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3 years ago

Can I have some help with this? 5b+4+10=2

Mathematics

2 answers:

My name is Ann [436]3 years ago

7 0

$5b+4+10=2 \\\\5b=-12\\\\b=-\dfrac{12}{5}$

Scrat [10]3 years ago

5 0

Hello!

$5b+4+10=2$

Add the numbers with 10+4=14.

$5b+14=2$

Then subtract by 14 from both sides.

$5b+14-14=2-14$

Simplify

$5b=-12$

You can also divide by 5 from both sides.

$\frac{5b}{5}=\frac{-12}{5}$

Answer

$b=-\frac{12}{5}$

Hope this helps!

Thank you for posting your question at here on Brainly.

Have a great day!

-Charlie

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Consider the region bounded by 4y=x^2 and 2y=x.

gayaneshka [121]

Answer:

a) ⅓ units²

b) 4/15 pi units³

c) 2/3 pi units³

Step-by-step explanation:

4y = x²

2y = x

4y = (2y)²

4y = 4y²

4y² - 4y = 0

y(y-1) = 0

y = 0, 1

x = 0, 2

Area

Integrate: x²/4 - x/2

From 0 to 2

(x³/12 - x²/4)

(8/12 - 4/4) - 0

= -⅓

Area = ⅓

Volume:

Squares and then integrate

Integrate: [x²/4]² - [x/2]²

Integrate: x⁴/16 - x²/4

x⁵/80 - x³/12

Limits 0 to 2

(2⁵/80 - 2³/12) - 0

-4/15

Volume = 4/15 pi

About the x-axis

x² = 4y

x² = 4y²

Integrate the difference

Integrate: 4y² - 4y

4y³/3 - 2y²

Limits 0 to 1

(4/3 - 2) - 0

-2/3

Volume = ⅔ pi

7 0

3 years ago

Find the volume of a trough 5 meters long whose ends are equilateral triangles, each of whose

yawa3891 [41]

Answer:

$V=5\sqrt{3}\ m^3$

Step-by-step explanation:

we know that

The volume of a trough is equal to

$V=BL$

where

B is the area of equilateral triangle

L is the length of a trough

step 1

Find the area of equilateral triangle B

The area of a equilateral triangle applying the law of sines is equal to

$B=\frac{1}{2} b^{2} sin(60\°)$

where

$b=2\ m$

$sin(60\°)=\frac{\sqrt{3}}{2}$

substitute

$B=\frac{1}{2}(2)^{2} (\frac{\sqrt{3}}{2})$

$B=\sqrt{3}\ m^{2}$

step 2

Find the volume of a trough

$V=BL$

we have

$B=\sqrt{3}\ m^{2}$

$L=5\ m$

substitute

$V=(\sqrt{3})(5)$

$V=5\sqrt{3}\ m^3$

3 0

4 years ago

What's the answer please, I need help

mezya [45]

Answer:

Ano pong sa Sagutin diyan

Step-by-step explanation:

hindi ko op maintiddihan sorry po

3 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 "will be able to travel a total distance of over 5000 miles without an engine failure" 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 random variable normally distributed (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 normal distribution 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:

We will use the concept of the standard normal distribution, 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 z-score.
A z-score is a kind of standardized value which tells us the distance of a raw score from the mean in standard deviation units. The formula for it is: $\\ z = \frac{x - \mu}{\sigma}$ . Where x is the value for the raw score (in this case x = 5000 miles).
The values for probabilities for the standard normal distribution are tabulated in the standard normal table (available in Statistics books and on the Internet). We will use the cumulative standard normal table (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 below 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:

$\\ P(z$

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

$\\ P(z$

Then

$\\ P(z1.25)$

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 "will be able to travel a total distance of over 5000 miles without an engine failure" 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 12 trucks, and this is a problem of the sampling distribution of the means.

In this case, we have samples from a normally distributed data, 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.

$\\ P(z$