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2 years ago

11

Amelia made 3 1/3 pounds of trail mix. If she puts 2/3 of a pound into each bag, how many bags can Amelia fill? Write your answe

r as a fraction or as a whole or mixed number. bags

1 answer:

Neporo4naja [7]2 years ago

7 0

Answer:

6

Step-by-step explanation:

Because ⅔ goes into 3 ⅓ 6 times.

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At a school district there are 55 employees. If next year the number of employees increases by

Dmitry_Shevchenko [17]

Answer:

66

Step-by-step explanation:

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3 years ago

Brainlist answer!!Can I get some help with this, please!!

Gemiola [76]

A) The question would be "Does cell phone use time correlate to brain cancer?". The population is all cell phone users. The sample is the 469 people with brain cancer, and the 422 healthy people.

B) This sample size may not portray the entire population of users because the positive results may have been fabricated by factors other than cell phone use. You would have to figure out the population size and determine the correct sample size.

C) This study may not portray cell phone use correlation over a large population efficiently because of the sample size and the predetermined conclusion that cell phone use causes cancer.

4 0

2 years ago

Simplify the following expression.( Fraction form)<br><br> 7-5/6 * 7-7/6

Ipatiy [6.2K]

Answer:

1295/36

Step-by-step explanation:

the statement tell us:

(7-(5/6))*(7-(7/6))

we have:

(7-(5/6))=((6*7)-5)/6=(42-5)/6=37/6

and we have:

(7-(7/6))=((6*7)-7)/6=(42-7)/6=35/6

we need multiply both terms:

(37/6)*(35/6)=(37*35)/(6*6)

finally we have

1295/36

5 0

2 years ago

What is the ratio of the number of days in the week that do not begin with the letter T to the total numbers of days in the week

Rom4ik [11]

Answer:

The answer is 5 over 7

Step-by-step explanation:

5 0

2 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

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

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

Part a

We want this probability:

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

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

2 years ago