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3 years ago

Find the first,fourth,and tenth terms of the arithmetic sequence described by the given rule. A(n) = 12+(n-1)(3)

1 answer:

4 0

The answers are:
the first term: 21
the fourth term: 22
the tenth term: 39

A(n) = 12+(n - 1)(3)

n = 1
A(1) = 12 + (1 - 1)(3) = 12 + 0*3 = 12 + 0 = 12

n = 4
A(4) = 12 + (4 - 1)(3) = 12 + 3*3 = 12 + 9 = 21

n = 10
A(10) = 12 + (10 - 1)(3) = 12 + 9*3 = 12 + 27 = 39

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Test scores: Scores on a statistics exam had a mean of 75 with a standard deviation of 5. Scores on a calculus exam had a mean o

beks73 [17]

Answer:

CV for statistics exam = 15%

CV for calculus exam = 19%

Since the CV for calculus exam is higher, it has a greater spread relative to the mean than the statistics exam.

Step-by-step explanation:

To find coefficient variation we use the formula:

CV = (SD/mean) * 100

CV for the statistics exam:

where; SD= 5

mean= 75

CV = ( 5/75) *100

= 0.15 or 15%

CV for calculus exam

SD = 11

Mean= 58

CV= (11 /58) * 100

= 0.19 or 19%

8 0

3 years ago

1 Point

I am Lyosha [343]

Option C

The ratio for the volumes of two similar cylinders is 8 : 27

<h3>Solution:</h3>

Let there are two cylinder of heights "h" and "H"

Also radius to be "r" and "R"

$\text { Volume of a cylinder }=\pi r^{2} h$

Where π = 3.14 , r is the radius and h is the height

Now the ratio of their heights and radii is 2:3 .i.e

$\frac{\mathrm{r}}{R}=\frac{\mathrm{h}}{H}=\frac{2}{3}$

Ratio for the volumes of two cylinders

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{\pi r^{2} h}{\pi R^{2} H}$

Cancelling the common terms, we get

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{\mathrm{r}}{R}\right)^{2} \times\left(\frac{\mathrm{h}}{\mathrm{H}}\right)$

Substituting we get,

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{2}{3}\right)^{2} \times\left(\frac{2}{3}\right)$

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{2 \times 2 \times 2}{3 \times 3 \times 3}$

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{8}{27}$

Hence, the ratio of volume of two cylinders is 8 : 27

7 0

3 years ago

Describe the diagram below using correct geometric names

bagirrra123 [75]

Angles PQT and RQS are vertical angles and angles PQR and TQS are vertical angles.

angle PQS and TQR are straight angles

angles TQP and PQR form a linear pair. they are also adjacent by default. the other angles have these same features.

that should bout do it. hope it helped!

7 0

3 years ago

For 30 minutes you do a combination of walking and jogging. At the end of your workout your pedometer displays a total of 2.5 mi

yKpoI14uk [10]

Walking for 10 minutes

4 0

3 years ago

I NEED HELP!! answer from the image below m∠BCD =

nevsk [136]

Answer:

28 degrees

Step-by-step explanation:

If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent.The corresponding angles are the ones at the same location at each intersection.

In the figure ABD and EDF are corresponding angles. So they are equal

So equation the angle ABD and EDF , we get

(3x+4) = (7x-20)

Group the like terms,

3x-7x = -20-4

-4x = -24

$x = \frac{-24}{-4}$

x = 6

Thus BCD will be,

(6x - 8)

=>(6(6)-8)

=>(36-8)

=> 28 degrees

4 0

3 years ago