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2 years ago

5

A rectangle has an area of 18cm2 (squared). The lengths of the sides are whole numbers. How long could the sides of the rectangl

e be?
Give all the examples you can think of

This is mathswatch i need help on it. the deadline for this work is on 27/05/2020

1 answer:

marishachu [46]2 years ago

7 0

Answer:

9 cm, 2 cm

or

6 cm , 3 cm

Step-by-step explanation:

There are two possibilities of the lengths of the sides of rectangle

$18= 9\times 2\\18=6\times 3\\$

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rusak2 [61]

ANSWER TO QUESTION 1

The set of natural numbers that are factors of 20 are natural numbers that divides 20 exactly without a remainder.

These natural numbers are;

{ $1,2,4,5,10,20$ }

Therefore the correct answer is D

ANSWER TO QUESTION 2

Given that the universal set U has elements

{ $1,2,3,4,5,6,7,8,9,10$ }

and the set P also has elements

{ $2,4,6,8,10$ }

The elements of P' are the elements that are in the universal set but are not in set P.

These elements are;

{ $1,3,5,7,9$ }

The correct answer is C

ANSWER TO QUESTION 3

The distance is given by the absolute value equation

$d=|130-25t|$

If the is 15 feet away then $d=15$

This implies that;

$15=|130-25t|$

We now split the absolute value sign to obtain,

$15=130-25t$

or

$-15=130-25t$

We now solve for t to obtain;

$15-130=-25t$

or

$-15-130=-25t$

$\frac{115}{25}=t$

or

$\frac{145}{25}=t$

This gives us

$4.6=t$

or

$5.8=t$

The correct answer is C

ANSWER TO QUESTION 4

Given

P={ $0,1,2,3,4,56,7,8,9$ }

and

V={ $2,4,6,810,1216,18,20$ }

P $\cap$ V={2,4,6,8}

P\V are elements in P only .

This means that we have to take out the elements of the intersection of the two sets from P.

P\V ={0,1,3,5,5,7,9}

The correct answer is B

ANSWER TO QUESTION 5

The correct answer is B.

In listing the elements of a set, we do not repeat the element. In order words we list only distinct elements.

Blueberry is repeating in both sets but we have to list it once.

ANSWER TO QUESTION 6.

The correct answer is 220

We were told that 250 play basket ball

If 120 play both sports, then

250-120=130

play only basketball

Now the number of people who play only baseball

$350-250=100$

But the number of people who play baseball

$120+100=220$

see diagram

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2 years ago

Cheryl wants to purchase a par value $500 bond from Agram Imports. Agram imports bonds have a market rate of 102.114 and offer 8

viva [34]

According to my study, current yield is calculated by using the formula Interest or dividends divided by the current price of the security or simple the current price. So the current price in this given question is 102.114 and the interest is 8.1%, we will simply divide the two; 8.1 / 102.114 = 0.07732310947 or approximately .077. The answer is A.

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2 years ago

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good job on increasing our team sales from 90 to 130 units per day.” wow i never thought that we could increase our sales by

artcher [175]

The sales was increased by 44.4%.

Step-by-step explanation:

Lets do this cumulative
90 of 10 % = 9 which is 99
90 of 10 % = 9 which is 108
90 of 10% = 9 which is 117
90 of 10% = 9 which is 126
90 of 4 % = 3.6 which is 129.6
90 of 0.044% = 0.0396 which is 129.99
This comes to nearly 44.44 to be closer there was an increase.
Always rule 1 approximation use 50% of 90 if it is above 130.
Down it to 40% of 90 see if the sales is lower than 130.
It should always start with 50,40,30....10 percents.
Alternative should be 10,1 or decimals.
Approximation using decimals rounding up becomes simple.

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2 years ago

See You Later Based on a Harris Interactive poll, 20% of adults believe in reincarnation. Assume that six adults are randomly se

REY [17]

Answer:

a) There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b) 0.0064% probability that all of the selected adults believe in reincarnation.

c) There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d) Since $P(X \geq 5) < 0.05$ , 5 is a significantly high number of adults who believe in reincarnation in this sample.

Step-by-step explanation:

For each of the adults selected, there are only two possible outcomes. Either they believe in reincarnation, or they do not. 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.

In this problem we have that:

$n = 6, p = 0.2$

a. What is the probability that exactly five of the selected adults believe in reincarnation?

This is P(X = 5).

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

$P(X = 5) = C_{6,5}.(0.2)^{5}.(0.8)^{1} = 0.0015$

There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b. What is the probability that all of the selected adults believe in reincarnation?

This is P(X = 6).

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

$P(X = 6) = C_{6,6}.(0.2)^{6}.(0.8)^{0} = 0.000064$

There is a 0.0064% probability that all of the selected adults believe in reincarnation.

c. What is the probability that at least five of the selected adults believe in reincarnation?

This is

$P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564$

There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d. If six adults are randomly selected, is five a significantly high number who believe in reincarnation?

5 is significantly high if $P(X \geq 5) < 0.05$

We have that

$P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564 < 0.05$

Since $P(X \geq 5) < 0.05$ , 5 is a significantly high number of adults who believe in reincarnation in this sample.

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2 years ago

Bryan wants to ship three medium-sized boxes. Two companies offer different rates.

nikitadnepr [17]

I think th answer is a

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2 years ago

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