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3 years ago

4 decimals whose sum is 2.35

Mathematics

1 answer:

Sedbober [7]3 years ago

6 0

Answer:

One way you can solve this problem would be to take 2.35/4

2.35/4= .5875

Then you can add .5875 + .5875 +.5875 +.5875

Which equals 2.24

.5875 + .5875 +.5875 +.5875

Hope this helps ;)

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Answer:

Answer is at the bottom!!

Step-by-step explanation:

y = - \frac{2}{5} x - 2

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

rearrange 2x + 5y = 10 into this form

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Read 2 more answers

Evaluate the expression 3√64

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Answer:

Step-by-step explanation:

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3 years ago

In one city, the probability that a person will pass his or her driving test on the first attempt is 0.68. Eleven people are sel

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Answer:

0.0308 = 3.08% probability that among these 11 people, the number passing the test is between 2 and 4 inclusive

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they pass the test on their first attempt, or they do not. Each person is independent of any other person. 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.

In one city, the probability that a person will pass his or her driving test on the first attempt is 0.68.

This means that $p = 0.68$

Eleven people are selected at random from among those taking their driving test for the first time.

This means that $n = 11$

What is the probability that among these 11 people, the number passing the test is between 2 and 4 inclusive

This is

$P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)$ .

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

$P(X = 2) = C_{11,2}.(0.68)^{2}.(0.32)^{9} = 0.0009$

$P(X = 3) = C_{11,3}.(0.68)^{3}.(0.32)^{8} = 0.0057$

$P(X = 4) = C_{11,4}.(0.68)^{4}.(0.32)^{7} = 0.0242$

$P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0057 + 0.0242 = 0.0308$ .

0.0308 = 3.08% probability that among these 11 people, the number passing the test is between 2 and 4 inclusive

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3 years ago

4 decimals whose sum is 2.35​

4 decimals whose sum is 2.35