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2 years ago

The diagram shows several points and lines.

Mathematics

1 answer:

Ghella [55]2 years ago

4 0

Answer:

Option B And Option E

Step-by-step explanation:

three or more points are said to be collinear if they lie on a single straight line.

In the given figure :

Points J, K, and Q are collinear.

Through any two given points only one line can be drawn.

In the given figure:

There is only one line that can be drawn through points L and P.

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A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d

tankabanditka [31]

The selection of r objects out of n is done in

$C(n, r)= \frac{n!}{r!(n-r)!}$ many ways.

The total number of selections 10 that we can make from 6+7=13 students is

$C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}= 286$
thus, the sample space of the experiment is 286

A.
"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.

 $C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35$

P(all 6 girls chosen)=35/286=0.12

B.
"The probability that a randomly chosen team has 3 girls and 7 boys."

with the same logic as in A, the number of groups were all 7 boys are in, is

 $C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20$

so the probability is 20/286=0.07

C.
"The probability that a randomly chosen team has either 4 or 6 boys."

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is 0.12.

case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.

 $C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105$

the probability is 105/286=0.367

since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"The probability that a randomly chosen team has 5 girls and 5 boys."

selecting 5 boys and 5 girls can be done in

 $C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126$

many ways,

so the probability is 126/286=0.44

6 0

2 years ago

Read 2 more answers

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equ

kompoz [17]

Answer:

0.62% probability that a random sample of 16 bulbs will have an average life of less than 775 hours.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution.

Problems of normally distributed samples 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.

In this problem, we have that:

$\mu = 800, \sigma = 40, n = 16, s = \frac{40}{\sqrt{16}} = 10$

Find the probability that a random sample of 16 bulbs will have an average life of less than 775 hours.

This probability is the pvalue of Z when $X = 775$ . So

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

$Z = \frac{775 - 800}{10}$

$Z = -2.5$

$Z = -2.5$ has a pvalue of 0.0062. So there is a 0.62% probability that a random sample of 16 bulbs will have an average life of less than 775 hours.

5 0

2 years ago

A sports statistician was interested in the relationship between game attendance (in thousands) and the number of

Luda [366]

Answer:

The predicted number of wins for a team that has an attendance of 2,100 is 25.49.

Step-by-step explanation:

The regression equation for the relationship between game attendance (in thousands) and the number of wins for baseball teams is as follows:

$\hat y = 4.9\cdot x + 15.2$

Here,

y = number of wins

x = attendance (in thousands)

Compute the number of wins for a team that has an attendance of 2,100 as follows:

$\hat y = 4.9\cdot x + 15.2$

$=(4.9\times 2.1)+15.2\\=10.29+15.2\\=25.49$

Thus, the predicted number of wins for a team that has an attendance of 2,100 is 25.49.

4 0

2 years ago

Which number, when substituted for x, makes the equation 4x+8=164x+8=16 true?

ra1l [238]

X = 2

4x + 8 = 16
-8 -8
----------------
4x= 8
-- --
4 4
------------
x=2

7 0

2 years ago

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Given: Circle X with a Radius r and circle Y with radius s Prove: Circle x is similar to circle y

lana [24]

Two figures are similar if one is the scaled version of the other.

This is always the case for circles, because their geometry is fixed, and you can't modify it in anyway, otherwise it wouldn't be a circle anymore.

To be more precise, you only need two steps to prove that every two circles are similar:

Translate one of the two circles so that they have the same center
Scale the inner circle (for example) unit it has the same radius of the outer one. You can obviously shrink the outer one as well

Now the two circles have the same center and the same radius, and thus they are the same. We just proved that any two circles can be reduced to be the same circle using only translations and scaling, which generate similar shapes.

Recapping, we have:

Start with circle X and radius r
Translate it so that it has the same center as circle Y. This new circle, say X', is similar to the first one, because you only translated it.
Scale the radius of circle X' until it becomes $s$ . This new circle, say X'', is similar to X' because you only scaled it

So, we passed from X to X' to X'', and they are all similar to each other, and in the end we have X''=Y, which ends the proof.

8 0

2 years ago

Read 2 more answers