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3 years ago

Find area triangle of 15 ft and 18 ft The area of the triangle is. Ft2

Mathematics

1 answer:

Harrizon [31]3 years ago

5 0

Answer:

a= 135

Step-by-step explanation:

18*15=135

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-3x + 3y = 4 -x+ y = 3

Natalka [10]

Answer:

No solution

Step-by-step explanation:

-3x + 3y = 4

-x + y = 3

Solve the equation for x:

Move the variable to the right side and change its sign

-x + y = 3

-x = 3 - y

Change the signs on both sides of the equation

-x = 3 - y

x = -3 + y

Substitute the given value of x into the equation -3x + 3y = 4

-3x + 3y = 4

x = -3 + y

-3(-3+y)+3y=4

Solve the equation for y

-3(-3+y)+3y=4

y = 0

There is no solution for y.

And since there's no solution for y, the system has no solution.

6 0

3 years ago

1.Ordenar de menor a mayor los siguientes números

Nataliya [291]

1. -16, -9, -3, 0, 5, 8, 25

2. 26, 13, 7, 6, 2, -1, -3, -50

Espero que esto ayude!

8 0

3 years ago

Read 2 more answers

Solve for x.

Zigmanuir [339]

X/2 >= -4
x >= -4 * 2
x >= -8

6 0

3 years ago

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2.a. If the distri

zalisa [80]

Answer:

$P(\= X \ge 51 ) =0.0062$

$P(\= X \ge 51 ) = 0$

Step-by-step explanation:

From the question we are told that

The mean value is $\mu = 50$

The standard deviation is $\sigma = 1.2$

Considering question a

The sample size is n = 9

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{9} }$

=> $\sigma_x = 0.4$

Generally the probability that the sample mean hardness for a random sample of 9 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_{x}} \ge \frac{51 - 50 }{0.4 } )$

$\frac{\= X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ \= X )$

$P(\= X \ge 51 ) = P( Z \ge 2.5 )$

=> $P(\= X \ge 51 ) =1- P( Z < 2.5 )$

From the z table the area under the normal curve to the left corresponding to 2.5 is

$P( Z < 2.5 ) = 0.99379$

=> $P(\= X \ge 51 ) =1-0.99379$

=> $P(\= X \ge 51 ) =0.0062$

Considering question b

The sample size is n = 40

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{40} }$

=> $\sigma_x = 0.1897$

Generally the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_x} \ge \frac{51 - 50 }{0.1897 } )$

=> $P(\= X \ge 51 ) = P(Z \ge 5.2715 )$

=> $P(\= X \ge 51 ) = 1- P(Z < 5.2715 )$

From the z table the area under the normal curve to the left corresponding to 5.2715 and

=> $P(Z < 5.2715 ) = 1$

$P(\= X \ge 51 ) = 1- 1$

=> $P(\= X \ge 51 ) = 0$

5 0

3 years ago

There are about 6 billion people on Earth. If they all lined up and held hands, how long of a line would they form in km? (Assum

Zina [86]

If they all lined up and held hands, the line will be 6 billion meter apart.

3 0

4 years ago