Calculate the x-intercepts of the quadratic y=9xsquared+3x

The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand

Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean strength of 50 items = 74.28

$\sigma$ = population standard deviation = 2.15

n = sample of items = 50

$\mu$ = population mean tensile strength after machine was adjusted

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $74.28-1.96 \times {\frac{2.15}{\sqrt{50} } }$ , $74.28+1.96 \times {\frac{2.15}{\sqrt{50} } }$ ]

= [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.

8 0

3 years ago

Joel is making sauces at his restaurant.He has 8 gallons of chicken broth. He puts it into 5 pots so each has the same amount. H

GarryVolchara [31]

Answer:1 r3

Step-by-step explanation:8 Divided by five

4 0

3 years ago

Find the volume of a square pyramid with base edges of 48 cm and a slant

SOVA2 [1]

The height of this pyramid will be:

√(26^2 - (48/2)^2) = 10cm

So the volume is: V = 1/3 * 48^2 * 10 = 7680 cm^3

4 0

3 years ago

Read 2 more answers

Graph the system of

9966 [12]

Answer:

(-0.2, 2.8)

General Formulas and Concepts:

Algebra I

Reading a Cartesian plane
Coordinates (x, y)
Solving systems of equations by graphing

Step-by-step explanation:

Where the 2 lines intersect is the solution to the systems of equations.

6 0

3 years ago

Please help me, I don't know how to do this...

Lunna [17]

Answer:

A. 5

Step-by-step explanation:

The triangle will be a right triangle when the side lengths satisfy the Pythagorean theorem: the sum of the squares of the legs is equal to the square of the hypotenuse.

(x +1)^2 +x^2 = (√61)^2

x^2 +2x +1 +x^2 = 61 . . . eliminate parentheses

2x^2 +2x = 60 . . . . . . .subtract 1, collect terms

x^2 +x -30 = 0 . . . . . divide by 2, subtract 30

(x +6)(x -5) = 0 . . . . factor the quadratic

x = -6 or +5

The solution is x = 5.

_____

Side lengths cannot be negative. Solution values are values of x that make the factors zero. x+6=0 when x=-6, for example.

8 0

2 years ago