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4 years ago

Skew lines, rays, and segments will sometimes intersect? true or false

Mathematics

1 answer:

MissTica4 years ago

6 0

Although skew lines do not intersect, rays and segments sometimes do, so i would say true?

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Solve the simultaneous equations C^2+d^2=5 and 3c+4d=2

Naily [24]

If 3c+4d=2 then c=(2-4d)/3
And if you put that into the other equation
${\frac{2 - 4d}{3}}^{2} + {d}^{2} = 5$
Then you multiply everything by 3, getting

${(2 - 4d)}^{2} + 3 {d}^{2} = 15$
And then

$2 - 4d + \sqrt{3} d = \sqrt{15}$
Leaving
$d( \sqrt{3} - 4) = \sqrt{15} - 2$
I dont have a calc with me right now but you just pot this into the calculator:
$d = \frac{ \sqrt{15} - 2 }{ \sqrt{3} - 4}$
And when you get the number you insert it into the nest eqation to get c

$c = \frac{2 - 4d}{3}$
So basically

$c = \frac{2 - 4 \frac{ \sqrt{15} - 2}{ \sqrt{3} - 4} }{3}$

6 0

3 years ago

Read 2 more answers

What are the name of these angles? Are<br> they congruent or supplementary?

kumpel [21]

They are supplementary

4 0

3 years ago

2y=4x−2 simplify the equation

frozen [14]

Answer:

y=2x-1

Step-by-step explanation:

4 0

3 years ago

Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that 104 eligible vot

crimeas [40]

Answer:

The probability that exactly 27 of 104 eligible voters voted is 0.057 = 5.7%.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

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

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this case, assume that 104 eligible voters aged 18-24 are randomly selected.

This means that $n = 104$ .

Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted.

This means that $p = 0.22$

Mean and standard deviation:

$\mu = 104*0.22 = 22.88$

$\sigma = \sqrt{104*0.22*0.78} = 4.2245$

Probability that exactly 27 voted

By continuity continuity, 27 consists of values between 26.5 and 27.5, which means that this probability is the p-value of Z when X = 27.5 subtracted by the p-value of Z when X = 26.5.

X = 27.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{27.5 - 22.88}{4.2245}$