Help if you can pls. if you can’t see the pic don’t bother answering.

The distance from Earth to the moon is estimated to be 2.52 x 105 miles. The distance from Earth to the sun is approximately 400

chubhunter [2.5K]

Answer: $9\times10^7\ miles$

Step-by-step explanation:

Given: The estimated distance from Earth to the moon $=2.25\times10^5\ miles$

If the distance from Earth to the sun is approximately 400 times farther.

Then, the distance from Earth to the sun $=400\times2.25\times10^5$

$=900\times10^5\ miles\\=9\times10^7\ miles$

Therefore, the approximate distance from Earth to the sun= $9\times10^7\ miles$

7 0

4 years ago

For the data set 62, 73, 74, 75, 76, the mean is 72. what is the mean absolute deviation.

Charra [1.4K]

Answer is in a pho $^{}$ to. I can only uplo $^{}$ ad it to a file host $^{}$ ing service. link below!

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6 0

3 years ago

IN URGENT NEED OF HELP. PLEASE HELP.

Firlakuza [10]

Answer:

533 sq. ft.

Step-by-step explanation:

First find the area of the rectangle on the bottom:

22*13=286

Then in your head move the triangle on the right to the other side to create a triangle (if this doesn't make sense tell me in a comment)

It will turn into a rectangle that is 13 by 19 so find the area of that:

13*19=247

Then add the areas together:

286+247=533

6 0

4 years ago

For each situation described below, set up and solve an equation to answer the question.

Roman55 [17]

Q4) x+4 = 15
15-4 = x
x = 11

Q5)

18 + 12 = x

x - 12 = 18

30 - 12 = 18

x = 30

I hope this helped, please give this the Brainliest answer! :3

3 0

4 years ago

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it

lubasha [3.4K]

Answer:

a. 10.4 problems are expected to be resolved today, with a standard deviation of 1.44 problems.

b. 0.2457 = 24.57% probability 10 of the problems can be resolved today

c. 0.5137 = 51.37% probability 10 or 11 of the problems can be resolved today

d. 0.5017 = 50.17% probability more than 10 of the problems can be resolved today

Step-by-step explanation:

For each case, there are only two possible outcomes. Either they are solved, or they are not. Cases are independent of each other. 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 expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

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

It can resolve customer problems the same day they are reported in 80% of the cases.

This means that $p = 0.8$

Sample of 13 cases:

This means that $n = 13$

a. How many of the problems would you expect to be resolved today? What is the standard deviation?

Expected:

$E(X) = np = 13*0.8 = 10.4$

Standard deviation:

$\sqrt{V(X)} = \sqrt{13*0.8*0.2} = 1.44$

10.4 problems are expected to be resolved today, with a standard deviation of 1.44 problems.

b. What is the probability 10 of the problems can be resolved today?

This is $P(X = 10)$ . So

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

$P(X = 10) = C_{13,10}.(0.8)^{10}.(0.2)^{3} = 0.2457$

0.2457 = 24.57% probability 10 of the problems can be resolved today.

c. What is the probability 10 or 11 of the problems can be resolved today?

This is $p = P(X = 10) + P(X = 11)$ . So

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

$P(X = 10) = C_{13,10}.(0.8)^{10}.(0.2)^{3} = 0.2457$

$P(X = 11) = C_{13,11}.(0.8)^{11}.(0.2)^{2} = 0.2680$

$p = P(X = 10) + P(X = 11) = 0.2457 + 0.2680 = 0.5137$

0.5137 = 51.37% probability 10 or 11 of the problems can be resolved today.

d. What is the probability more than 10 of the problems can be resolved today?

This is

$P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13)$ . So

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

$P(X = 11) = C_{13,11}.(0.8)^{11}.(0.2)^{2} = 0.2680$

$P(X = 12) = C_{13,12}.(0.8)^{12}.(0.2)^{1} = 0.1787$

$P(X = 13) = C_{13,13}.(0.8)^{13}.(0.2)^{0} = 0.0550$

$P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) = 0.2680 + 0.1787 + 0.0550 = 0.5017$

0.5017 = 50.17% probability more than 10 of the problems can be resolved today

6 0

3 years ago