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Rufina [12.5K]
2 years ago
10

The distance of planet Mercury from the Sun is approximately 5.8 ⋅ 10^7 kilometers, and the distance of Earth from the Sun is 1.

5 ⋅ 10^8 kilometers. About how many more kilometers is the distance of Earth from the Sun than the distance of Mercury from the Sun? (1 point)
4.3 ⋅ 10^7 kilometers
9.2 ⋅ 10^7 kilometers
9.2 ⋅ 10^8 kilometers
5.7 ⋅ 10^9 kilometers
Mathematics
2 answers:
Sphinxa [80]2 years ago
7 0
Or a more easier way to say is its b. Goodluck
egoroff_w [7]2 years ago
6 0

Answer:

The answer is 9.2 * 10^7

Step-by-step explanation:

(1.5*10^8)-(5.8*10^7)=92,000,000

92,000,000= 9.2* 10^7

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What is the solution to (3 x 102)x(5 x 105) written in scientic notation​
fiasKO [112]

Answer:

1.5 + 10^8.

Step-by-step explanation:

(3 x 10^2) x (5 x 10^5)

= 3 x 5 * 10^2 x 10^5

= 15 x  10^(2 + 5)

= 15 x 10^7

= 1.5 + 10^8.

3 0
3 years ago
Proving the Congruent Supplements Theorem
AleksandrR [38]

Answer:

Congruent Supplements Theorem : If two angles are supplements of the same angles (or of congruent angles), then the two angles are congruent.

4 0
3 years ago
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What is r i 4.5+10r=-6.5 with work pls
Evgen [1.6K]

Answer:

-11/10 or -1.1

Step-by-step explanation:

To start, let's get rid of those pesky decimals by multiplying everything by 2. Then we get 9 + 20r = -13. Now, let's isolate the r by subtracting 9 from both sides to get 20r = -22. Finally, we need to divide everything by 20 to get r = -11/10 or -1.1.

8 0
3 years ago
In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were o
ad-work [718]

The question is incomplete. Here is the complete qeustion.

In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were obtained: 0.10 0.13 0.16 0.15 0.14 0.008 0.15

(a) Construct a 99% confidence interval for the mean nitrogen-oxide emissions of all cars.

(b) If the EPA requires that nitrogen-oxide emissions be less than 0.165 g/mi, based on the 99% confidence interval in (a), can we safely conclude that this requirement is being met?

Answer: (a) 0.089 ≤ μ ≤ 0.171

(b) No

Step-by-step explanation:

(a) To determine the confidence interval, first calculate the mean (X) and standard deviation (s) of the sample

X = \frac{0.1+0.13+0.16+0.15+0.14+0.08+0.15}{7}

X = 0.13

s = \sqrt{\frac{(0.1-0.13)^{2} + (0.13 - 0.13)^{2} + ... + (0.15 - 0.13)^{2}}{7-1} }

s = 0.029

The degrees of freedom is

N - 1 = 7 - 1 = 6

And since the confidence is of 99%:

α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005

The t-test statistics for t_{6,0.005} is 3.707

(Value found in the t-distribution table)

Now, calculate Error:

E = t_{6,0.005} . \frac{s}{\sqrt{N} }

E = 3.707. \frac{0.029}{\sqrt{7} }

E = 0.041

The interval will be:  

0.13 - 0.041 ≤ μ ≤ 0.13+0.041

0.089 ≤ μ ≤ 0.171

(b) No, because according to the interval, the nitrode-oxide emissions range from 0.089 to 0.171, which is greater than required by EPA.

7 0
3 years ago
A state end-of-grade exam in American History is a multiple-choice test that has 50 questions with 4 answer choices for each que
Assoli18 [71]

Answer:

Q1) The student has a 0.01% probability of passing the test.

Q2) She has a 99.91% probability of passing in the test.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he gets it correct, or he gets it wrong. So we solve this problem using the binomial probability distribution.

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}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And \pi is the probability of X happening.

For this problem, we have that:

Question 1.

There are 50 questions, so n = 50.

The student is going to guess each question, so he has a \pi = \frac{1}{4} = 0.25 probability of getting it right.

He needs to get at least 25 question right.

So we need to find P(X \geq 25).

Using a binomial probability calculator, with n = 50 and \pi = 0.25 we get that P(X \geq 25) = 0.0001.

This means that the student has a 0.01% probability of passing the test.

Question 2.

Now, we need to find P(X \geq 25) with \pi = 0.70. So P(X \geq 25) = 0.9991

She has a 99.91% probability of passing in the test.

7 0
3 years ago
Read 2 more answers
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