HELP PLS FIRST RIGHT ANSWER BRAINLIEST

A rectangle has a perimeter of (20x+12y). If one side of the rectangle is (3x-4y), write the expression for the other side

Nady [450]

Answer:

(7x + 10y)

Step-by-step explanation:

To find this add (3x - 4y) to itself to calculate to lengths of the shorter sides.

(3x - 4y) + (3x - 4y) = 6x - 8y

Subtract this from (20x + 12y)

(20x + 12y) - (6x - 8y) = 14x + 20y Divide this by two to get the length of one side

14x + 20y / 2 = 7x + 10y

If this answer is correct, please make me Brainliest!

3 0

4 years ago

The circulation of a newsletter decreased from 6000 to 3710. Find the percent of decrease in circulation to the nearest percent.

max2010maxim [7]

It would be approximately a decrease of 38%

6 0

3 years ago

In the figure below, BD and AC are diameters of circle P.

elena-14-01-66 [18.8K]

Answer:

$arc\ DC=30^o$

Step-by-step explanation:

The picture of the question in the attached figure

The point P is the center of the circle (the drawing is not a scale )

step 1

Find the measure of angle ∠CPD

we know that

$m\angle CPD+150^o=180^o$ ---> by supplementary angles (form a linear pair)

$m\angle CPD=180^o-150^o=30^o$

step 2

Find the measure of arc DC

we know that

Central angle is the angle that has its vertex in the center of the circumference and the sides are radii of it

so

$arc\ DC=m\angle CPD$ ----> by central angle

therefore

$arc\ DC=30^o$

5 0

3 years ago

Three six-sided fair dice are rolled. The six sides of each die are numbered 1; 2; : : : ; 6. Let A be the event that the first

valkas [14]

Answer:

1.) [A, B, C]

2.) [A', B', C']

3.) [A', B, C] or [A, B', C] or [A, B, C'] or [ A', B', C] or [A', B, C'] or [A, B', C'] or [A', B', C']

4.) A', B', C] or [A', B, C'] or [A, B', C'] or [A', B, C] or [A, B', C] or [A, B, C'] or [A, B, C]

Step-by-step explanation:

If A is the event that the first die shows an even number, then A' is the event that first die DOES NOT show an even number.

If B is the event that the Second die shows and even number, then B' is the event that second die DOES NOT show an even number.

If C is the event that the third die shows an even number, then C' is the event that third die DOES NOT show an even number.

Hence, our possible events are A, A', B, B', C, C'.

For question 1, in the event that all three dice show and even number, the expression of the event = [A, B, C]

For question two, in the event that no die shows an even number, the expression of the event = [A', B', C']

For question 3,In the event that at least one die shows an odd number, the expression of the event = [A', B, C] or [A, B', C] or [A, B, C'] or [ A', B', C] or [A', B, C'] or [A, B', C'] or [A', B', C']

For question 4,in the event that at most two dice show odd numbers, the expression of the event = [A', B', C] or [A', B, C'] or [A, B', C'] or [A', B, C] or [A, B', C] or [A, B, C'] or [A, B, C]

6 0

3 years ago

A Packaging Company produces boxes out of cardboard and has a specified weight of 16 oz. A random sample of 36 boxes yielded a s

user100 [1]

Answer:

The margin of error is of 0.73 oz.

The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.005 = 0.995$ , so Z = 2.575.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.575\frac{1.7}{\sqrt{36}} = 0.73$

The margin of error is of 0.73 oz.

The lower end of the interval is the sample mean subtracted by M. So it is 15.3 - 0.73 = 14.57 oz.

The upper end of the interval is the sample mean added to M. So it is 15.3 + 0.73 = 16.03 oz.

The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.

4 0

3 years ago