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3 years ago

Find the common ratio of the sequence -164, -82, -41, -20.5

Mathematics

1 answer:

Aleks [24]3 years ago

5 0

Answer:

r = $\frac{1}{2}$

Step-by-step explanation:

The common ratio r of a geometric sequence is

r = $\frac{a_{2} }{a_{1} }$ = $\frac{a_{3} }{a_{2} }$ = $\frac{a_{4} }{a_{3} }$ = ......

r = $\frac{-82}{-164}$ = $\frac{-41}{-82}$ = $\frac{1}{2}$

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Jane wants to save money to buy a motorcycle. She invests in an ordinary annuity that earns 4.2% interest, compounded quarterly.

Sauron [17]

Answer:

she need to pay is $550.40

Step-by-step explanation:

given data

interest = 4.2 % compounded quarterly = 0.042 / 4 = 0.0105

future value = $7000

time = 3 year = 3 × 4 = 12 months

to find out

How much money she need to pay

solution

we will apply here formula for future value for compound quarterly

that is

future value = principal × $\frac{(1+r)^{t} -1 }{r}$ .............1

put here all these value

future value = principal × $\frac{(1+r)^{t} -1 }{r}$

7000 = principal × $\frac{(1+0.0105)^{12} -1 }{0.0105}$

principal = 550.40

so she need to pay is $550.40

3 0

3 years ago

Alice purchased paint in a bucket with a radius of 9 in. and a height of 9.5 in. The paint cost $0.09 per in3.

ratelena [41]

Volume = (pi)( $radius^{2}$ (height)
$Cost = ($0.09)(3.14)(9*9)(9.5) = $217.4607

3 0

3 years ago

From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ

bulgar [2K]

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given by $P(x)=C(n,x)p^x(1-p)^{n-x}$ where $C(n,x)=\frac{n!}{x!(n-x)!}$
Here we're given
p=0.10 [ success = defective ]
n=3

(a) x=0
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,0)0.1^0(1-0.1)^{3-0}$
$=1(1)(0.729)$
$=0.729$

(b) x=2
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,2)0.1^2(1-0.1)^{3-2}$
$=3(0.01)(0.9)$
$=0.027$

(c) x ≥ 2
$P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}$
$=P(2)+P(3)$
$=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}$
$=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}$
$=3(0.01)(0.9)+1(0.001)1$
$=0.027+0.001$
$=0.028$

8 0

3 years ago

How many quarters are in $3.50

laila [671]

You only need 14 quaters

8 0

3 years ago

Find the angle of depression from the top of a lighthouse, 260 feet above water to a ship that's 270 feet offshore?

expeople1 [14]

Answer: OPTION C.

Step-by-step explanation:

Observe the triangle ABC attached.

Notice that the angle of depression is represented with $\alpha$ .

Knowing that the top of a lighthouse is 260 feet above water and the ship is 270 feet offshore, you can find the value of $\alpha$ by using arctangent:

$\alpha= arctan(\frac{opposite}{adjacent})$

In this case you can identify that:

$opposite=260\\adjacent=270$

Therefore, substuting values into $\alpha= arctan(\frac{opposite}{adjacent})$ , you get that the angle of depression is:

$\alpha= arctan(\frac{260}{270})\\\\\alpha=43.92\°$

4 0

3 years ago