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Anna71 [15]
3 years ago
13

How do you convert units in length?

Mathematics
1 answer:
Vlada [557]3 years ago
6 0
Just look it up in the internet i think it will tell you
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If students sit 4 per row on a rollercoaster, 6 students are left without a seat.
dlinn [17]

An expression is defined as a set of numbers, variables, and mathematical operations. The total number of students is 54.

<h3>What is an Expression?</h3>

In mathematics, an expression is defined as a set of numbers, variables, and mathematical operations formed according to rules dependent on the context.

Let the number of columns in the rollercoaster be represented by x.

If students sit 4 per row on a rollercoaster, 6 students are left without a seat. Therefore, the number of students can be written as,

Total number of students = 4x + 6

If they sit 6 per row, there are 3 rows left empty. Therefore, the total number of students can be written as,

Total number of students = 6(x-3)

The total number of students can be written as,

4x + 6 = 6(x - 3)

4x + 6 = 6x - 18

2x = 24

x = 12

Now, the total number of students in the class are,

4x + 6 = 4(12) + 6 = 54

Hence, the total number of students are 54.

Learn more about Expression:

brainly.com/question/13947055

#SPJ1

3 0
2 years ago
In its monthly report, the local animal shelter states that it currently has 24 dogs and 18 cats available for adoption. Eight o
andrezito [222]

Answer with Step-by-step explanation:

We are given that

Total dogs=24

Total cats=18

Total animals=24+18=42

Male dogs=8

Male cats=6

Total male animals=8+6=14

Total female animals=42-14=28

Female cats=18-6=12

Dogs female=24-8=16

a.We have to find the probability that the pet is male, given that it is cat.

It means we have to find P(Male/cat)  

Conditional probability: P(A/B)=\frac{P(A\cap B)}{P(B)}

P(Cat)=\frac{18}{42}

P(male cat)=\frac{6}{42}

P(Male/cat)=\frac{\frac{6}{42}}{\frac{18}{42}}

P(male/cat)=\frac{6}{18}=0.33

b.P(female cat)=\frac{12}{42}

P(female)=\frac{28}{42}

P(Cat/Female)=\frac{\frac{12}{42}}{\frac{28}{42}}=\frac{12}{28}=0.43

c.P(Dog)=\frac{24}{42}

P(dog female)=\frac{16}{42}

P(female/dog)=\frac{\frac{16}{42}}{\frac{24}{42}}

P(female/dog)=\frac{16}{24}=0.67

5 0
3 years ago
Danielle works as a cashier in a bookstore she works 35 hours a week and orange $11 an hour each week Daniel spend $300 and save
exis [7]
 Her estimate is not accurate because if you multiply 30 by 11 you would get 330. Then you multiply that by 52and you would get $10560 a year.
6 0
3 years ago
Read 2 more answers
Evaluate the expression when g=8 and h=7. <br> 4g-h
Lelechka [254]
25 because 4(8)-7 = 32-7 = 25
3 0
3 years ago
Read 2 more answers
According to the National Bridge Inspection Standard (NBIS), public bridges over 20 feet in length must be inspected and rated e
slamgirl [31]

Answer:

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Step-by-step explanation:

For each bridge, there are only two possible outcomes. Either it has rating of 4 or below, or it does not. The probability of a bridge being rated 4 or below is independent from other bridges. So we use the binomial probability distribution to solve this problem.

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.

For the year 2020, the engineers forecast that 9% of all major Denver bridges will have ratings of 4 or below.

This means that p = 0.09

Use the forecast to find the probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Either less than 4 have a rating of 4 or below, or at least 4 does. The sum of the probabilities of these events is 1.

So

P(X < 4) + P(X \geq 4) = 1

We want P(X \geq 4)

So

P(X \geq 4) = 1 - P(X < 4)

In which

P(X < 4) = 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_{12,0}.(0.09)^{0}.(0.91)^{12} = 0.3225

P(X = 1) = C_{12,1}.(0.09)^{1}.(0.91)^{11} = 0.3827

P(X = 2) = C_{12,2}.(0.09)^{2}.(0.91)^{10} = 0.2082

P(X = 3) = C_{12,3}.(0.09)^{3}.(0.91)^{9} = 0.0686

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3225 + 0.3827 + 0.2082 + 0.0686 = 0.982

Finally

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.982 = 0.0180

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

6 0
3 years ago
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