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Flura [38]
2 years ago
13

Timothy explains to his brother that the model y=4mt has a constant of variation of 4 and is an example of _____ variation.

Mathematics
1 answer:
Lunna [17]2 years ago
7 0
Do you have any Answer choices??
You might be interested in
The label on the car's antifreeze container claims to protect the car between −40°C and 140°C. To covert Celsius temperature to
sergejj [24]

Answer:

-40 < t < 284

Step-by-step explanation:

The antifreeze protects the car between −40°C and 140°C.

Using t for temperature, the compound inequality of the Celsius temperature range is:

-40 < t < 140

The conversion formula is given to find degrees C in given degrees F. We can solve the formula for F, so we get degrees F in terms of degrees C.

C = (5/9)(F - 32)     <------ <em>conversion formula from deg F to deg C</em>

Solve for F:

(9/5)C = F - 32

(9/5)C + 32 = F

F = (9/5)C + 32    <------ <em>conversion formula from deg C to deg F</em>

Now we convert -40 deg C to deg F using the formula we just derived.

F = (9/5)C + 32

F = (9/5)(-40) + 32

F = -72 + 32

F = -40

-40 deg C = -40 deg F

(This is not a mistake or a typo. -40 deg C really is the same as -40 deg F.)

Now we convert 140 deg C to deg F using the formula we just derived.

F = (9/5)C + 32

F = (9/5)(140) + 32

F = 252 + 32

F = 284

140 deg C = 284 deg F

Now we rewrite the compound inequality with Fahrenheit temperatures.

-40 < t < 284

4 0
1 year ago
Quadrilateral CAMP below is a rhombus. the length PQ is (x+2) units, and the length of QA is (3x-14) units. Which statements bes
valentinak56 [21]

Answer:

<h3>Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles.</h3><h3>The value of x is 8.</h3>

Step-by-step explanation:

Given that Quadrilateral CAMP below is a rhombus. the length PQ is (x+2) units, and the length of QA is (3x-14) units

From the given Q is the middle point, which cut the diagonal PA into 2 equal halves.

By the definition of rhombus, diagonals meet at right angles.

Implies that PQ = QA

x+2 = 3x - 14

x-3x=-14-2

-2x=-16

2x = 16

dividing by 2 on both sides, we will get,

x =\frac{16}{2}

<h3>∴ x=8</h3><h3>Since Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles we can equate x+2 = 3x-14 to find the value of x.</h3>

The line segment \overrightarrow{PA}=\overrightarrow{PQ}+\overrightarrow{QA}

\overrightarrow{PA}=x+2+3x-14

=4x-12

=4(8)-12 ( since x=8)

=32-12

=20

<h3>∴ \overrightarrow{PA}=20 units</h3>
3 0
2 years ago
Orange M&amp;M’s: The M&amp;M’s web site says that 20% of milk chocolate M&amp;M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0
2 years ago
Assume that two marbles are drawn without replacement from a box with 1 blue, 3 white, 2 green, and 2 red marbles. find the prob
navik [9.2K]
Before I answer the question i am going to start with the blue Marble you would have a 1/8 chance of drawing the blue marble from the box and after that you would have a 3/7 of drawing a white marble because you did not replace the blue marble and I hope this help.
8 0
1 year ago
Which graph represents the function of f(x) = the quantity of 9 x squared plus 9 x minus 18, all over 3 x plus 6
lora16 [44]

Answer:

The answer is the option C

graph of 3x minus 3, with discontinuity at negative 2, negative 9

Step-by-step explanation:

we have

f(x)=\frac{9x^{2}+9x-18}{3x+6}

Simplify

f(x)=9\frac{(x^{2}+x-2)}{3(x+2)}

f(x)=3\frac{(x^{2}+x-2)}{(x+2)}

Step 1

Convert to a factored form the numerator

x^{2}+x-2=0    

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}+x=2

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}+x+0.25=2+0.25

x^{2}+x+0.25=2.25

Rewrite as perfect squares

(x+0.5)^{2}=2.25

Square root both sides

x+0.5=(+/-)1.5

x=-0.5(+/-)1.5

x=-0.5+1.5=1

x=-0.5-1.5=-2

so

x^{2}+x-2=(x-1)(x+2)  

Step 2

Simplify the function f(x)

f(x)=3\frac{(x^{2}+x-2)}{(x+2)}=3\frac{(x-1)(x+2)}{(x+2)}

The domain of the function f(x) is all real numbers except the number x=-2

Because the denominator can not be zero

f(x)=3\frac{(x-1)(x+2)}{(x+2)}=3(x-1)=3x-3  

f(x)=3x-3  ------> with a discontinuity at x=-2

f(-2)=3(-2)-3=-9

The discontinuity is at point (-2,-9)

the answer in the attached figure

8 0
2 years ago
Read 2 more answers
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