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3 years ago

A hall measures 12 metres long,

Mathematics

1 answer:

Ilya [14]3 years ago

7 0

$\huge\sf\red{Answer : - }$

<h2>Given :- </h2>

Length of the hall = 12m

wide of the hall (Breadth) = 10m

Height of the hall = 4m

Surface area of the walls (Four walls area)

<h2>Formula :- </h2>

Four walls area of the hall = 2h(l+b)

where,

h = height

l = length

b = breadth

<h2>Solution : - </h2>

By substituting values according to the formula,

➙ 2 × 4m(12m + 10m)

➙ 8m(22m)

➠ 176m²

It's option B

<h3><u>Hence, The surface area of the walls is (B) 176m²</u></h3>

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3 years ago

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2 years ago

When a scientist conducted a genetics experiments with peas, one sample of offspring consisted of 943 peas, with 717 of them hav

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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.

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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>.