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3 years ago

24% of 289= Estimate using a rate per 100

Mathematics

1 answer:

ss7ja [257]3 years ago

4 0

use the equation is over of equals percent over 100 so it would x over 289 equals 24 over 100. and then you cross multiply and then find x.

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I’m stuck can some one help me

abruzzese [7]

Answer:

3) Midpoint is (-4,0.5)

Option A is correct.

4) Midpoint is (2.5,0)

Option B is correct.

5) The factors are (x+4)(x-7)

Option C is correct.

6) The factors are (x+4)(x+2)

Option A is correct.

Step-by-step explanation:

Question 3

Find midpoint of the following:

(2,-7), (-10,8)

The formula used to find midpoint is: $Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )$

We have $x_1=2, y_1=-7, x_2=-10,y_2=8$

Putting values and finding midpoint

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )\\Midpoint=(\frac{2-10}{2},\frac{-7+8}{2} )\\Midpoint=(\frac{-8}{2},\frac{1}{2} )\\Midpoint=(-4,0.5 )$

So, Midpoint is (-4,0.5)

Option A is correct.

Question 4

Find midpoint of the following:

(2,-10), (3,10)

The formula used to find midpoint is: $Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )$

We have $x_1=2, y_1=-10, x_2=3,y_2=10$

Putting values and finding midpoint

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )\\Midpoint=(\frac{2+3}{2},\frac{-10+10}{2} )\\Midpoint=(\frac{5}{2},\frac{0}{2} )\\Midpoint=(2.5,0 )$

So, Midpoint is (2.5,0)

Option B is correct.

Question 5

Factor each completely

$x^2-3x-28$

We will break the middle term and find factors

$x^2-3x-28\\=x^2-7x+4x-28\\Taking\:common\\=x(x-7)+4(x-7)\\=(x+4)(x-7)$

So, the factors are (x+4)(x-7)

Option C is correct.

Question 6

Factor each completely

$x^2+6x+8$

We will break the middle term and find factors

$x^2+6x+8\\=x^2+4x+2x+8\\=x(x+4)+2(x+4)\\=(x+4)(x+2)$

So, the factors are (x+4)(x+2)

Option A is correct.

7 0

2 years ago

GUYS PLEASE HELP HELPING ME GIVES U 50 POINTS PLEASE HELP ME Y = 1/1 x + 1

Semenov [28]

Is it a fraction? Or what

6 0

3 years ago

Read 2 more answers

A design engineer wants to construct a sample mean chart for controlling the service life of a halogen headlamp his company prod

tekilochka [14]

Answer:

C) 515 hours.

D) 500 hours

c) sample 3

Step-by-step explanation:

1. Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample Service Life (hours)

1 2 3

495 525 470

500 515 480

505 505 460

500 515 470

∑2000 2060 1880

x1`= ∑x1/n1= 2000/4= 500 hours

x2`= ∑x2/n2= 2060/4= 515 hours

x3`= ∑x3/n3= 1880/4= 470 hours

2. The mean of the sampling distribution of sample means for whenever service life is in control is 500 hours . It is the given mean in the question and the limits are determined by using μ ± σ , μ±2 σ or μ ± 3 σ.

In this question the limits are determined by using μ ± σ .

3. Upper control limit = UCL = 520 hours

Lower Control Limit= LCL = 480 Hours

Sample 1 mean = x1`= ∑x1/n1= 2000/4= 500 hours

Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample 3 mean = x3`= ∑x3/n3= 1880/4= 470 hours

This means that the sample mean must lie within the range 480-520 hours but sample 3 has a mean of 470 which is out of the given limit.

We see that the sample 3 mean is lower than the LCL. The other two means are within the given UCL and LCL.

This can be shown by the diagram.

8 0

2 years ago

What is the value of x in the solution to the following system of equations?

andrew-mc [135]

The answer is
x _ 2y = 2
3x + y = 6
3x _ 6y = 6 (1
3x + y = 6 (2
(1 - (2
-7y=0, y=0, so x - 2.0=2 implies x=2

7 0

3 years ago

Read 2 more answers

Part A and Part B Plzzzz

LenKa [72]

Answer:

Part A:

Length A=5

Length B=4

Part B:

The volume of the trough is 88

Step-by-step explanation:

7 0

3 years ago