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3 years ago

6

Colin is designing a series of paper chains. The second paper chain has 9 links. The third chain has 13 links, and the fourth ch

ain has 17 links. If all of Colin's chains follow the same sequence, how many links were there in colin's first chain?

1 answer:

Kay [80]3 years ago

3 0

The first chain is 5 because 17-13 equals 4 and 9-4 equals 5 so that's the answer

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2115-0830 in millitary time please and thank you so much if i get it right

neonofarm [45]

Answer:

2115 = 9:15 PM

0830 = 8:30 AM

Step-by-step explanation:

12 - hour conversion.

7 0

3 years ago

Simplify 7x²-x²<br> how to simplify this maths question

kondaur [170]

Hi student, let me help you out! :)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We are asked to simplify the expression $\large\pmb{7x^2-x^2}$ .

$\triangle~\fbox{\bf{KEY:}}$

Combine like terms! :)

That's exactly what we need to do - combine like terms.

So, we subtract:

$\star~\large\pmb{7x^2-x^2}$

Which gives us:

$\star~\large\pmb{6x^2}$

Hope it helps you out! :D

Ask in comments if any queries arise.

#StudyWithBrainly

~Just a smiley person helping fellow students :)

6 0

2 years ago

Two mechanics worked on a car. The first mechanic charged $115 per hour, and the second mechanic charged $50 per hour. The mecha

anygoal [31]

115*20=2300.
50*5=250

First mechanic 20 hours
Second mechanic 5 hours

8 0

3 years ago

The heights of a certain type of tree are approximately normally distributed with a mean height p = 5 ft and a standard

arsen [322]

Answer:

A tree with a height of 6.2 ft is 3 standard deviations above the mean

Step-by-step explanation:

⇒ $1^s^t$ statement: A tree with a height of 5.4 ft is 1 standard deviation below the mean(FALSE)

an X value is found Z standard deviations from the mean mu if:

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

In this case we have: $\mu=5\ ft$ $\sigma=0.4\ ft$

We have four different values of X and we must calculate the Z-score for each

For X =5.4\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.4-5}{0.4}=1$

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.

⇒ $2^n^d$ statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean. (FALSE)

For X =4.6 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{4.6-5}{0.4}=-1$

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean .

⇒ $3^r^d$ statement:A tree with a height of 5.8 ft is 2.5 standard deviations above the mean (FALSE)

For X =5.8 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.8-5}{0.4}=2$

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.

⇒ $4^t^h$ statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean. (TRUE)

For X =6.2\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3$

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.

6 0

3 years ago

ABC is a tangent to the circle below. O is the centre of the circle. Work out the size of angle 0. Give reasoning to justify you

Delvig [45]

Answer:

Θ = 46°

Step-by-step explanation:

the angle between a tangent and a radius at the point of contact is 90° , so

∠ ABO = 90°

since OB = OD ( radii of circle ) then Δ BOD is isosceles and

∠ OBD = ∠ ODB = 22°

the exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

∠ AOB is an exterior angle of the triangle , then

∠ AOB = 22° + 22° = 44°

the sum of the 3 angles in Δ AOB = 180° , then

Θ + 44° + 90° = 180°

Θ + 134° = 180° ( subtract 134° from both sides )

Θ = 46°

6 0

2 years ago