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2 years ago

X-Y= 3 work out the value of 2x -2y

Mathematics

2 answers:

maw [93]2 years ago

3 0

Answer:

Step-by-step explanation:

Let x=4

Let y=1

4-1=3

8-2=6

LekaFEV [45]2 years ago

3 0

The answer is six (6)

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-1(x+5)=3[x+(2x-1)]<br> solved and justified each step

madreJ [45]

Answer:

Step-by-step explanation:

$-1(x+5)=3[x+2x-1)]\\\\\text{for}\ -1(x+5):\ \text{distribtutive property}\\\text{for}\ [x+(2x-1)]:\ \text{associative property}\\\\(-1)(x)+(-1)(5)=3[(x+2x)-1]\\-x-5=3(3x-1)\\\\\text{for}\ 3(3x-1):\ \text{distributive property}\\\\-x-5=(3)(3x)+(3)(-1)\\-x-5=9x-3\\\\\text{for the equation}:\ \text{addition property of equality}\\\\-x-5=9x-3\qquad\text{add 5 to both sides}\\-x-5+5=9x-3+5\\-x=9x+2\\\\\text{for the equation:}\ \text{subtraction property of equality}\\\\-x=9x+2\qquad\text{subtract}\ 9x\ \text{from both sides}$

$-x-9x=9x-9x+2\\-10x=2\\\\\text{for the equation:}\ \text{division property of equality}\\\\-10x=2\qquad\text{divide both sides by -10}\\\dfrac{-10x}{-10}=\dfrac{2}{-10}\\x=-0.2$

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2 years ago

Weights were recorded for all nurses at a particular hospital, the mean weight for an individual nurse was 135 lbs. with a stand

Aloiza [94]

Given:
population mean, μ =135
population standard deviation, σ = 15
sample size, n = 19

Assume a large population, say > 100,
we can reasonably assume a normal distribution, and a relatively small sample.
The use of the generally simpler formula is justified.

Estimate of sample mean
$\bar{x}=\mu=135$

Estimate of sample standard deviation
$\s=\sqrt{\frac{\sigma^2}{n}}$
$=\sqrt{\frac{15^2}{19}}=3.44124$ to 5 decimal places.

Thus, using the normal probability table,
$P(125$
$=P(\frac{125-135}{3.44124}$
$=P(-2.90593$
$=P(Z$
$=P(Z$

Therefore
The probability that the mean weight is between 125 and 130 lbs
P(125<X<130)=0.0731166-0.0018308
=0.0712858

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2 years ago