Find the sum of the first 8 terms of the geometric sequence if the first term is 9 and the common ratio is -3.

Answers

Answer 1

Answer:
hmmmm let's see $\bf \textit{sum of a finite geometric sequence}=S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\n\n\begin{cases}n=n^(th)\ term\na_1=\textit{first term}\nr=\textit{common ratio}\end{cases}\qquad thus\implies S_8=9\left( \cfrac{1-(-3)^8}{1-(-3)} \right)$

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X²+4x+20=0a=1 b=20 c=4
a=1 b=4 c=-20
a=1 b =4 c = 20
a=1 b=-4 c=0

2x²+6x+5=0

-2x²+x+5 identify the coefficients

Answers

Step-by-step explanation:

The general form of the quadratic equation is :

$ax^2+bx+c$

The given equation is x²+4x+20=0

On comparing the given equation and the general equation, we find that

a = 1, b = 4 and c = 20

Hence, the correct option is a=1 b =4 c = 20.

If the equation is 2x²+6x+5=0, then

a = 2, b = 6 and c = 5

If the equation is -2x²+x+5

a = -2, b = 1 and c = 5

Find the perimeter of the figure.
Please consider helping!

Answers

Answer:

Perimeter is 56

Step-by-step explanation:

In all, 54 children were assigned to the Montessori school and 115 to other schools at age three. When the children were five, parents of 38 of the Montessori children and 34 of the others could be located and agreed to and subsequently participated in testing. This information reveals a possible source of bias in the comparison of outcomes. Explain why. (Round your answers to the nearest whole number.) About % of Montessori parents participated in the study, compared to about % of the other parents. (c) One of the many response variables was score on a test of ability to apply basic mathematics to solve problems. Here are summaries for the children who took this test:

Answers

Full question attached

Answer:

Option C: because the percent of children compared differ too much

Explanation:

Sample Proportion of students from Montessori = students that participated from Montessori /total number of students assigned to Montessori

=30/54= 0.56

Sample proportion of students from other schools = students that participated I'm the study /total number of students assigned to other schools

=25/112

=0.22

We observe that there is a large difference(0.56-0.22) between the two sample proportions hence the bias

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historically, the failure rate for LED light bulbs that the company manufactures is 5%. Suppose a random sample of 10 LED light bulbs is selected. What is the probability that a) None of the LED light bulbs are defective? b) Exactly one of the LED light bulbs is defective? c) Two or fewer of the LED light bulbs are defective? d) Three or more of the LED light bulbs are not defective?

Answers

Answer:

a) There is a 59.87% probability that none of the LED light bulbs are defective.

b) There is a 31.51% probability that exactly one of the light bulbs is defective.

c) There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) There is a 100% probability that three or more of the LED light bulbs are not defective.

Step-by-step explanation:

For each light bulb, there are only two possible outcomes. Either it fails, or it does not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

In which $C_(n,x)$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_(n,x) = (n!)/(x!(n-x)!)$

And p is the probability of X happening.

In this problem we have that:

$n = 10, p = 0.05$

a) None of the LED light bulbs are defective?

This is P(X = 0).

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

$P(X = 0) = C_(10,0)*(0.05)^(0)*(0.95)^(10) = 0.5987$

There is a 59.87% probability that none of the LED light bulbs are defective.

b) Exactly one of the LED light bulbs is defective?

This is P(X = 1).

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

$P(X = 1) = C_(10,1)*(0.05)^(1)*(0.95)^(9) = 0.3151$

There is a 31.51% probability that exactly one of the light bulbs is defective.

c) Two or fewer of the LED light bulbs are defective?

This is

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 2) = C_(10,2)*(0.05)^(2)*(0.95)^(8) = 0.0746$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5987 + 0.3151 + 0.0746 0.9884$

There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) Three or more of the LED light bulbs are not defective?

Now we use p = 0.95.

Either two or fewer are not defective, or three or more are not defective. The sum of these probabilities is decimal 1.

$P(X \leq 2) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 0) = C_(10,0)*(0.95)^(0)*(0.05)^(10)\cong 0$

$P(X = 1) = C_(10,1)*(0.95)^(1)*(0.05)^(9) \cong 0$

$P(X = 2) = C_(10,1)*(0.95)^(2)*(0.05)^(8) \cong 0$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0$

$P(X \geq 3) = 1 - P(X \leq 2) = 1$

There is a 100% probability that three or more of the LED light bulbs are not defective.

Final answer:

The question relates to binomial distribution in probability theory. The probabilities calculated include those of none, one, two or less, and three or more LED bulbs being defective out of a random sample of 10.

Explanation:

This question relates to the binomial probability distribution. A binomial distribution is applicable because there are exactly two outcomes in each trial (either the LED bulb is defective or it's not) and the probability of a success remains consistent.

a) In this scenario, 'none of the bulbs being defective' means 10 successes. The formula for probability in a binomial distribution is p(x) = C(n, x) * [p^x] * [(1-p)^(n-x)]. Plugging in the values, we find p(10) = C(10, 10) * [0.95^10] * [0.05^0] = 0.5987 or 59.87%.

b) 'Exactly one of the bulbs being defective' implies 9 successes and 1 failure. Following the same formula, we get p(9) = C(10, 9) * [0.95^9] * [0.05^1] = 0.3151 or 31.51%.

c) 'Two or less bulbs being defective' means 8, 9 or 10 successes. We add the probabilities calculated in (a) and (b) with that of 8 successes to get this probability. Therefore, p(8 or 9 or 10) = p(8) + p(9) + p(10) = 0.95.

d) 'Three or more bulbs are not defective' means anywhere from 3 to 10 successes. As the failure rate is low, it's easier to calculate the case for 0, 1 and 2 successes and subtract it from 1 to find this probability. This gives us p(>=3) = 1 - p(2) - p(1) - p(0) = 0.98.

Learn more about Binomial Probability here:

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Find the value of the expression below.

Answers

Answer:

what expression

Step-by-step explanation:

theres nothing attached bruv

I need help on this. ive been struggling. TwTlateral area
surface area
volume

Answers

Answer:

Lateral area = 352 Square m

Surface area = 756. 3 square m

Volume = 1408 cubic m

Step-by-step explanation:

$lateral \: area = 2\pi \: rh \n = 2 * (22)/(7) * (16)/(2) * 7 \n = 22 * 16 \n = 352 \: {m}^(2) \n \n surface \: area = 2\pi \: r(h + r) \n = 2 * 3.14 * 8(7 + 8) \n = 6.28 * 8 * 15 \n = 6.28 * 120 \n = 756.3 \: {m}^(2) \n \n volume = \pi {r}^(2) h \n = (22)/(7) * {8}^(2) * 7 \n = 22 * 64 \n = 1408 \: {m}^(3) \n$

Answer:

total surface area Stot = 753.9816 m2

lateral surface area Slat = 351.85808 m2

top surface area Stop = 201.06176 m2

bottom surface area Sbot = 201.06176 m2

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A quality control inspector has drawn a sample of 15 light bulbs from a recent production lot. Suppose that 60% of the bulbs in the lot are defective. What is the expected value of the number of defective bulbs in the sample