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Iteru [2.4K]
3 years ago
12

there are 5 more than twice as many students taking Algebra 1 then taking algebra 2. If there are 44 students taking Algebra 2,

what is the least number of students who could be taking Algebra 1? The least number of students is?
Mathematics
2 answers:
Temka [501]3 years ago
8 0

Answer:

93

Step-by-step explanation:

Key :

A1 = Algebra 1

A2 = Algebra 2

Alright so basically lets first look at the info they gave us :

We have 5 more than twice as many students taking A1 than we do A2.

We have 44 students taking A2.

And we need to find the least amount of students that could be taking A1.

So we need to take the amount of students taking A2 (44) and double it to find the amount taking A1.

So we can do 44 x 2 = 88 to get this.

But the problem also states there is 5 more then twice the number of students taking A2.

So we have that 88 but now we just need to add 5 to make up for them telling us that in the problem.

So :

88 + 5 = 93

Our final answer and least amount of students taking A1 is 93 students.

lakkis [162]3 years ago
5 0

Answer:

93 students

Step-by-step explanation:

44 x 2 = 88

88 + 5 = 93

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A 2005 survey found that 7% of teenagers (ages 13 to 17) suffer from an extreme fear of spiders (arachnophobia). At a summer cam
bagirrra123 [75]

Answer:

Probability that at least one of them suffers from arachnophobia is 0.5160.

Step-by-step explanation:

We are given that a 2005 survey found that 7% of teenagers (ages 13 to 17) suffer from an extreme fear of spiders (arachnophobia).

Also, At a summer camp there are 10 teenagers sleeping in each tent.

Firstly, the binomial probability is given by;

P(X=r) =\binom{n}{r}p^{r}(1-p)^{n-r} for x = 0,1,2,3,....

where, n = number of trials(teenagers) taken = 10

           r = number of successes = at least one

          p = probability of success and success in our question is % of

                the teenagers suffering from arachnophobia, i.e. 7%.

Let X = Number of teenagers suffering from arachnophobia

So, X ~ Binom(n= 10,p=0.07)

So, probability that at least one of them suffers from arachnophobia

= P(X >= 1) = 1 - probability that none of them suffers from arachnophobia

= 1 - P(X = 0) = 1 - \binom{10}{0}0.07^{0}(1-0.07)^{10-0}

                       = 1 - (1 * 1 * 0.93^{10} ) = 1 - 0.484 = 0.5160 .

Therefore, Probability that at least one of them suffers from arachnophobia is 0.5160 .

3 0
3 years ago
Tickets for a concert cost eight dollars for the main floor and six dollars for the balcony If 1125 tickets were sold and the ti
cricket20 [7]

Answer:

650 tickets for the main floor and 475 tickets for the balcony were sold.

Step-by-step explanation:

From the information given, you can write the following equations:

x+y=1125 (1)

8x+6y=8050 (2), where:

x is the number of tickets for the main floor

y is the number of tickets for the balcony

First, you can solve for x in (1):

x=1125-y (3)

Second, you can replace (3) in (2):

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Third, you can replace the value of y in (3) to find x:

x=1125-y

x=1125-475

x=650

According to this, the answer is that 650 tickets for the main floor and 475 tickets for the balcony were sold.

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