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3 years ago

Which of the following numbers completes the sequence 3, 8, 14, 21, 29,

1 answer:

5 0

I think that the number that would complete the number sequence is 38. My reasoning is that as the numbers increase, there is a sort of pattern to them. 3 goes to 8, increasing by 5. 8 goes to 14, increasing by 6. 12 goes to 21, increasing by 7. 21 increases to 29 by 8, so logically, 29 would increase to 38 by 9, following the current pattern of increase.

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F(x) = 3x + 5<br> Work out a if f(a)= -13

vekshin1

Answer:

a = - 6

Step-by-step explanation:

Given

f(x) = 3x + 5 , then

f(a) = 3a + 5 and f(a) = - 13 , then equating the right sides

3a + 5 = - 13 ( subtract 5 from both sides )

3a = - 18 ( divide both sides by 3 )

a = - 6

6 0

3 years ago

Read 2 more answers

Question Which sequence of transformations maps pre-image ABC to image A′B′C′ ? Triangle ABC was reflected over the y-axis and t

Stells [14]

The sequence of transformations was Triangle ABC was reflected over the y-axis and then translated 2 units up and 2 units to the right.

<h3>What is a transformation?</h3>

Transformation is the movement of a point from its initial location to a new location. Types of transformation are reflection, translation, rotation and dilation.

Rigid transformation is the transformation that does not change the shape or size of a figure. Examples of rigid transformations are <em>translation, reflection and rotation</em>.

Find out more on transformation at: brainly.com/question/4289712

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7 0

2 years ago

There are 4 people at a party. Consider the random variable X=’number of people having the same birthday ’ (match only month, N=

yulyashka [42]

Answer:

S = {0,2,3,4}

P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001

Mean = 0.879

Standard Deviation = 1.033

Step-by-step explanation:

Let the number of people having same birth month be = x

The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)

The number of ways of distributing their birthdays = 12⁴

The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)

P(X = 0) = $\frac{12P4}{12^{4} }$

P(X=0) = 0.573

P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)

P(X=2) = $\frac{3C2 * 12P2}{12^{4} } + \frac{4C2 * 12P3}{12^{4} }$

P(X=2) = 0.401

P(X=3) = $\frac{4C3 * 12P2}{12^{4} }$

P(x=3) = 0.025

P(X=4) = $\frac{12}{12^{4} }$

P(X=4) = 0.001

Mean, $\mu = \sum xP(x)$

$\mu = (0*0.573) + (2*0.401) + (3*0.025) + (4*0.001)\\\mu = 0.879$

Standard deviation, $SD = \sqrt{\sum x^{2} P(x) - \mu^{2}} \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2} * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}$