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Korvikt [17]
3 years ago
5

The table represents the equation y = 2 – 4x. A 2-column table with 5 rows. The first column is labeled x with entries negative

2, negative 1, 0, 1, 2. The second column is labeled y with entries 10, blank, 2, negative 2, negative 6. Use the drop-down menus to complete the statements. The x-values are the . The y-values are the . The missing value in the table for x = –1 is y = .
Mathematics
2 answers:
kherson [118]3 years ago
7 0

The missing value for x=-1 is y=6

Explanation:

It is given that the equation for the table is y=2-4x

The table has 2 column with 5 rows.

Thus, we have,

x       y

-2      10

-1       ---

0        2

1        -2

2       -6

We need to determine the value of y when x=-1

The value of y can be determined by substituting x=-1 in the equation y=2-4x

Thus, we have,

y=2-4(-1)

Multiplying the term within the bracket, we have,

y=2+4

Adding the terms, we have,

y=6

Thus, the value of y when x=-1 is 6.

Hence, the missing value for x=-1 is y=6

STALIN [3.7K]3 years ago
7 0

Answer: the x-values are the <u>INPUTS</u>

the y-values are the <u>OUTPUTS</u>

the missing value in the table for x = - 1 is <u>y = 6</u>

Step-by-step explanation: I got it right on edge 2020.

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schepotkina [342]

Answer:

y = 1.6 x

y = 24

Step-by-step explanation:

The general equation of a direct variation is

y = k x

In this case

x = 10

y = 16

y = k*x

16 = 10*k

16/10 =  k

k = 1.6

y = 1.6x

================

Let x = 15

y = 1.6 * 15

y = 24

7 0
3 years ago
What is the probability that she chose the four students so that exactly ONE of them is a junior?
8090 [49]

Answer:

  B)  0.246

Step-by-step explanation:

The desired probability is the sum of the probabilities of each of the ways one junior can be chosen. If we let a, b, c, d represent the ratio of juniors in classes 1 to 4, then this probability is ...

  p(1 junior) = (a)(1-b)(1-c)(1-d) + (1-a)(b)(1-c)(1-d) + (1-a)(1-b)(c)(1-d) + (1-a)(1-b)(1-c)(1-d)

We can simplify this a little bit to ...

  p(1 junior) = (1-a)(1-b)(1-c)(1-d) × (a/(1-a) +b/(1-b) +c/(1-c) +d/(1-d))

___

So, for {a, b, c, d} = {9/15, 12/15, 6/15, 3/15} = {3/5, 4/5, 2/5, 1/5} the probability of interest is ...

  p(1 junior) = (2/5·1/5·3/5·4/5) × (3/2 + 4/1 + 2/3 + 1/4) = 24/625 × 77/12

  p(1 junior) = 154/625 ≈ 0.246

_____

<em>Comment on the sum we're computing</em>

Each of the probabilities in the first sum above is ...

  p(junior in given class) · p(not a junior in the other 3 classes)

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3 years ago
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Answer:

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Step-by-step explanation:

8 0
3 years ago
O TRIGONOMETRIC FUNCTIONS
grandymaker [24]

Answer:

r = 22.9 in.

Step-by-step explanation:

Central angle in radians, ∅ = π/4

Length of arc, S = 18 in.

radius (r) = ?

Length of arc, C = r∅

Plug in the values and find r

18 = r*π/4

18 = (πr)/4

Multiply both sides by 4

4*18 = πr

72 = πr

Divide both sides by π

72/π = r

r = 22.9183118

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5 0
3 years ago
Solve the following quadratic equations by extracting square roots.Answer the questions that follow.
Vikentia [17]

Answer:

1.  x=±4

2. t=±9

3. r=±10

4. x=±12

5. s=±5

Step-by-step explanation:

1. x^2 = 16

Taking square root on both sides

\sqrt{x^2}=\sqrt{16}\\\sqrt{x^2}=\sqrt{(4)^2}\\

x=±4

2. t^2=81

Taking square root on both sides

\sqrt{t^2}=\sqrt{81}\\\sqrt{t^2}=\sqrt{(9)^2}

t=±9

3. r^2-100=0

r^{2}-100=0\\r^2 =100\\Taking\ Square\ root\ on\ both\ sides\\\sqrt{r^2}=\sqrt{100}\\\sqrt{r^2}=\sqrt{(10)^2}

r=±10

4. x²-144=0

x²=144

Taking square root on both sides

\sqrt{x^2}=\sqrt{144}\\\sqrt{x^2}=\sqrt{(12)^2}

x=±12

5. 2s²=50

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s=±5 ..

4 0
3 years ago
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