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3 years ago

10

A student skipped a step when she tried to convert 11 Hours to seconds and she got the following incorrect result: 11 Hours (60

seconds/1 Minute)=660 seconds what conversion ratio did she skip in this multiple-step equation?

2 answers:

irinina [24]3 years ago

8 0

The ratio she skipped was the ratio the converted the hours into minutes

lisabon 2012 [21]3 years ago

6 0

Answer:

The skipped conversion ratio is 60 minutes/1 hour.

Step-by-step explanation:

Given : A student skipped a step when she tried to convert 11 Hours to seconds and she got the following incorrect result: 11 Hours (60 seconds/1 Minute)=660 seconds.

To find : What conversion ratio did she skip in this multiple-step equation?

Solution :

We have to convert 11 hours into seconds.

We know, 1 hour = 60 minutes

1 minute = 60 seconds

So, $1\text{ hour}=60\times 60\text{ minutes}$

$11\text{ hour}=60\times 60\times 11\text{ minutes}$

$11\text{ hour}=39600\text{ minutes}$

In the question, student converted 1 minute into 60 seconds but forgot to convert 1 hour into 60 minutes.

Therefore, The skipped conversion ratio is 60 minutes/1 hour.

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If UV = 6 and TV = 12, what is TU?

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Answer:

TU = 6

Step-by-step explanation:

Using the Segment Addition Postulate, we know that TU + UV = TV, and since UV = 6 and TV = 12, we know that TU + 6 = 12, therefore, TU = 6.

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Can someone please find the volume of this shape and show work for how you did it??

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$221.4m^{3}$

Step-by-step explanation:

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1) Multiply 12 by 4.1.

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$49.2/2=24.6$

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3 years ago

A computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

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Answer:

(a) 0.8836

(b) 0.6096

(c) 0.3904

Step-by-step explanation:

We are given that a computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

(a) <u>Two computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 2 computers

r = number of success = both 2

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

<em>LET X = Number of computers that are classified as ancient</em>

So, it means X ~ $Binom(n=2, p=0.94)$

Now, Probability that both computers are ancient is given by = P(X = 2)

P(X = 2) = $\binom{2}{2}\times 0.94^{2} \times (1-0.94)^{2-2}$

= $1 \times 0.94^{2} \times 1$

= 0.8836

(b) <u>Eight computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = all 8

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

<em>LET X = Number of computers that are classified as ancient</em>

So, it means X ~ $Binom(n=8, p=0.94)$

Now, Probability that all eight computers are ancient is given by = P(X = 8)

P(X = 8) = $\binom{8}{8}\times 0.94^{8} \times (1-0.94)^{8-8}$

= $1 \times 0.94^{8} \times 1$

= 0.6096

(c) <u>Here, also 8 computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = at least one

p = probability of success which is now the % of computers

that are classified as cutting dash edge, i.e; p = (1 - 0.94) = 0.06

<em>LET X = Number of computers classified as cutting dash edge</em>

So, it means X ~ $Binom(n=8, p=0.06)$

Now, Probability that at least one of eight randomly selected computers is cutting dash edge is given by = P(X $\geq$ 1)

P(X $\geq$ 1) = 1 - P(X = 0)

= $1 - \binom{8}{0}\times 0.06^{0} \times (1-0.06)^{8-0}$

= $1 - [1 \times 1 \times 0.94^{8}]$

= 1 - $0.94^{8}$ = 0.3904

Here, the probability that at least one of eight randomly selected computers is cutting dash edge is 0.3904 or 39.04%.

For any event to be unusual it's probability is very less such that of less than 5%. Since here the probability is 39.04% which is way higher than 5%.

So, it is not unusual that at least one of eight randomly selected computers is cutting dash edge.

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