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2 years ago

2

Mathematics

1 answer:

svet-max [94.6K]2 years ago

5 0

Answer:

B. 4200

Step-by-step explanation:

A suitable calculator can add the terms for you. (See attached)

The first term is 69, and the common difference is 75-69 = 6. The general term is ...

an = a1 +d(n -1)

an = 69 +6(n -1)

Then the 28th term is ...

a28 = 69 +6(27) = 231

The average term is (a28 +a1)/2 = (231 +69)/2 = 150.

The sum is the number of terms multiplied by the average term:

sum = 28×150= 4200

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PLZ HELP !! Which of the points (s, t) is not one of the vertices of the shaded region of the

stira [4]

Answer:

A(0,10)

Step-by-step explanation:

Given the 4 inequalities:

s ≥ 12 - 0.5t (1)

s ≥ 10 -t (2)

s ≤ 20-t (3)

s ≥ 0

t ≥ 0

Let analyse all 4 possible answer:

A(0, 10)

Let substitute this point into (1) we have: 10 ≥ 12 -0.5*0 Wrong

We do not choose A

B (0, 20)

Let substitute this point into (1) we have: 20 ≥ 12 -0.5*0 True

Let substitute this point into (2) we have: 20 ≥ 10 - 0 True

Let substitute this point into (3) we have: 20 ≤ 20 - 0 True

We choose B as the vertex

C. (4, 16)

Let substitute this point into (1) we have: 16 ≥ 12 -0.5*4 True

Let substitute this point into (2) we have: 16 ≥ 10 - 4 True

Let substitute this point into (3) we have: 16 ≤ 20 - 4 True

We choose C as the vertex

D. (16,4)

Let substitute this point into (1) we have: 4 ≥ 12 -0.5*16 True

Let substitute this point into (2) we have: 4 ≥ 10 - 16 True

Let substitute this point into (3) we have: 4 ≤ 20 - 16 True

We choose B as the vertex

Hence, the point A(0,10) is not one of the vertices of the shaded region of the set of inequalities.

3 0

3 years ago

g A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of pa

adoni [48]

Complete Question

A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of parts had a mean of 1.6 millimeters with a standard deviation of 0.03 millimeters. what standard deviation will be needed to achieve a process capability index f 2.0?

Answer:

The value required is $\sigma = 0.0133$

Step-by-step explanation:

From the question we are told that

The upper specification is $USL = 1.68 \ mm$

The lower specification is $LSL = 1.52 \ mm$

The sample mean is $\mu = 1.6 \ mm$

The standard deviation is $\sigma = 0.03 \ mm$

Generally the capability index in mathematically represented as

$Cpk = min[ \frac{USL - \mu }{ 3 * \sigma } , \frac{\mu - LSL }{ 3 * \sigma } ]$

Now what min means is that the value of CPk is the minimum between the value is the bracket

substituting value given in the question

$Cpk = min[ \frac{1.68 - 1.6 }{ 3 * 0.03 } , \frac{1.60 - 1.52 }{ 3 * 0.03} ]$

=> $Cpk = min[ 0.88 , 0.88 ]$

$Cpk = 0.88$

Now from the question we are asked to evaluated the value of standard deviation that will produce a capability index of 2

Now let assuming that

$\frac{\mu - LSL }{ 3 * \sigma } = 2$

$\frac{ 1.60 - 1.52 }{ 3 * \sigma } = 2$

=> $0.08 = 6 \sigma$

=> $\sigma = 0.0133$

$\frac{ 1.68 - 1.60 }{ 3 * 0.0133 }$

=> $2$

Hence

$Cpk = min[ 2, 2 ]$

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So $\sigma = 0.0133$ is the value of standard deviation required

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3 years ago

1.What are the zeros of the polynomial function?

Lorico [155]

Let's to the first example:

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a = 1
b = 9
c = 20

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Delta = 9^2 - 4.(1).(20)

Delta = 81 - 80

Delta = 1

x = [ -b +/- √(Delta) ]/2a

Replacing the data:

x = [ -9 +/- √1 ]/2

x' = (-9 -1)/2 <=> - 5

Or

x" = (-9+1)/2 <=> - 4
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Already the second example:

f(x) = x^2 -4x -60

Ussing the formula of basckara again

a = 1
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c = -60

Delta = b^2 -4ac

Delta = (-4)^2 -4.(1).(-60)

Delta = 16 + 240

Delta = 256

Then, following:

x = [ -b +/- √(Delta)]/2a

Replacing the information

x = [ -(-4) +/- √256 ]/2

x = [ 4 +/- 16]/2

x' = (4-16)/2 <=> -6

Or

x" = (4+16)/2 <=> 10
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Now we are going to the 3 example

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Isolating 14x , but changing the sinal positive to negative

x^2 - 14x + 24 = 0

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a = 1
b = -14
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Delta = b^2 -4ac

Delta = (-14)^2 -4.(1).(24)

Delta = 196 - 96

Delta = 100

Then we stayed with:

x = [ -b +/- √Delta ]/2a

x = [ -(-14) +/- √100 ]/2

We wiil have two possibilities

x' = ( 14 -10)/2 <=> 2

Or

x" = (14 +10)/2 <=> 12
________________

To the last example will be the same thing.

f(x) = x^2 - x -72

a = 1
b = -1
c = -72

Delta = b^2 -4ac

Delta = (-1)^2 -4(1).(-72)

Delta = 1 + 288

Delta = 289

Then we are going to stay:

x = [ -b +/- √Delta]/2a

x = [ -(-1) +/- √289]/2

x = ( 1 +/- 17)/2

We will have two roots

That's :

x = (1 - 17)/2 <=> -8

Or

x = (1+17)/2 <=> 9

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3 years ago

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers