I need the answer please

Mathematics

2 answers:

Law Incorporation [45]2 years ago

5 0

So, 9.8 x 10^3 .
:)))

yan [13]2 years ago

3 0

Answer:

98 and 2 next to the ten

Step-by-step explanation:

Simple math 10 sqaured is 10 times 10 times then you multiply 98.

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Solve the system of equations using substitution y=x-1 and y=-2x+5

pav-90 [236]

Since the two equations equal y, set them equal to each other.

x-1=-2x+5

From there, solve for x.

First get x on one side, by using the addition property of equality.

x-1=-2x+5
3x-1=5

Isolate x by adding 1.
3x=6

Lastly get x all by itself by dividing each side by 3.
x=2

You can now substitute your x-value, 2, into one of the equations (or both, if you wish; either one will result in the same answer.)

y=x-1
y=2-1
y=1

OR

y=-2x+5
y=-2(2)+5
y=-4+5
y=1

Final answer:
x=2 and y=1

Any questions or anything you would like me to clarify, feel free to ask :)

7 0

2 years ago

Quality control studies for Speedy Jet Computer Printers show the lifetime of the printer follows a normal distribution with mea

Anika [276]

Answer:

About 3 years Speedy Jet printers be guaranteed.

Step-by-step explanation:

Given : Quality control studies for Speedy Jet Computer Printers show the lifetime of the printer follows a normal distribution with

mean $\mu = 4$ and standard deviation $\sigma = 0.78$

If the company wishes to replace no more than 10% of the printers.

To find : How long should Speedy Jet printers be guaranteed ?

Solution :

Using Z- table,

Z value corresponding to 10% or 0.10 = -1.28.

Applying z-score formula,

$z=\frac{x-\mu}{\sigma}$

The value of x is given by,

$x=(z\times \sigma)+\mu$

Substitute the values,

$x=(-1.28\times 0.78)+4$

$x=-0.9984+4$

$x=3.0016$

So, approx 3 years.

Therefore, About 3 years Speedy Jet printers be guaranteed.

8 0

2 years ago

What is the derivative of the function below?

11111nata11111 [884]

$g(x)=4x-\dfrac{1}{4x}=4x-(4x)^{-1}\\\\\\\\g'(x)=\left[4x-(4x)^{-1}\right]'=(4x)'-\left[(4x)^{-1}\right]'=4-[-1(4x)^{-2}]\cdot4\\\\=4+\dfrac{1}{(4x)^2}\cdot4=4+\dfrac{4}{4^2x^2}=\boxed{4+\dfrac{1}{4x^2}}\\\\Used:\\\\(f(x)-g(x))'=f'(x)-g'(x)\\\\.[f(g(x))]'=f'(g(x))\cdot g'(x)\\\\(x^n)'=nx^{n-1}$