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3 years ago

Which option is equal to 7 1/5

Mathematics

2 answers:

masya89 [10]3 years ago

8 0

Answer:

<h2><em><u>D. is the answer</u></em></h2>

Step-by-step explanation:

Question:

7^1/5

The number given has an exponent of a fraction: fraction exponent = 1/5

<h2><em><u>So, when you have a fraction - you always have a square root - Important!!</u></em></h2>

Since the top is one, the number 7 stays the same. = 7^1 = 7

The bottom is a 5. This means it is to the fifth root.

Answer = D

Hope this helped,

Kavitha

lisov135 [29]3 years ago

3 0

Answer: If 36/7 is one of the options, choose that one.

If the question involves an exponent, you should use the "caret" which is ^ found above the 6 on a keyboard. [Shift + 6]. That helps avoid confusion.

Step-by-step explanation: 7 is equal to 35/5 because 7×5=35

Add 1/5 and you end up with 36/5. A Common rational number.

7^(1/5) = the 5th root of 7. A very small irrational number!

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Your flight has been delayed: At Denver International Airport, 85% of recent flights have arrived on time. A sample of flights i

Morgarella [4.7K]

Answer:

a) $P(X = n) = (0.85)^{n}$

b) $P(X = x) = C_{n,x}.(0.85)^{x}.(0.15)^{n-x}$

$P(X \geq x) = P(X = x) + P(X = x+1) + ... + P(X = n)$

For each value, we use

$P(X = x) = C_{n,x}.(0.85)^{x}.(0.15)^{n-x}$

d) If $x > E(X) + \sqrt{V(X)}$ , yes, otherwise no

Step-by-step explanation:

For each flight, there are only two possible outcomes. Either it arrived on time, or it did not. Flights are independent. This means that 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 expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

A measure is said to unusual if it is more than 2.5 standard deviations from the mean.

85% of recent flights have arrived on time.

This means that $p = 0.85$

(a) Find the probability that all of the flights were on time.

This is $P(X = n)$ , in which n is the size of the sample studied.

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

$P(X = n) = C_{n,n}.(0.85)^{n}.(0.15)^{n-n}$

$P(X = n) = (0.85)^{n}$

(b) Find the probability that exactly x of the flights were on time.

Exactly x of the flights: So

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

$P(X = x) = C_{n,x}.(0.85)^{x}.(0.15)^{n-x}$

(c) Find the probability that x or more of the flights were on time.

This is

$P(X \geq x) = P(X = x) + P(X = x+1) + ... + P(X = n)$

For each value, we use

$P(X = x) = C_{n,x}.(0.85)^{x}.(0.15)^{n-x}$

(d) Would it be unusual for x or more of the flights to be on time?

If $x > E(X) + \sqrt{V(X)}$ , yes, otherwise no

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11 I believe it is because you add 32 and 12

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