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katovenus [111]
3 years ago
14

Fifteen pencils are sold for $750. what is the cost for 1 pencil ?​

Mathematics
2 answers:
Alborosie3 years ago
7 0

Answer:

50$

Step-by-step explanation:

750/15=50

(do u mean 7.50 cuz if u did its 0.50 cents)

aniked [119]3 years ago
3 0
One pencils cost $50
Because you to do 750/15 to get you X
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makvit [3.9K]

Is a number negative



5 0
4 years ago
Find the midpoint of the line using the midpoint formula.<br> Midpoint of AB:______
dimaraw [331]

Answer: MP=(1,1)

Step-by-step explanation:

MP= (\frac{x_1+x_2}{2})(\frac{y_1+y_2}{2})

MP= (\frac{-2+4}{2})(\frac{-3+5}{2}) -2+4=2  -3+5=2

MP= (\frac{2}{2})(\frac{2}{2}) 2/2=1

MP= (1,1)    

8 0
3 years ago
On Monday, the theater club sells 14 tickets for the school play. On Tuesday, they sell 10 tickets. On those two days, they sell
Margarita [4]

Answer:

<h2>31 tickets</h2>

Step-by-step explanation:

Step one:

number of tickets sold

monday=14

tuesday=10

the number of tickets sold for Monday and Tuesday combined is

14+10= 24

step two:

we want to find how many tickets make 30%

so 30/100*24

0.3*24

=7.2 tickets,

Approximately 7 ticket

Therefore the total ticket is

14+10+7

=31 tickets

7 0
3 years ago
Given f(x)=x-8 find f(1)+10
Ronch [10]

Answer:

3

Step-by-step explanation:

replace all x variables with 1 and solve that equation ; f(x)=x-8 so.... f(x) = 1-8

there for f(1)=-7

now, add -7 and 10 for your answer:

f(1)+10= 3

6 0
3 years ago
When grading an exam, 90% of a professor's 50 students passed. If the professor randomly selected 10 exams, what is the probabil
Damm [24]

Using the binomial distribution, it is found that there is a:

a) 0.9298 = 92.98% probability that at least 8 of them passed.

b) 0.0001 = 0.01% probability that fewer than 5 passed.

For each student, there are only two possible outcomes, either they passed, or they did not pass. The probability of a student passing is independent of any other student, hence, the binomial distribution is used to solve this question.

<h3>What is the binomial probability distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 90% of the students passed, hence p = 0.9.
  • The professor randomly selected 10 exams, hence n = 10.

Item a:

The probability is:

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)

In which:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 8) = C_{10,8}.(0.9)^{8}.(0.1)^{2} = 0.1937

P(X = 9) = C_{10,9}.(0.9)^{9}.(0.1)^{1} = 0.3874

P(X = 10) = C_{10,10}.(0.9)^{10}.(0.1)^{0} = 0.3487

Then:

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1937 + 0.3874 + 0.3487 = 0.9298

0.9298 = 92.98% probability that at least 8 of them passed.

Item b:

The probability is:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Using the binomial formula, as in item a, to find each probability, then adding them, it is found that:

P(X < 5) = 0.0001

Hence:

0.0001 = 0.01% probability that fewer than 5 passed.

You can learn more about the the binomial distribution at brainly.com/question/24863377

3 0
2 years ago
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