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2 years ago

What is 4/7 times 6/7?

Mathematics

2 answers:

Zinaida [17]2 years ago

6 0

Answer:

24/49

Step-by-step explanation:

There are no common factors to cancel out so just multiply the fractions.

gtnhenbr [62]2 years ago

3 0

Answer:

10,-0000 FRRS LIKE IF ITS NOT THEN

Step-by-step explanation:

EEEEEE

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》》what is democratic country?~~

nika2105 [10]

Answer:

A democratic country has a system of government in which the people have the power to participate in decision-making. ... In some democracies citizens help make decisions directly by voting on laws and policy proposals (direct democracy).

8 0

2 years ago

Read 2 more answers

What is the answer to this math problem 4-3= 8-9=

ivann1987 [24]

Answer:

4-3=1

8-9=-1

Step-by-step explanation:

I hope this helps

3 0

2 years ago

Harvey decided to make chocolate pudding pies for a celebration. He had to make 4 pies. Each pie needed 2 quarts of milk. How ma

Blababa [14]

Answer:

He will need 16 pints.

Step-by-step explanation:

2 pints per quart

2 quarts = 4 pints

4 x number of pies (4)

3 0

2 years ago

Read 2 more answers

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

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:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

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

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

2 years ago

Plese answer the puzzle please

____ [38]

Look at the attached picture⤴

Hope this will help u...

8 0

3 years ago