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love history [14]
3 years ago
8

A candle is 10 in. tall after burning for 2 hours. After 3 hours, it is 8 1/2 in. tall. a. Write a linear equation to model the

height h of the candle after burning t hours. b. Predict how tall the candle will be after burning 6 hours.
Mathematics
1 answer:
GrogVix [38]3 years ago
3 0
The candle was 13 inches tall when it was first lit. It burns at a speed of 1.5 in. every hour. After burning for 6 hours it will be 4 inches tall.
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Which statement is true about the value of the expression below?<br> PLEASE HELP
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Show that sinxtanx=1-cos^2x/cosx
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LHS: = \sin x \tan x =  \frac{\sin x \sin x }{\cos x} =  \frac{\sin^2 x}{\cos x} (using \tan x =  \frac{\sin x}{\cos x})

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7 0
3 years ago
When a scientist conducted a genetics experiments with peas, one sample of offspring consisted of 943 peas, with 717 of them hav
pshichka [43]

Using the normal approximation to the binomial distribution, it is found that:

a) 0.242 = 24.2% probability of getting 717 or more peas with red flowers.

b) Since Z < 2, 717 peas with red flowers is not significantly high.

c) Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

For each pea, there are only two possible outcomes. Either they have a red flower, or they do not. The probability of a pea having a red flower is independent of any other pea, which means that the binomial distribution is used to solve this question.

Binomial distribution:

Probability of x successes on n trials, with p probability.

Normal distribution:

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.
  • If Z > 2, the result is considered <u>significantly high</u>.

If np \geq 10 and n(1-p) \geq 10, the binomial distribution can be approximated to the normal with:

\mu = np

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

In this problem:

  • 943 peas, thus, n = 943
  • 3/4 probability of being red, thus p = \frac{3}{4} = 0.75.

Applying the approximation:

\mu = np = 943(0.75) = 707.25

\sigma = \sqrt{np(1-p)} = \sqrt{943(0.75)(0.25)} = 13.297

Item a:

Using continuity correction, this probability is P(X \geq 717 - 0.5) = P(X \geq 716.5), which is <u>1 subtracted by the p-value of Z when X = 716.5</u>.

Then:

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

Z = \frac{716.5 - 707.25}{13.297}

Z = 0.7

Z = 0.7 has a p-value of 0.758.

1 - 0.758 = 0.242

0.242 = 24.2% probability of getting 717 or more peas with red flowers.

Item b:

Since Z < 2, 717 peas with red flowers is not significantly high.

Item c:

Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

A similar problem is given at brainly.com/question/25212369

6 0
3 years ago
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