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3 years ago

Find the fifth term of a sequence whose first term is 8 and common ratio is 3/2

Mathematics

1 answer:

Arlecino [84]3 years ago

3 0

Answer:

Option c. 81/2

Step-by-step explanation:

Fifth term: a5=?

First term: a1=8

Common ratio: r=3/2

an=a1*r^(n-1)

with n=5

a5=a1*r^(5-1)

a5=a1*r^4

Replacing the known values:

a5=8*(3/2)^4

a5=8*(3)^4/(2)^4

a5=8*81/16

Simplifying: Dividing the numerator and denominator by 8:

a5=(8*81/8)/(16/8)

a5=81/2

Answer: Option c. 81/2

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CD is a perpendicular bisector of . Which of the following could be the slopes of the two lines if plotted on a coordinate grid?

NISA [10]

Step-by-step explanation:

A is the correct answer of this question

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3 years ago

A chemist recorded the change in temperature of a cooling liquid as −25.5°F over a 1.5 minute period. The temperature fell at a

Strike441 [17]

The rate of change in temperature is −17°F/min

<h3>Rate of change of a function</h3>

The formula for calculating the rate of change of a function is expressed as:

Rate of change = rise/run

This is also known as the slope of the function

Given the following parameters

Temperature = −25.5°F

Cooling time = 1.5mins

Determine the rate of change

Rate = −25.5°F/1.5min

Rate = −17°F/min

Hence the rate of change in temperature is −17°F/min

Learn more on rate of change here: brainly.com/question/8728504

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2 years ago

A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Kar

ch4aika [34]

Answer:

We conclude that the mean waiting time is less than 10 minutes.

Step-by-step explanation:

We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.

Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.8 minutes with a standard deviation of 2.5 minutes.

Let $\mu$ = mean waiting time for bus number 14.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 10 minutes {means that the mean waiting time is more than or equal to 10 minutes}

Alternate Hypothesis, $H_A$ : $\mu$ < 10 minutes {means that the mean waiting time is less than 10 minutes}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean waiting time = 7.8 minutes

s = sample standard deviation = 2.5 minutes

n = sample of different occasions = 18

So, test statistics = $\frac{7.8-10}{\frac{2.5}{\sqrt{18} } }$ ~ $t_1_7$

= -3.734

The value of t test statistics is -3.734.

Now, at 0.01 significance level the t table gives critical value of -2.567 for left-tailed test.

Since our test statistic is less than the critical value of t as -3.734 < -2.567, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean waiting time is less than 10 minutes.

5 0

3 years ago

Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 b

jeka94

Answer:

Step-by-step explanation:

From the given information; Let's assume that R should represent the set of all possible outcomes generated from a bit string of length 10 .

So; as each place is fitted with either 0 or 1

$\mathbf{|R|= 2^{10}}$

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0

Now;

if a 0 bit and a 1 bit are equally likely

The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

$\mathbf{P(E) = \dfrac{|E|}{|R|}}$

so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:

$\mathbf{{|E|}=1 }$ ; $\mathbf{|R|= 2^{10}}$

$\mathbf{P(E) = \dfrac{1}{2^{10}}}$

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3 years ago

Is it possible for the area of a figure to change after it is translated? Explain.

Mariulka [41]

It is not possible for the area to change with a translation because translations are simply movements. They are not changing anything about the object's form, like something such as scaling does.

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3 years ago

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