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3 years ago

Ria is performing an experiment in chemistry class with a certain

Mathematics

1 answer:

icang [17]3 years ago

4 0

Answer:

Part A: Growing

Part B: 100

Part C: 8100

Step-by-step explanation:

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4x what = 364 please put the calculation

NNADVOKAT [17]

Answer:

Step-by-step explanation:

4 x 91 = 364

To work this out, we must do the inverse operation. The inverse (opposite) of x (times or multiply) is ÷ (divide).

So, we do 364 ÷ 4 = 91

Hope this helps!

- profparis

7 0

2 years ago

The sum of twice a number and 12 is no less than 24

noname [10]

The answer is 12 + 12 = 24

3 0

3 years ago

48 passengers in 16 cars = passengers per car

Eduardwww [97]

Answer:

Step-by-step explanation:

48/16=3

5 0

3 years ago

Read 2 more answers

The probability that an Oxnard University student is carrying a backpack is .70. If 10 students are observed at random, what is

Rus_ich [418]

Answer:

35.03% probability that fewer than 7 will be carrying backpacks

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they are carrying a backpack, or they are not. The probability of a student carrying a backpack is independent from other students. So we use the binomial probability distribution 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.

The probability that an Oxnard University student is carrying a backpack is .70.

This means that $p = 0.7$

If 10 students are observed at random, what is the probability that fewer than 7 will be carrying backpacks

This is $P(X < 7)$ when $n = 10$ . So

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

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

$P(X = 0) = C_{10,0}.(0.7)^{0}.(0.3)^{10} = 0.000006$

$P(X = 1) = C_{10,1}.(0.7)^{1}.(0.3)^{9} = 0.0001$

$P(X = 2) = C_{10,2}.(0.7)^{2}.(0.3)^{8} = 0.0014$

$P(X = 3) = C_{10,3}.(0.7)^{3}.(0.3)^{7} = 0.0090$

$P(X = 4) = C_{10,4}.(0.7)^{4}.(0.3)^{6} = 0.0368$

$P(X = 5) = C_{10,5}.(0.7)^{5}.(0.3)^{5} = 0.1029$

$P(X = 6) = C_{10,6}.(0.7)^{6}.(0.3)^{4} = 0.2001$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.000006 + 0.0001 + 0.0014 + 0.0090 + 0.0368 + 0.1029 + 0.2001 = 0.3503$

35.03% probability that fewer than 7 will be carrying backpacks

4 0

4 years ago

Can you answer this???

Andreas93 [3]

Answer:

no sorry

Step-by-step explanation:

4 0

3 years ago