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erik [133]
3 years ago
12

The area of the triangle is 24 feet.

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
4 0
Does anybody know the answer 
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A steel worker dropped his wrench off the top of a new high rise building. How far will the wrench fall in 3 seconds?
rosijanka [135]
Using  S = ut + 1/2gt²,   from rest, u = 0

S = 1/2gt²,               g ≈ 9.8 m/s²

S = <span>1/2 *9.8*3²

</span>S = 0.5<span> *9.8*3*3 
</span>
<span>S ≈ 44.1 m</span>
6 0
3 years ago
You bike 1 mile the first day of your​ training, 1.3 miles the second​ day, 1.9 miles the third​ day, and 3.1 miles the fourth d
JulsSmile [24]

Answer:

19.9 miles

Step-by-step explanation:

In this problem we have:

d_1=1 mi is the distance travelled during the 1st day

d_2=1.3 mi is the distance travelled during the 2nd day

d_3=1.9 mi is the distance travelled during the 3rd day

d_4=3.1 mi is the distance travelled during the 4th day

We notice that the difference between the distance travelled on the (n+1)-th day and the distance travelled on the n-th day doubles every day. In fact:

d_2-d_1=0.3\\d_3-d_2=2\cdot 0.3 = 0.6\\d_4-d_3=2\cdot 0.6 = 1.2

Which can be rewritten using the general formula:

d_{n+1}-d_n=2(d_n-d_{n-1})

This means that

d_{n+1}=d_n+2(d_n-d_{n-1})

By applying this formula recursively, we can find the 7th term, which is the distance travelled on the 7th day:

d_1=1\\d_2=1.3\\d_3=1.9\\d_4=3.1\\d_5=3.1+2\cdot 1.2=5.5\\d_6=5.5+2\cdot 2.4=10.3\\d_7=10.3+2\cdot 4.8=19.9 mi

So, the  distance travelled on the 7th day is 19.9 miles.

7 0
3 years ago
2. The time between engine failures for a 2-1/2-ton truck used by the military is
OLEGan [10]

Answer:

A truck "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" with a probability of 0.89435 or about 89.435%.

For a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

Step-by-step explanation:

We have here a <em>random variable</em> <em>normally distributed</em> (the time between engine failures). According to this, most values are around the mean of the distribution and less are far from it considering both extremes of the distribution.

The <em>normal distribution</em> is defined by two parameters: the population mean and the population standard deviation, and we have each of them:

\\ \mu = 6000 miles.

\\ \sigma = 800 miles.

To find the probabilities asked in the question, we need to follow the next concepts and steps:

  1. We will use the concept of the <em>standard normal distribution</em>, which has a mean = 0, and a standard deviation = 1. Why? With this distribution, we can easily find the probabilities of any normally distributed data, after obtaining the corresponding <em>z-score</em>.
  2. A z-score is a kind of <em>standardized value</em> which tells us the <em>distance of a raw score from the mean in standard deviation units</em>. The formula for it is: \\ z = \frac{x - \mu}{\sigma}. Where <em>x</em> is the value for the raw score (in this case x = 5000 miles).
  3. The values for probabilities for the standard normal distribution are tabulated in the <em>standard normal table</em> (available in Statistics books and on the Internet). We will use the <em>cumulative standard normal table</em> (see below).

With this information, we can solve the first part of the question.

The chance that a truck will be able to travel a total distance of over 5000 miles without an engine failure

We can "translate" the former mathematically as:

\\ P(x>5000) miles.

The z-score for x = 5000 miles is:

\\ z = \frac{5000 - 6000}{800}

\\ z = \frac{-1000}{800}

\\ z = -1.25

This value of z is negative, and it tells us that the raw score is 1.25 standard deviations <em>below</em> the population mean. Most standard normal tables are made using positive values for z. However, since the normal distribution is symmetrical, we can use the following formula to overcome this:

\\ P(z

So

\\ P(z

Consulting a standard normal table available on the Internet, we have

\\ P(z

Then

\\ P(z1.25)

\\ P(z1.25)

However, this value is for P(z<-1.25), and we need to find the probability P(z>-1.25) = P(x>5000) (Remember that we standardized x to z, but the probabilities are the same).

In this way, we have

\\ P(z>-1.25) = 1 - P(z

That is, the complement of P(z<-1.25) is P(z>-1.25) = P(x>5000). Thus:

\\ P(z>-1.25) = 1 - 0.10565

\\ P(z>-1.25) = 0.89435  

In words, a truck "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" with a probability of 0.89435 or about 89.435%.

We can see the former probability in the graph below.  

The chance that a fleet of a dozen trucks will have an average time-between-failures of 5000 miles or more

We are asked here for a sample of <em>12 trucks</em>, and this is a problem of <em>the sampling distribution of the means</em>.

In this case, we have samples from a <em>normally distributed data</em>, then, the sample means are also normally distributed. Mathematically:

\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})

In words, the samples means are normally distributed with the same mean of the population mean \\ \mu, but with a standard deviation \\ \frac{\sigma}{\sqrt{n}}.

We have also a standardized variable that follows a standard normal distribution (mean = 0, standard deviation = 1), and we use it to find the probability in question. That is

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ z \sim N(0, 1)

Then

The "average time-between-failures of 5000" is \\ \overline{x} = 5000. In other words, this is the mean of the sample of the 12 trucks.

Thus

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ z = \frac{5000 - 6000}{\frac{800}{\sqrt{12}}}

\\ z = \frac{-1000}{\frac{800}{\sqrt{12}}}

\\ z = \frac{-1000}{230.940148}

\\ z = -4.330126

This value is so low for z, that it tells us that P(z>-4.33) is almost 1, in other words it is almost certain that for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is almost 1.

\\ P(z

\\ P(z

\\ P(z

The complement of P(z<-4.33) is:

\\ P(z>-4.33) = 1 - P(z or practically 1.

In conclusion, for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

7 0
3 years ago
GIVING BRAINLEIST HURRY PLEASE ‼️‼️‼️‼️
Olenka [21]

Answer:

I would say NO

Step-by-step explanation:

because if you look at those triangles, the one on they have two matching angles and two one side. The other angle in the triangle on the left isn't marked or anything so we can assume and the one on the right has no marked side. I hope I made a little sense. I don't know how to explain but I guess in short you could say that you don't have enough information to determine if they're the same

AGAIN!! I CAN'T PROMISE IF THIS IS RIGHT!!

7 0
3 years ago
HELP AGAINNNNN !!!!!!!!!!!!!!!!!!!
luda_lava [24]

Answer: A

Step-by-step explanation: 5*5=25  it is negative so the answer is -25

6 0
3 years ago
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