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GrogVix [38]
2 years ago
4

Without any equipment, you can see stars that are 2{,}800{,}0002,800,0002, comma, 800, comma, 000 light-years away. By looking t

hrough a small telescope, you can see stars that are 3{,}112{,}000{,}0003,112,000,0003, comma, 112, comma, 000, comma, 000 light-years away. Approximately how many times as far can you see using a small telescope as without any equipment?
Mathematics
1 answer:
DiKsa [7]2 years ago
3 0

Answer:

1111 times

Step-by-step explanation:

Without any equipment, you can see stars that are 2,800,000 light-years away.

Using a small telescope, you can see  stars that are 3,112,000,000 light years away.

To determine approximately how many times as far you can see using a small telescope as without any equipment, we divide the number of light years.

That is:

\dfrac{\text{Distance Covered Using a telesope}}{\text{Distance Covered without any equipment}} =\dfrac{3,112,000,000}{2,800,000}\\=\dfrac{3112}{2.8} \\=1111.43\\\approx 1111

Therefore, you can see approximately 1111 times as far using a small telescope as without any equipment.

You might be interested in
It is common in many industrial areas to use a filling machine to fill boxes full of product. This occurs in the food industry a
Sphinxa [80]

Answer:

(a) P(B) = 0.008, (b) P(A∩B) = 0, (c) Yes, A and B are mutually exclusive events, (d) P(A∪B)=0.948, (e) 0.948, (f) 0.06

Step-by-step explanation:

We have three different posibilities

A: fill to specification

B: underfill

C: overfill

in probability the sum of the different events which are mutually exclusive should sum to 1, so, we should have

(a) P(B) = 1 - P(A)-P(C) = 1-0.940-0.052=0.008

(b) P(A∩B)=probability that the machine fill to specification and underfill = 0 because a machine can't fill to specification and underfill at the same time

(c) Yes, A and B are mutually exclusive events, because a machine can't fill to specification and underfill at the same time

(d) Because A and B are mutually exclusive events we should have that

P(A∪B)=P(A)+P(B)=0.940+0.008=0.948

(e) The probability that the machine does not overfill is the same that the probability that the machine fill to specification plus the probability that the machine underfill, i.e, the probability that the machine does not overfill is P(A)+P(B)=0.948, because does not overfill is equivalent either to fill to specification or to underfill.

(f) The probability that the machine either overfill or underfills is

P(C∪B)=P(C)+P(B)=0.052+0.008=0.06 because C and B are mutually exclusive events.

5 0
2 years ago
Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers.
In-s [12.5K]

Answer:

The probability that Albert's sample of 64 will have a mean between 13.5 and 16.5 minutes is 0.9973.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

X \sim N(\mu=15,\sigma=4)  

And let \bar X represent the sample mean, the distribution for the sample mean is given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

On this case  \bar X \sim N(15,\frac{4}{\sqrt{64}})

Solution to the problem

We are interested on this probability

P(13.5  

If we apply the Z score formula to our probability we got this:

P(13.5

=P(\frac{13.5-15}{\frac{4}{\sqrt{64}}}

And we can find this probability on this way:

P(-3

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(-3

The probability that Albert's sample of 64 will have a mean between 13.5 and 16.5 minutes is 0.9973.

7 0
2 years ago
Wat is 1000 divided by 350
pochemuha
2.8571 i think im not sure
7 0
2 years ago
You have a coupon for a local craft store that is for 40% off one item. You would like to purchase a glue gun for $8.99 and a ba
Yanka [14]

Answer:

$26.06

Step-by-step explanation:

You would like to purchase a glue gun for $8.99 and a basket for $25.99. The highest-priced item is a basket for $25.99. So, the coupon will take the discount off for this item.

The new price of the basket is

\$25.99\cdot (1-0.4)=\$25.99\cdot 0.6=\$15.594

The total cost of your buying  is

\$8.99+\$15.594=\$25.584

There will be a 6% sales tax added to the price of all items. So, the end cost is

\$25.584\cdot (1+0.06)=\$25.584\cdot 1.06=\$26.05904\approx \$26.06

4 0
2 years ago
The​ life, in​ years, of a certain type of electrical switch has an exponential distribution with an average life β=44. If 100 o
Bond [772]

Answer:

0.9999

Step-by-step explanation:

Let X be the random variable that measures the time that a switch will survive.

If X has an exponential distribution with an average life β=44, then the probability that a switch will survive less than n years is given by

\bf P(X

So, the probability that a switch fails in the first year is

\bf P(X

Now we have 100 of these switches installed in different systems, and let Y be the random variable that measures the the probability that exactly k switches will fail in the first year.

Y can be modeled with a binomial distribution where the probability of “success” (failure of a switch) equals 0.0225 and  

\bf P(Y=k)=\binom{100}{k}(0.02247)^k(1-0.02247)^{100-k}

where  

\bf \binom{100}{k} equals combinations of 100 taken k at a time.

The probability that at most 15 fail during the first year is

\bf \sum_{k=0}^{15}\binom{100}{k}(0.02247)^k(1-0.02247)^{100-k}=0.9999

3 0
2 years ago
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