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2 years ago

THIS ABOUT TO BE OVERDUE PLS HELP ME

Mathematics

1 answer:

Drupady [299]2 years ago

3 0

Variable , constant, variable

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What is a fact family? (Saying "it's a family of facts" will not be accepted")

Mrac [35]

Answer: Fact families are 3 numbers that make up a set of related math facts

Step-by-step explanation:

.An example of a fact family for multiplication and division is 2, 3, and 6: 2 X 3 = 6, 3 X 2 = 6, 6 2 = 3, and 6 3 = 2.

7 0

2 years ago

Read 2 more answers

renu is 2 times older than suman. Anita is 9 years older than suman and 6 years younger than renu. jif some of their ages is 69,

SashulF [63]

Answer:

suman is 15 and renu is 30 anita is 24

Step-by-step explanation:

15+30+24+69

3 0

3 years ago

A company produces birthday cards and get well cards at a ratio of 6 to 5 how many birthday cards can ve expected if 60 get well

Alik [6]

The number of birthday cards made is 72.

<h3>Description of ratios</h3>

Ratio expresses the relationship between two numbers. It shows the frequency of the number of times that one value is contained in another number. The sign used to represent ratios is :

<h3>Determining the total cards made. </h3>

Total card = (sum of ratios x number of get well cards) / ratio of get well cards

(11 x 60) / 5 = 132

<h3>Number of birthday cards made </h3>

6/11 x 132 = 72 cards

To learn more about ratios, please check: brainly.com/question/25927869

4 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

On your wedding day you leave for the church 34 minutes before the ceremony is to begin, which should be plenty of time since th

Schach [20]

Answer:

Average speed is 37.35 mi/h

Step-by-step explanation:

given data

leave = 34 minutes before

church distance = 12.0 miles

average speed first 17 minutes = 5.0 mi/h

solution

so we find Total distance travel in first 17 minutes = speed × time

Total distance travel in first 17 minutes = 5 × $\frac{17}{60}$

Total distance travel in first 17 minutes = 1.416 mi

and

Distance Remaining = 12 - 1.416 = 10.584 mi

Time Remaining = 34 - 17 min = 17 min

remaining distance Average speed = $\frac{distance}{time}$

Average speed = $\frac{10.584}{\frac{17}{60}}$

Average speed is 37.35 mi/h

3 0

2 years ago