In lancaster, the library is 9 miles due south of the courthouse and 4 miles due west of the

Jane think of a number . She divides it by six and adds two. She gets an answer of 9. What number was she thinking of?

torisob [31]

The unknown number is X; x/6+2=9; X/6=7; X=42; The answer is 42

3 0

3 years ago

Find the other endpoint of the line segment with the given endpoint and midpoint. Endpoint: (-2, -6) Midpoint: (-5, 1)

FromTheMoon [43]

Answer:

(-8,8)

Step-by-step explanation:

The midpoint $M(x_M,y_M)$ of the segment AB with endpoints $A(x_A,y_A)$ and $B(x_B,y_B)$ has coordinates

$x_M=\dfrac{x_A+x_B}{2}\\ \\y_M=\dfrac{y_A+y_B}{2}$

In your case, $A(-2,-6), \ M(-5,1),$ then

$-5=\dfrac{-2+x_B}{2}\Rightarrow -2+x_B=-10,\ x_B=-10+2=-8\\ \\1=\dfrac{-6+y_B}{2}\Rightarrow -6+y_B=2,\ y_B=2+6=8$

3 0

3 years ago

Help please and sorry it’s not clear

aliina [53]

Answer:

C

Step-by-step explanation:

8 0

2 years ago

A statement of an employee’s biweekly earnings is given below. What is the employee’s gross pay?

Olin [163]

The employee's gross pay
$726.80

3 0

3 years ago

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

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

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]

= [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0

3 years ago