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4 years ago

Consider the expression below.

Mathematics

1 answer:

Elena L [17]4 years ago

5 0

Since 3 doesn’t have any term goes with it then 3 is a constant number. Hope this will help!

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Suppose that scores on a test are normally distributed with a mean of 80 and a standard deviation of 8. Answer the questions bel

Studentka2010 [4]

Answer:

(a) 84.2

(b) 10.6

Step-by-step explanation:

To solve this questions we can use the standardization formula, where we know that if $X\sim N(\mu,\sigma^2)$ then $Z=\frac{X-\mu}{\sigma} \sim N(0,1)$

So for

(a) we know that the z score for the 70th percentile is 0.524, so using the normalization equation we have

$\frac{X-\mu}{\sigma}=0.524$

$X=0.524*8+80=84.192$

(b) We can procede as above and get

$P(X$

3 0

4 years ago

Read 2 more answers

The square of the sum of two numbers is 144 and the sum of of their square is 180, find the number... Please I really need the a

mr Goodwill [35]

Answer:

The numbers are either ±13.35 and ±1.35 or ±1.35 and ±13.3

Step-by-step explanation:

Let the two numbers be 'x' and 'y'.

Given:

The squares of the sum of two numbers = 144

The sum of their squares = 180

The sum of the numbers = $x+y$

The square of the sum of numbers = $(x+y)^2$

Square of first number = $x^2$

Square of second number = $y^2$

The sum of their squares = $x^2+y^2$

Now, as per question:

$(x+y)^2=144---(1)\\\\x^2+y^2=180---(2)$

Expanding equation (1) using the formula $(a+b)^2=a^2+b^2+2ab$

This gives,

$(x+y)^2=144\\\\x^2+y^2+2xy=144\\\\\textrm{Plug in the value of }x^2+y^2\textrm{ from equation (2)}\\\\180+2xy=144\\\\2xy=144-180\\\\2xy=-36\\\\xy=-\frac{36}{2}\\\\xy=-18\\\\y=-\frac{18}{x}---(3)$

Plug in the value of 'y' from equation (3) in equation (1). This gives,

$x^2+(-\frac{18}{x})^2=180\\\\x^2+\frac{324}{x^2}=180\\\\x^4+324=180x^2\\\\x^4-180x^2+324=0\\\\\textrm{On solving, we get:}\\\\x^2=178.2\ or\ x^2=1.8\\\\\textrm{Square root both sides, we get:}\\\\x=\pm13.35\ or\ x=\pm1.35$