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stellarik [79]
3 years ago
6

Today Ian wants to run less than 7/12 mile which of the following distances is less than 7/12?

Mathematics
2 answers:
Viefleur [7K]3 years ago
8 0
X=1 or x=2 because math is math
dybincka [34]3 years ago
6 0

A fraction with a denominator of 4 that is less than 7/12 miles is 1/4.Let the fraction be  x/4. Since this fraction must be less than 7/12 miles, we can write the inequality,  x/4 < 7/12.We now solve the inequality for x to obtain,  x<7/12(4).

This simplifies to x<7/3

This implies that x<2 1/3

This is distance so x is positive.

So we can choose x=1 or x=2.

Therefore the possible fractions include 1/4 or 2/4.

In the simplest form, the most appropriate fraction is 1/4





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a person on Mars only weighs 38% of their weight on Earth if Carla weighs 78 pounds on Earth how much would you weigh on Mars to
Bumek [7]

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

                  Hello!

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

❖ Carla would weigh 30 pounds.

38/100 x 78 = 29.64

29.64 rounded to the nearest pound is 30.

So, Carla would weigh 30 pounds.

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

8 0
3 years ago
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Need the answer and steps please
Ne4ueva [31]

Answer:

coool

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
An automobile manufacturer finds that 1 in every 2500 automobiles produced has a particular manufacturing defect. ​(a) Use a bin
Advocard [28]

Answer:

a) 0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.

b) 0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

Step-by-step explanation:

Binomial 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.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

To use the Poisson approximation for the binomial, we have that:

\mu = np

1 in every 2500 automobiles produced has a particular manufacturing defect.

This means that p = \frac{1}{2500} = 0.0004

a) Use a binomial distribution to find the probability of finding 4 cars with the defect in a random sample of 7000 cars.

This is P(X = 4) when n = 7000. So

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

P(X = 4) = C_{7000,4}.(0.0004)^{4}.(0.9996)^{6996} = 0.1558

0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.

(b) The Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of p.

Using the approximation:

\mu = np = 7000*0.0004 = 2.8. So

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 4) = \frac{e^{-2.8}*(2.8)^{4}}{(4)!} = 0.1557

0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

6 0
3 years ago
The graph of y = ax 2 + bx + c is shown below. Determine the solution set of 0 = ax 2 + bx + c.
Pavlova-9 [17]
On the graph, it's shown that the function has two real roots. This means that there will be two solutions. So the answer is 1. {0,4}
4 0
3 years ago
Read 2 more answers
5. Find a counterexample to show that the following con-
Bess [88]

Answer:  see below

<u>Step-by-step explanation:</u>

Any value x such that |x| < 1 will make the conjecture false

<em>In simpler words, let x be a fraction between 0 and 1.</em>

One example: Let x = \dfrac{1}{2}

Then x⁴ = \bigg(\dfrac{1}{2}\bigg)^4

             = \dfrac{1}{16}

\dfrac{1}{16} < \dfrac{1}{2}

so the conjecture is false.

7 0
3 years ago
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