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3 years ago

Drew was earning $14

Mathematics

1 answer:

ohaa [14]3 years ago

7 0

Answer:

the answer is B

Step-by-step explanation:

x1.07 means an increase of 7%

so answer is 14 x 1.07, B

please mark as brainliest tyyyyy

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Jack spends 1 1/2 as long on his homework as Jill. Last week Jill spent 9 1/2 hours doing homework. How long did Jack spend doin

alina1380 [7]

The amount of time Jill spent to do her homework
= 9 1/2 hours. which equals to 570 minutes.

Which means, the amount of times Jack needed to do his is
= 570 minutes x 1.5
= 855 minutes which equals to 14 1/4 hours

hope this helps

7 0

3 years ago

In ΔFGH, what is the measure of < H?

kogti [31]

X + x - 1 + 43 = 180. Solve for x. 2x +42 = 180 and 2x = 138. Therefore, x = 69. If x = 69, then angle H is 68 and angle G is 69. 68 + 69 + 43 = 180.

3 0

3 years ago

This is Matrix for pre calc

gavmur [86]

The given system of equations in augmented matrix form is

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]$

If you need to solve this, first get the matrix in RREF:

Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]$

Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]$

Add 164(row 3) to -91(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]$

Multiply row 4 by 1/13080:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]$

Add -153(row 4) to row 3:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]$

Multiply row 3 by -1/91:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add 6(row 3) and -8(row 4) to row 2:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 2 by 1/5:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

$\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 1 by 1/3:

$\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

So the solution to this system is $(w,x,y,z)=(1,-2,4,-3)$ .

6 0

3 years ago

Which choice is equivalent to the product below?

Crank

Answer:

2√21

Step-by-step explanation:

1) Simplify.

√84

2) Simplify √84 to 2√21.

2√21

8 0

2 years ago

You work with the office of city planning which is currently evaluating a new contract for a major highway. The chosen contracto

Ipatiy [6.2K]

Answer:

0.5488 = 54.88% probability that there are no defects in the completed highway.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

The portion of highway that your city is building is 30 miles long.

The mean is one defect per 50 miles. Since the highway is 30 miles, we have that:

$\mu = \frac{30}{50} = 0.6$

What is the probability that there are no (0) defects in the completed highway?

This is P(X = 0). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.6}*0.6^{0}}{(0)!} = 0.5488$

0.5488 = 54.88% probability that there are no defects in the completed highway.

7 0

3 years ago