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2 years ago

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"The Highest Common Factor (HCF) of my two numbers is 3 The Lowest Common Multiple (LCM) of my two numbers is 45" (b) Write down

two numbers that James could be thinking of.

1 answer:

Mila [183]2 years ago

6 0

Answer:

Step-by-step explanation:

15 and 3

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Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set

SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

$E(X)=np$

Compute the expected number of orange milk chocolate M&M’s in a bag of 55 candies as follows:

$E(X)=np$

$=55\times 0.20\\=11$

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

$P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}$

$=0.1968$

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0

2 years ago

What is the distance between (3, 5.25) and (3, –8.75)? 6 units 8.25 units 11.75 units 14 units

evablogger [386]

By using distance formula :

$\text{Distance formula,} \bold{ \boxed{ Distance = \sqrt{( x_{2} -x_{1})^{2}+(y_{2} - y_{1})^{2}) }}}$

Given points = ( 3 , 5.25 ) and ( 3 , - 8.75 )

$\bold{Taking \: \: \: x_{1}=3 \: \: , \: \: x_{2}= 3 \: \: , \: \: y_{1}= 5.25 \: \: , \: \: y_{2}= -8.75}$

On applying formula, we get

$Distance = \sqrt{ ( x_{2}-x_{1})^{2}+(y_{2}-y_{1})^2} \\ \\ \\ Distance = \sqrt{ ( 3 - 3 )^{2} + ( - 8.75 - 5.25 )^{2}} \\ \\ \\ Distance = \sqrt{ ( 0 )^{2} + ( - 14)^{2}} \\ \\ \\ Distance = \sqrt{ ( - 14 )^{2}} \\ \\ \\ Distance = \sqrt{ 14^{2}} \:\:\:\:\:\:\:\:\:\:\: \: \: \: \: \: \: \: \: | \bold{ ( - 14 )^{2} = 14^{2}} \\ \\ \\ Distance = {14}^{2 \times \frac{1}{2} } \\ \\ \\ Distance = {14}^{1} \\ \\ \\ Distance = 14 \: units$

Hence, Option D is correct.

4 0

2 years ago

Read 2 more answers

After years of maintaining a steady population of 32,000, the population of a town begins to grow exponentially. After 1 year an

galben [10]

Answer:

$y=32000(1+0.08)^x$

Step-by-step explanation:

Exponential growth function is $y=a(1+r)^x$

Where 'a' is the initial population

r is the rate of growth and x is the time period in years

a steady population of 32,000. So initial population is 32,000

an increase of 8% per year. the rate of increase is 8% that is 0.08

a= 32000 and r= 0.08

Plug in all the values in the general equation

$y=a(1+r)^x$

$y=32000(1+0.08)^x$

$y=32000(1+0.08)^x$

4 0

2 years ago

Read 2 more answers

a shopkeeper sold goods for rs 2400 and made a profit of 25% in the process. find his profit per cent if he had sold his goods f

miv72 [106K]

case 1,

Let the CP be ₹x,

SP = ₹2400

Profit = SP – CP

= 2400 – x

Profit % = {(2400–x)/ x} × 100%

According to the question,

{(2400–x)/ x} × 100 = 25

=> (2400–x)/ x= 25 /100

=> 100(2400–x) = 25x [ cross multiplication]

=> 240000 – 100x = 25x

=> 240000 = 25x + 100x

=> 240000 = 125x

=> 240000/125 = x

=> x = 1920

So, CP = ₹1920

case 2,

SP = ₹2040

Profit = SP – CP

= 2040 – 1920

= ₹120

profit % = 120/1920 × 100%

= 16%

<h3>Thus, his profit would be 16% if he had sold his goods for ₹2040.</h3>

6 0

2 years ago

The heights of fully grown sugar maple trees are normally distributed, with a mean of 87.5 feet and a standard deviation of 6.25

Julli [10]

Answer:

20.33%

Step-by-step explanation:

We have that the mean (m) is equal to 87.5, the standard deviation (sd) 6.25 and the sample size (n) = 12

They ask us for P (x <86)

For this, the first thing is to calculate z, which is given by the following equation:

z = (x - m) / (sd / (n ^ 1/2))

We have all these values, replacing we have:

z = (86 - 87.5) / (6.25 / (12 ^ 1/2))

z = -0.83

With the normal distribution table (attached), we have that at that value, the probability is:

P (z <-0.83) = 0.2033

The probability is 20.33%

6 0

2 years ago