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3 years ago

2 to the 3rd power times 4 divided by 2 plus 2

Mathematics

1 answer:

marin [14]3 years ago

6 0

Answer: 18

Step-by-step explanation: 2 x 2 x 2= 8

8 x 4 = 32

32 / 2 = 16

16 + 2 = 18

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The first two terms of a geometric sequence are a1 = 1/3 and a2 = 1/6. What is a8, the eighth term?

SIZIF [17.4K]

Answer:

B. 1/384

Step-by-step explanation:

a1=1/3

a2 = 1/6

The constant factor is 1/2

a1=2/3*1/2

a2=2/3*1/2^2

a8=2/3*(1/2)^8

2/3*1/256

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3 years ago

Read 2 more answers

Name the plane represented by each surface of the box.<br><br> the top

nydimaria [60]

Answer:

Hello the required image is missing attached below is the missing box

answer : FHGE

Step-by-step explanation:

The plane representing the top surface of the box is FHGE. there are still other planes there but do not represent the entire top surface of the Box and they are : FGE, HGE, FHG and FHE. This is because a plane can be represented by three to four Noncollinear points on a solid like a box.

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3 years ago

A car traveling at a constant speed travels 1000 ft in 12 seconds. how long in hours will it take to travel 175 miles?

artcher [175]

9514 1404 393

Answer:

3.08 hours

Step-by-step explanation:

We assume you're interested in the time in hours.

$175\text{ mi}\times\dfrac{12\text{ s}}{1000\text{ ft}}\times\dfrac{5280\text{ ft}}{1\text{ mi}}\times\dfrac{1\text{ h}}{3600\text{ s}}=\dfrac{175\times12\times5280}{1000\times3600}\text{ h}=\boxed{3.08\text{ h}}$

It will take 3.08 hours to travel 175 miles at that speed.

7 0

3 years ago

600 can be written as 2^a x b x c^d. where a, b, c and d are all prime numbers. find the values of a, b, c and d

Sergeu [11.5K]

Answer:

easy one

Step-by-step explanation:

first factorise 600 as 2*2*2*3*5*5

then write as 2^3 * 3*5^2= 2^a*b*c^d

equating corresponding elements as a=3

b=1and c= 2 is the answer

8 0

3 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

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

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

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

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

3 years ago