}2+\lambda x^2=0\\\\\dfrac{xyz}2+2\lambda y^2=0\\\\\dfrac{xyz}2+3\lambda z^2=0\\\\x^2+2y^2+3y^2=96\end{cases}$

So by adding the first three equations together, you end up with

$\dfrac32xyz+\lambda(x^2+2y^2+3z^2)=0$

and the fourth equation allows you to write

$\dfrac32xyz+96\lambda=0\implies \dfrac{xyz}2=-32\lambda$

Now, substituting this into the first three equations in the most recent system yields

$\begin{cases}-32\lambda+\lambda x^2=0\\-32\lambda+2\lambda y^2=0\\-32\lambda+3\lambda z^2=0\end{cases}\implies\begin{cases}x=\pm4\sqrt2\\y=\pm4\\z=\pm4\sqrt{\dfrac23}\end{cases}$

So we found a grand total of 8 possible critical points. Evaluating $f(x,y,z)=xyz$ at each of these points, you find that $f(x,y,z)$ attains a maximum value of $\dfrac{128}{\sqrt3}$ whenever exactly none or two of the critical points' coordinates are negative (four cases of this), and a minimum value of $-\dfrac{128}{\sqrt3}$ whenever exactly one or all of the critical points' coordinates are negative.

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3 years ago