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Alinara [238K]
3 years ago
8

Use synthetic division and the Remainder Theorem to find f(4) if f(x) = – 2x3 + 9x2

Mathematics
1 answer:
Elis [28]3 years ago
6 0

Step-by-step explanation:

Let's see. f(4) makes the divisor x-4

4|-2 9 -6 7

| 8 -68 296

-2 17 -74 2072

The quotient is -2x²+17x-74 with a remainder of 2072

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vova2212 [387]
The second answer is correct
8 0
3 years ago
Keisha, Scott, and Ryan have a total of $101 in their wallets. Keisha has $6 more than Scott. Ryan has 3 times what Scott has. H
Yuri [45]

Answer:

Keisha =$25

Scott =$19

Ryan = $57

Step-by-step explanation:

Let the amount each have be represented by A, B and C

Keisha = A

Scott = B

Ryan = C

Keisha , Scott and Ryan have a total of $101.

That’s

A + B + C = $101

Keisha has $6 more than Scott

A = 6 + B

Ryan has 3 times Scott

C = 3B

Substitute 6+B for A and 3B for C in the first equation.

That’s

A + B + C = 101

6 + B + B + 3B = 101

6 + 5B = 101

Subtract 6 from both sides

6 - 6 + 5B = 101 - 6

5B = 95

Divide both sides by 5

B = 95/5

B = 19

Scott has $19

Recall Keisha has $6 more than Scott B.

That’s A = 6 + B

A = $6 + $19

A = $25

Keisha has $25

Also, Ryan C has 3 times what Scott B has .

That’s

C = 3 x B

C = 3 x 19

C = $57

Therefore, Keisha has $25, Scott has $19 and Ryan has $57

3 0
3 years ago
What is the measure of 0 in radians?
Allushta [10]

Answer:

Thus, the expression to find the measure of θ in radians is θ = π÷3

Step-by-step explanation:

Given that the radius of the circle is 3 units.

The arc length is π.

The central angle is θ.

We need to determine the expression to find the measure of θ in radians.

Expression to find the measure of θ in radians:

The expression can be determined using the formula,

where S is the arc length, r is the radius and θ is the central angle in radians.

Substituting S = π and r = 3, we get;

Dividing both sides of the equation by 3, we get;

3 0
3 years ago
7.2 Given a test that is normally distributed with a mean of 100 and a standard deviation of 10, find: (a) the probability that
kompoz [17]

Answer:

a)

<em>The probability that a single score drawn at random will be greater than 110  </em>

<em>P( X > 110) = 0.1587</em>

<em>b) </em>

<em>The probability that a sample of 25 scores will have a mean greater than 105</em>

<em>  P( x> 105) = 0.0062</em>

<em>c) </em>

<em>The probability that a sample of 64 scores will have a mean greater than 105</em>

<em> P( x⁻> 105)  = 0.002</em>

<em></em>

<em>d) </em>

<em> The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105</em>

<em>    P( 95 ≤ X≤ 105) = 0.9544</em>

<em></em>

Step-by-step explanation:

<u><em>a)</em></u>

Given mean of the Normal distribution 'μ'  = 100

Given standard deviation of the Normal distribution 'σ' = 10

a)

Let 'X' be the random variable of the Normal distribution

let 'X' = 110

Z = \frac{x-mean}{S.D} = \frac{110-100}{10} =1

<em>The probability that a single score drawn at random will be greater than 110</em>

<em>P( X > 110) = P( Z >1)</em>

                = 1 - P( Z < 1)

               =  1 - ( 0.5 +A(1))

               = 0.5 - A(1)

               = 0.5 -0.3413

              = 0.1587

b)

let 'X' = 105

Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{25} } } = 2.5

<em>The probability that a single score drawn at random will be greater than 110</em>

<em>  P( x> 105) = P( z > 2.5)</em>

<em>                    = 1 - P( Z< 2.5)</em>

<em>                    = 1 - ( 0.5 + A( 2.5))</em>

<em>                   = 0.5 - A ( 2.5)</em>

<em>                  = 0.5 - 0.4938</em>

<em>                  = 0.0062</em>

<em>The probability that a single score drawn at random will be greater than 105</em>

<em>  P( x> 105) = 0.0062</em>

<em>c) </em>

let 'X' = 105

Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{64} } } =  4

<em>The probability that a single score drawn at random will have a mean greater than 105</em>

<em>  P( x> 105) = P( z > 4)</em>

<em>                    = 1 - P( Z< 4)</em>

<em>                    = 1 - ( 0.5 + A( 4))</em>

<em>                   = 0.5 - A ( 4)</em>

<em>                  = 0.5 - 0.498</em>

<em>                  = 0.002</em>

<em> The probability that a sample of 64 scores will have a mean greater than 105</em>

<em> P( x⁻> 105)  = 0.002</em>

<em>d) </em>

<em>Let  x₁ = 95</em>

Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{95-100}{\frac{10}{\sqrt{16} } } =  -2

<em>Let  x₂ = 105</em>

Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{16} } } =  2

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = P( -2≤z≤2)

                         = P(z≤2) - P(z≤-2)

                        = 0.5 + A( 2) - ( 0.5 - A( -2))

                      = A( 2) + A(-2)       (∵A(-2) =A(2)

                     =  A( 2) + A(2)  

                    = 2 × A(2)

                  = 2×0.4772

                  = 0.9544

<em> The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105</em>

<em>    P( 95 ≤ X≤ 105) = 0.9544</em>

<em>    </em>

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