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LiRa [457]
3 years ago
6

The distance between New York City, New York and Los Angeles, California is approximately 4485 kilometers. What is the distance

between these two major cities in miles? Use the conversion equivalency of 1 mile = 1.609 kilometers and round to the nearest mile.
Mathematics
2 answers:
nadezda [96]3 years ago
5 0
This is your best hope! 

2786.85 Miles
arsen [322]3 years ago
3 0

Answer:

The answer is 2788 miles.

Step-by-step explanation:

The distance between New York City, New York and Los Angeles, California is approximately 4485 kilometers.

The given conversion rate is:

1 mile = 1.609 kilometers

or 1.609 km = 1 mile.

So, 1 km = \frac{1}{1.609} miles

And 4485 km = \frac{1}{1.609}*4485 miles

= 2787.45 miles.

In nearest miles we can round this to 2788 miles.

You might be interested in
A wheat farmer cuts down the stalks of wheat and gathers them in 200 piles. The 200 gathered piles will be put on a truck. The t
denpristay [2]

Answer:

Check the explanation

Step-by-step explanation:

We want to estimate the total weight of grain on the field based on the data on a simple random sample of 5 piles out of 200. The population and sample sizes are N=200 & n=5 respectively.

1) Let Y_1,Y_2,...,Y_{200} be the weight of grain in the 200 piles and y_1,y_2,...,y_{5} be the weights of grain in the pile from the simple random sample.

We know, the sample mean is an unbiased estimator of the population mean. Therefore,

\widehat{\mu}=\overline{y}=\frac{1}{5}\sum_{i=1}^{5}y_i=\frac{1}{5}(3.3+4.1+4.7+5.9+4.5)=4.5

where \mu is the mean weight of grain for all the 200 piles.

Hence, the total grain weight of the population is

\widehat{Y}=Y_1+Y_2+...+Y_{200}

=200\times \widehat{\mu}\: \: \: =200\times 4.5\: \: \: =900\, lbs

2) To calculate a bound on the error of estimates, we need to find the sample standard deviation.

The sample standard deviation is

 S=\sqrt{\frac{1}{5-1}\sum_{i=1}^{5}(y_i-\overline{y})^2}\: \: \: =0.9486

Then, the standard error of \widehat{Y} is

\sigma_{\widehat{Y}} =\sqrt{\frac{N^2S^2}{n}\bigg(\frac{N-n}{N}\bigg)}\: \: \:=83.785

Hence, a 95% bound on the error of estimates is

[\pm z_{0.025}\times \sigma_{\widehat{Y}}]\: \: \: =[\pm 1.96\times 83.875]\: \: \: =[\pm 164.395]

3) Let x_1,x_2,...,x_5 denotes the total weight of the sampled piles.

Mean total weight of the sampled piles is

\overline{x}=\frac{1}{5}\sum_{i=1}^{5}x_i=45

The sample ratio is

r=\frac{\overline{y}}{\overline{x}}=\frac{4.5}{45}=0.1 , this is also the estimate of the population ratio R=\frac{\overline{Y}}{\overline{X}} .

Therefore, the estimated total weight of grain in the population using ratio estimator is

\widehat{Y}_R\: \: =r\times 8800\: \: =0.1\times 8800\: \: =880\, lbs

4) The variance of the ratio estimator is

var(r)=\frac{N-n}{N}\frac{1}{n}\frac{1}{\mu_x^2}\frac{\sum_{i=1}^{5}(y_i-rx_i)^2}{n-1}   , where \mu_x=8800/200=44lbs

=\frac{200-5}{200}\, \frac{1}{5}\: \frac{1}{44^2}\, \frac{0.2}{5-1}=0.000005

Hence, the standard error of the estimate of the total population is

\sigma_R=\sqrt{X^2 \: var(r)}\: \: \: =\sqrt{8800^2\times 0.000005}\: \: \:=21.556

Hence, a 95% bound on the error of estimates is

[\pm z_{0.025}\times \sigma_{R}]\: \: \: =[\pm 1.96\times 21.556]\: \: \: =[\pm 42.25]

8 0
4 years ago
Read 2 more answers
A rectangle is 14 inches long and 6 inches wide what is its area
Burka [1]
The answer would be 14 by 6, which is 84 inches squared.
7 0
3 years ago
Read 2 more answers
Given that the radius of circle A is 4 cm, and the diameter of circle B is 6 cm, what can be concluded about the two circles. So
Alinara [238K]
All circles are similar.

If they were congruent, they would have the same radius's/diameter's.

Circle B = diameter of 6

Circle A = radius of 4

Diameter = 2 * Radius

Circle A = diameter of 4 * 2 = 8

So they have different diameter's, therefore they are not congruent. So they are similar only.
8 0
3 years ago
PLEASE HELP ME WITH THIS!!!!!​
nadezda [96]

For the given experiment of spinning a spinner and drawing a card, we will see that:

  • A) Yes, the events are independent.
  • B) P(3 and A) = 1/16
  • C) P(5 and C) = 0
  • D) P(4) = 1/4
  •     P(B or C) = 1/2
  • E) P(3 and not D) = 3/16

<h3>What are independent events?</h3>

We say that two events are independent if the outcome of one does not affect the outcome of the other.

In this case, the outcome of the spinner clearly does not affect the outcome of the card draw, so the events are independent.

<h3>How to get the probabilities?</h3>

B) First we want to get the probability of getting 3 and A.

The probability of getting a 3 when drawing a card is given by the quotient between the numbers of card with the number 3 (only one) and the total number of cards, so we have:

p = 1/4

Similar for the case of the spinner, the letter A appears twice, and there are a total of 8 letters, then the probability is:

q = 2/8 = 1/4

The joint probability is the product of the two individual probabilities, we have:

P(3 and A) = p*q = 1/4*1/4 = 1/16

C) The probability of getting a 5 and the letter C is 0, because there is no card with the number 5.

D) P(4) is the probability of drawing the card with the number 4, this is:

p = 1/4

P(B or C) is the probability of spinning the letter B or C. There are 2 B's and 2 C's, and a total of 8 letters, so the probability is:

q = (2 + 2)/8 = 1/2

E) P(1 and not D) is equal to: P(1 and A or B or C).

P(1) is 1/4.

P(A or B or C) is 6 over 8 (because there are 6 cards that are either an A, a B, or a C)

Then the joint probability is:

P(1 and not D) = (1/4)*(6/8) = 3/16

If you want to learn more about probability, you can read:

brainly.com/question/251701

8 0
2 years ago
A total of
netineya [11]
Let A be the number of Adult tickets and S, the Students'

A+ S = 273, but we know that S=2.A

Then A + 2A = 273 ; 3A =273; A =273/3 =91

So 91 tickets were sod to Adults & 182 Tickets to Students
6 0
3 years ago
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