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AysviL [449]
3 years ago
5

harpers house is 30 feet wide and 45 feet long on a scale drawing the house is 4 inches wide and 6 inches long and harpers bedro

om is 1.2 inches wide and 1.6 inches long what are the actual measurements of harpers bedroom. harpers bedroom is ------ feet wide and ------- feet long
Mathematics
2 answers:
Masteriza [31]3 years ago
8 0
Harpers bedroom is 9 feet wide and 12 feet long
anastassius [24]3 years ago
7 0

9 feet wide 12 feet long

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kiruha [24]
<span> 1/8,2/16,3/24 that should be the answer </span>
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3 years ago
On a math test, Jessica answered 20% questions incorrectly. She answered 16 questions correctly. How many total questions did sh
Tanzania [10]
The answer to the question is 36
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3 years ago
Read 2 more answers
The quality-control department of Starr Communications, the manufacturer of video-game cartridges, has determined from records t
notsponge [240]

Answer: (A) The probability that a cartridge purchased will have a video or audio defect is 1.9%

(B) The probability that a cartridge purchased will not have a video or audio defect is zero.

Step-by-step explanation: The data given shows that 1.2% (or 120) cartridges have video defects, 0.9% have audio defects (or 90) and 0.2% (or 20) have both audio and video defects.

The possible outcomes for all events (audio defects and video defects) is derived as 120 plus 90 which is equals 210 possibilities (or possible outcomes).

Therefore the probability of having an audio defect is calculated as follows;

P(Audio) = Number of required outcomes/Number of all possible outcomes

P(Audio) = 90/210

P(Audio) = 3/7

Also the probability of having a video defect is derived as follows;

P(Video) = Number of required outcomes/Number of all possible outcomes

P(Video) = 120/210

P(Video) = 4/7

However we should take note of the fact that 0.2% or 20 of the cartridges in the sample size has both audio and video defects. Hence the probability that a cartridge has both audio and video defects is calculated as;

P(Audio and Video) = Number of required outcomes/Number of all possible outcomes

P(Audio and Video) = 20/210

P(Audio and Video) = 2/21

To calculate the probability that a cartridge bought would have either an audio or a video defect would mean to add both probabilities together, but we MUST SUBTRACT the probability of having both an audio defect and video defect (that is P{Audio and Video}). The reason is that this is already included in both probabilities and we need to avoid double counting. Hence we have;

(A); P(Video OR Audio defect) = P(Audio) + P(Video) - P(Audio and Video)

P(Video OR Audio defect) = (3/7 + 4/7) - 2/21

P(Video OR Audio defect) = 1 - 2/21

P((Video OR Audio defect) = 19/21

Therefore the probability that a cartridge purchased will have a video or audio defect is 190, or better still 1.9%.

(B): From all possibilities shown, which is 210 possibilities of either events, we have determined that 120 will be the probability of having an audio defect and 90 will be the probability of having a video defect. Therefore the probability that a cartridge purchased will not fall into any of either possibilities is zero.

6 0
3 years ago
A fair coin is to be tossed 20 times. Find the probability that 10 of the tosses will fall heads and 10 will fall tails, (a) usi
lbvjy [14]

Using the distributions, it is found that there is a:

a) 0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

b) 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

c) 0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item a:

Binomial probability distribution

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

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 20 tosses, hence n = 20.
  • Fair coin, hence p = 0.5.

The probability is <u>P(X = 10)</u>, thus:

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

P(X = 10) = C_{20,10}.(0.5)^{10}.(0.5)^{10} = 0.1762

0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item b:

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.  
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • The binomial distribution is the probability of <u>x successes on n trials, with p probability</u> of a success on each trial. It can be approximated to the normal distribution with \mu = np, \sigma = \sqrt{np(1-p)}.

The probability of an exact value is 0, hence 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item c:

For the approximation, the mean and the standard deviation are:

\mu = np = 20(0.5) = 10

\sigma = \sqrt{np(1 - p)} = \sqrt{20(0.5)(0.5)} = \sqrt{5}

Using continuity correction, this probability is P(10 - 0.5 \leq X \leq 10 + 0.5) = P(9.5 \leq X \leq 10.5), which is the <u>p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.</u>

X = 10.5:

Z = \frac{X - \mu}{\sigma}

Z = \frac{10.5 - 10}{\sqrt{5}}

Z = 0.22

Z = 0.22 has a p-value of 0.5871.

X = 9.5:

Z = \frac{X - \mu}{\sigma}

Z = \frac{9.5 - 10}{\sqrt{5}}

Z = -0.22

Z = -0.22 has a p-value of 0.4129.

0.5871 - 0.4129 = 0.1742.

0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

A similar problem is given at brainly.com/question/24261244

6 0
3 years ago
The mother of Kriz Anthony was hospitalized and he needs an
Artyom0805 [142]

Answer: Kriz should pay $50,100 at the end of the term.

Step-by-step explanation:

If the interest is compunded quarterly, then the accumulated amount after t years will be:

A=P(1+\frac{r}{400})^{4t} , where P= principal value , r = rate of interest .

Given: P= 50,000 , r= 5% = 0.05, t= 4 years

Then,

A=50000(1+\dfrac{0.05}{400})^{4(4)}\\\\= 50000(1+0.000125)^{16}\\\\=50000(1.000125)^{16}\\\\=50000(1.002)\\\\=\$50,100

Hence, Kriz should pay $50,100 at the end of the term.

3 0
3 years ago
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