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svetoff [14.1K]

6 months ago

7

Students are measuring various objects around the classroom. A pencil measures 8 ⅜ inches while a pair of scissors measures 5 ½

inches. How much longer is the pencil than the pair of scissors?

1 answer:

Maslowich6 months ago

5 0

The pencil is longer than the pair of scissors by $2\frac{7}{8}$ inches.

A pencil measures $8\frac{3}{8}$ inches.

A pair of scissors measures $5\frac{1}{2}$ inches.

We need to find the difference between the measures of a pencil and the scissors.

We see that the pencil measure is greater than the pair of scissors.

<h3>What is a mix fraction?</h3>

A fraction is a part of a whole number. It has two parts - numerator and denominator.

Example: If you cut a cake into three slices each slice is 1/3.

Where 1 = numerator and 3 = denominator.

Four types of fractions:

Unit fraction - when the numerator is 1.

Proper fraction - When the numerator is less than the denominator.

Improper fraction - When the numerator is greater than the denominator.

Mixed fraction - When the fraction contains a whole number with a proper fraction.

Example: $3\frac{1}{3}$

The given measure of the pen and the pair of scissors is in mix fraction.

The pencil measure is longer than the pair of scissors measure so,

We will subtract the measure of the pair of scissors from the measure of the pen.

We have,

= $8\frac{3}{8} - 5\frac{1}{2}$

We can write the mix fraction into an improper fraction.

We multiply the denominator with the whole number and add this product with the numerator to get our numerator of the improper fraction. The denominator will remain the same as the mix fraction in the improper fraction.

So,

$8\frac{3}{8}=\frac{(8\times8)+3}{8}=\frac{64 + 3}{8} = \frac{67}{8}\\ 5\frac{1}{2} =\frac{(5\times2)+1}{2}= \frac{(10+1)}{2} = \frac{11}{2}$

= 67/8 - 11/2

= (67-44) / 8

= 23 / 8

Writing in mix fraction.

= $2\frac{7}{8}$

Thus the pencil is longer than the pair of scissors by $2\frac{7}{8}$ inches.

Learn more about mix fraction here:

brainly.com/question/13881046

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A large <em>p</em>-value (> 0.05) specifies fragile proof against the null hypothesis, so you fail to discard the null hypothesis.

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2 years ago

If 20% of the people in a community use the emergency room at a hospital in one year, find

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Answer:

a) 87.91% probability that at most three used the emergency room

b) 20.13% probability that exactly three used the emergency room.

c) 3.28% probability that at least five used the emergency room

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they use the emergency room, or they do not. The probability of a person using the emergency room is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Sample of 10 people:

This means that $n = 10$

20% of the people in a community use the emergency room at a hospital in one year

This means that $p = 0.2$

a) At most three used the emergency room

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

$P(X = 1) = C_{10,1}.(0.2)^{1}.(0.8)^{9} = 0.2684$

$P(X = 2) = C_{10,2}.(0.2)^{2}.(0.8)^{8} = 0.3020$

$P(X = 3) = C_{10,3}.(0.2)^{3}.(0.8)^{7} = 0.2013$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1074 + 0.2684 + 0.3020 + 0.2013 = 0.8791$

87.91% probability that at most three used the emergency room

b) Exactly three used the emergency room

$P(X = 3) = C_{10,3}.(0.2)^{3}.(0.8)^{7} = 0.2013$

20.13% probability that exactly three used the emergency room.

c) At least five used the emergency room

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

From 0 to 3, we already have in a).

$P(X = 4) = C_{10,4}.(0.2)^{4}.(0.8)^{6} = 0.0881$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.1074 + 0.2684 + 0.3020 + 0.2013 + 0.0881 = 0.9672$

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.9672 = 0.0328$

3.28% probability that at least five used the emergency room

4 0

2 years ago