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3 years ago

What is the answer to 4[5(12 3)-2]-7

Mathematics

1 answer:

AleksandrR [38]3 years ago

6 0

4[5(12*3)-2]-7 4[5(36)-2]-7 4[180-2]-7 4[178]-7 712-7= 705 :)

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An arts academy requires there to be 5 teachers for every 95 students and 4 tutors for every 40 students how many students does

Whitepunk [10]

Answer:

19 students per teacher and 10 students per tutor

Step-by-step explanation:

To find the amount of students per teacher and per tutor you have to divide

95 students /5 teachers

19 students per teacher

40 students /4 tutors

10 studetns pet tutor

5 0

2 years ago

Shaq painted 600 portraits in 8 hours at that rate how many portraits would he paint in 4 minutes?

Deffense [45]

Answer: In 4 minutes he will make 5 portraits

Step-by-step explanation:

$\large \boldsymbol{} 600p-8h=480 min \\\\ x \ p-4 min \\\\ x=\dfrac{600\cdot 4}{480} =\dfrac{2400}{480} =5 \ portraits \\\\ \rm Answer : 5$

3 0

3 years ago

The diagram shows a right-angled triangle.

Alexus [3.1K]

Step-by-step explanation:

there is no diagram so I'm just going to guess this however

sin x = 6/9

sin x = 2/3

sin x = 0.67

x = sin inverse of 0.67

x = 42.1

8 0

3 years ago

Given x-intercepts of -3 and 2 find and equation for a graph that goes through both points. (You do NOT have to graph the equati

lesya [120]

Answer:

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6 0

3 years ago

10 normal six sided dice are thrown.Find the probability of obtaining at least 8 failuresif a success is 5 or 6.

erastova [34]

Answer:

0.2992 = 29.92% probability of obtaining at least 8 failures.

Step-by-step explanation:

For each dice, there are only two possible outcomes. Either a failure is obtained, or a success is obtained. Trials are independent, which means that the binomial probability distribution is used to solve this question.

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.

A success is 5 or 6.

A dice has 6 sides, numbered 1 to 6. Since a success is 5 or 6, the other 4 numbers are failures, and the probability of failure is:

$p = \frac{4}{6} = 0.6667$

10 normal six sided dice are thrown.

This means that $n = 10$

Find the probability of obtaining at least 8 failures.

This is:

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

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

$P(X = 8) = C_{10,8}.(0.6667)^{8}.(0.3333)^{2} = 0.1951$

$P(X = 9) = C_{10,9}.(0.6667)^{9}.(0.3333)^{1} = 0.0867$

$P(X = 10) = C_{10,10}.(0.6667)^{10}.(0.3333)^{0} = 0.0174$

Then

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1951 + 0.0867 + 0.0174 = 0.2992$

0.2992 = 29.92% probability of obtaining at least 8 failures.

8 0

3 years ago