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3 years ago

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On Average, Mercury is about 57,000,000 Km from the sun, whereas Neptune is about 4.5 x 10^9 km from the sun. What is the differ

ence between Mercury’s and Neptune’s distances from the sun?

1 answer:

Amiraneli [1.4K]3 years ago

8 0

Answer:

4.443 x10^9

Step-by-step explanation:

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Convert 65.949 to expanded form. A) (6 + 5) + (9 × 1 10 ) + (4 × 1 100 ) + (9 × 1 1000 ) B) (6 × 5) + (9 × 1 10 ) + (4 × 1 100 )

muminat

The answer is option D.

6 0

3 years ago

Janet hikes a trail at a local forest each day. The trail is 3.6 miles long, and she has hiked 5 days in the past week. How many

a_sh-v [17]

Answer:

<em>7.2 miles</em>

Step-by-step explanation:

Janet hikes a trail 3.6 miles long each day. Considering the forth and back path, she trails 2*3.6= 7.2 miles.

Since she has hiked 5 days in the past week, she has hiked 5*7.2 = 36 miles.

The question does not specify if the path back is to be considered. If not, then the total distance traveled is 3.6*5 = 18 miles.

4 0

3 years ago

2. Tatyanna made $179.35 babysitting. Her sister, Nia, made $154.80 pet-sitting

Kitty [74]

Answer:

m=400-($179.35+$154.80)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Help now plsssssss!!!?

Elina [12.6K]

I think is c but I’m not sure

4 0

3 years ago

Suppose 58% of the population has a retirement account. If a random sample of size 570 is selected, what is the probability that

omeli [17]

Answer:

The probability that the proportion of persons with a retirement account will be less than 57%=31.561%

Step-by-step explanation:

We are given that

n=570

p=58%=0.58

We have to find the probability that the proportion of persons with a retirement account will be less than 57%.

q=1-p=1-0.58=0.42

By takin normal approximation to binomial then sampling distribution of sample proportion follow normal distribution.

Therefore, $\hat{p}\sim N(\mu,\sigma^2)$

$\mu_{\hat{p}}=p=0.58$

$\sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

$\sigma_{\hat{p}}=\sqrt{\frac{0.58\times 0.42}{570}}$

$\sigma_{\hat{p}}=0.02067$

Now,

$P(\hat{p}$

$P(\hat{p}$

$P(\hat{p}$

$P(\hat{p}$ =31.561%

Hence, the probability that the proportion of persons with a retirement account will be less than 57%=31.561%

3 0

3 years ago