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4 years ago

How to write an inequality stating the restrictions of x

Mathematics

1 answer:

inessss [21]4 years ago

6 0

For example
1>x>3
says x is less than 1 and greater than 3

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The complement of an angle is one-sixth the measure of the supplement of the angle. What is the measure of the complement angle?

cestrela7 [59]

Answer:

The measure of the complement angle is $18\°$

Step-by-step explanation:

Let

x-----> the angle

we know that

The complement of an angle is equal to $(90-x)\°$

The supplement of an angle is equal to $(180-x)\°$

we have

The complement of an angle is one-sixth the measure of the supplement of the angle

$(90-x)\°=(1/6)(180-x)\°$

solve for x

$(540-6x)\°=(180-x)\°$

$(6x-x)=(540-180)\°$

$(5x)=(360)\°$

$x=72\°$

Find the measure of the complement angle

$(90-x)\°$ ------> $(90-72)=18\°$

7 0

3 years ago

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Irina wants to build a fence around a rectangular vegetable garden so that it has a width of at least 10 feet. She can use a max

liraira [26]

The answer to your question is l = 20 ft; w = 10 ft

7 0

3 years ago

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Dina has a mass of 50 kilograms and is waiting at the top of a ski slope that’s 5.0 meters high. What is her potential energy

Bess [88]

Answer:

Step-by-step explanation:

Formula for potential energy is PE = m*g*h

where

m is the mass, in kg

g is the acceleration due to gravity, in meters per second squared, and

h is the height, in meters

The problem gives m = 50, g = 9.8, and h = 5. We plug that into the formula and get out answer. So:

$PE=mgh\\PE=(50)(9.8)(5)\\PE=2450$

Correct answer is C.

3 0

4 years ago

Read 2 more answers

What is the total r value for a wall made of 4 inches of wood, 2 inches of fiberglass, and 3 inches of foam? assume that the r v

Vladimir [108]

The total R value is computed from the R-values of the components in the stack as ...
(inches of wood)*(R-value of wood) +(inches of fiberglass)*(R-value of fiberglass) +(inches of foam)*(R-value of foam)
= (4 in)(2 /in) +(2 in)(4 /in) +(3 in)(5 /in)
= 8 +8 +15
= 31

The total R-value of the wall is 31.

8 0

3 years ago

Let X be the number of the cars being repaired at a repair shop. We have the following information:

worty [1.4K]

Answer:

(a) Sample Space

$S = \{0,1,2,3\}$

(b) PMF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

(c) CDF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

Step-by-step explanation:

Solving (a): The sample space

From the question, we understand that at most 3 cars will be repaired.

This implies that, the number of cars will be 0, 1, 2 or 3

So, the sample space is:

$S = \{0,1,2,3\}$

Solving (b): The PMF

From the question, we have:

$P(2) = P(1)$

$P(0) = P(3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$ can be represented as:

$P(1) + P(2) = 0.5[P(0) + P(3)]$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(1) + P(1) = 0.5[P(0) + P(0)]$

$2P(1) = 0.5[2P(0)]$

$2P(1) = P(0)$

$P(0)= 2P(1)$

Also note that:

$P(0) + P(1) + P(2) + P(3) = 1$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(0) + P(1) + P(1) + P(0) = 1$

$2P(1) + 2P(0) = 1$

Substitute $P(0)= 2P(1)$

$2P(1) + 2*2P(1) = 1$

$2P(1) + 4P(1) = 1$

$6P(1) = 1$

Solve for P(1)

$P(1) = \frac{1}{6}$

To calculate others, we have:

$P(2) = P(1)$

$P(2) = P(1) = \frac{1}{6}$

$P(0)= 2P(1)$

$P(0) =2 * \frac{1}{6}$ $P(0) =\frac{1}{3}$

$P(3) = P(0) =\frac{1}{3}$

Hence, the PMF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

See attachment (1) for histogram

Solving (c): The CDF ; F(x)

This is calculated as:

$F(x) = P(X \le x) =\sum\limit^{3}_{x_i \le x} P(x_i)$

For x = 0;

We have:

$P(X \le 0) = P(0)$

$P(X \le 0) = 1/3$

For x = 1

$P(X \le 1) = P(0) + P(1)$

$P(X \le 1) = 1/3 + 1/6$

$P(X \le 1) = 1/2$

For x = 2

$P(X \le 2) = P(0) + P(1) + P(2)$

$P(X \le 2) = 1/3 + 1/6 + 1/6$

$P(X \le 2) = 2/3$

For x = 3

$P(X \le 3) = P(0) + P(1) + P(2) + P(3)$

$P(X \le 3) = 1/3 + 1/6 + 1/6 + 1/3$

$P(X \le 3) = 1$

Hence, the CDF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

See attachment (2) for histogram

6 0

3 years ago