Solve for b b+4.22=7.08

A computer can be classified as either cutting dash edge or ancient. Suppose that 86% of computers are classified as ancient.

Leno4ka [110]

Answer:

(a) The probability that both computers are ancient is 0.7396

(b) The probability that all seven computers are ancient is 0.3479

(c) The probability that at least one of seven randomly selected computers is cutting dash edge is 0.6520.

Because the probability is about 65% is it not unusual that at least one of seven randomly selected computers is cutting dash edge, it's more likely than not.

Step-by-step explanation:

We know that 86% of computers are classified as ancient. This means, if one computer is chosen at random, there is an 86% chance that it will be classified as ancient.

$P(ancient)=0.86$

(a) To find the probability that two computers are chosen at random and both are ancient you must,

The probability that the first computer is ancient is $P(ancient)=0.86$ and the probability that the second computer is ancient is $P(ancient)=0.86$

These events are independent; the selection of one computer does not affect the selection of another computer.

When calculating the probability that multiple independent events will all occur, the probabilities are multiplied, this is known as the rule of product.

Let A be the event "the first computer is ancient" and B the event "the second computer is ancient".

$P(A\:and \:B)=P(A)\cdot P(B)=0.86\cdot 0.86=0.86^2= 0.7396$

(b) To find the probability that seven computers are chosen at random and all are ancient you must,

Following the same logic in part (a) we have

Let A be the event "the first computer is ancient",

B the event "the second computer is ancient",

C the event "the third computer is ancient",

D the event "the fourth computer is ancient",

E the event "the fifth computer is ancient",

F the event "the sixth computer is ancient", and

G the event "the seventh computer is ancient"

$P(A\:and \:B\:and \:C\:and \:D\:and \:E\:and \:F\:and \:G)=\\P(A)\cdot P(B)\cdot P(C)\cdot P(D)\cdot P(E)\cdot P(F)\cdot P(G) =(0.86)^7=0.3479$

(c) To find the probability that at least one of seven randomly selected computers is cutting dash edge you must

Use the concept of complement. The complement of an event is the subset of outcomes in the sample space that are not in the event.

Let C the event "the computer is cutting dash edge".

Let A the event "the seven computers are ancient".

$P(C)=1-P(A)=1-0.3479=0.6520$

Because the probability is about 65% is it not unusual that at least one of seven randomly selected computers is cutting dash edge, it's more likely than not.

5 0

3 years ago

In the past decades there have been intensive antismoking campaigns sponsored by both federal and private agencies. In one study

DIA [1.3K]

Answer:

$z=\frac{0.384-0.362}{\sqrt{0.374(1-0.374)(\frac{1}{4276}+\frac{1}{3908})}}=2.055$

The p value can be calculated from the alternative hypothesis with this probability:

$p_v =2*P(Z>2.055)=0.0399$

And the best option for this case would be:

C. between 0.01 and 0.05.

Step-by-step explanation:

Information provided

$X_{1}=1642$ represent the number of smokers from the sample in 1995

$X_{2}=1415$ represent the number of smokers from the sample in 2010

$n_{1}=4276$ sample from 1995

$n_{2}=3908$ sample from 2010

$p_{1}=\frac{1642}{4276}=0.384$ represent the proportion of smokers from the sample in 1995

$p_{2}=\frac{1415}{3908}=0.362$ represent the proportion of smokers from the sample in 2010

$\hat p$ represent the pooled estimate of p

z would represent the statistic

$p_v$ represent the value for the pvalue

System of hypothesis

We want to test the equality of the proportion of smokers and the system of hypothesis are:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

The statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{1642+1415}{4276+3908}=0.374$

Replacing the info given we got:

$z=\frac{0.384-0.362}{\sqrt{0.374(1-0.374)(\frac{1}{4276}+\frac{1}{3908})}}=2.055$

The p value can be calculated from the alternative hypothesis with this probability:

$p_v =2*P(Z>2.055)=0.0399$

And the best option for this case would be:

C. between 0.01 and 0.05.

7 0

3 years ago

(2t^(3)+3t-4)-(4t^(2)-6t)

Gnoma [55]

2

3

−

4

2

+

9

−

4

Step-by-step explanation:

3 0

3 years ago

The sum of Ted's and Leon's ages is

olganol [36]

Answer:

Ted is 6 years old

Leon is 7 years old

Andrew is 8 years old

Step-by-step explanation:

Ted+Leon=13 ⇒T+L=13 ⇒L=13-T

Leon+Andrew=15 ⇒L+A=15 ⇒A=15-L ⇒A=15-(13-T) because Leon =13-Ted

A=2+T

Ted+Andrew=14

T+A=14 ( substitute Andrew age from A=2+T)

T+2+T=14

2T=12

<h2>T=12/2=6 (Ted is 6 years old)</h2><h2>A=2+T=2+6=8 ( Andrew is 8 years old)</h2><h2>Leon=13-T=13-6=7 ( Leons is seven years old)</h2>

check: Ted+Leon=13 ⇒ 6+7=13

Leon + Andrew=15 ⇒7+8=15

Andrew+Ted=14 ⇒ 6+8=14

7 0

3 years ago

A Family purchased tickets to the movies in spent a total of $40.75. The family purchase five tickets there was a $.90 processi

muminat

Answer:

$7.25

Step-by-step explanation:

Total amount spent on the ticket = $40.75

Number of ticket purchased = 5

Processing fee per ticket = $0.90

Unknown:

Cost of each ticket = ?

Solution:

To solve this problem, we need to find the processing fee for the 5 tickets purchase;

Processing fee = number ticket x processing fee per ticket

Processing fee = 5 x 0.9 = $4.5

Now, the real cost a ticket = $40.75 - $4.5 = $36.25

So, cost per ticket is;

Cost = $\frac{36.25}{5}$ = $7.25

7 0

3 years ago