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3 years ago

Ariana’s annual(year) pay is $34,320. What is her average gross pay?

Mathematics

1 answer:

MArishka [77]3 years ago

4 0

Answer:

9. $2,860 per month

10. $660 per week

11. $94 per day (not 100% sure on this one, not sure if its getting techinical with working 5 days a week)

Step-by-step explanation:

9. $34,320 divide by number of months in a year (12)

10. $34,320 divide by number of weeks in a year (52)

11. $34,320 divide by number of days in a year (365)

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so each ticket costs 12 dollars multiply that bye 5 you get 60

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then all five girls get a soda which costs 4.75 so you would multiply 4.75 bye 5 which is 23.75

so in total you would get B.92.75

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3 years ago

Read 2 more answers

I was having trouble solving #25 of this packet. Can you help?

Murljashka [212]

Answer:

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3 years ago

What is the standard form equation of the line shown below? Graph of a line going through negative 2, 3 and 1, negative 3

V125BC [204]

Answer:

3. 2x + y = −1

Step-by-step explanation:

To find the equation of the line, we write it first in the slope-intercept form:

$y=mx+q$

where

m is the slope

q is the y-intercept

From the graph, we see that the line crosses the y-axis at y = -1, so the y-intercept is -1:

$q=-1$

Now we have to find the slope, by calculating the rate of change of the line through 2 points:

$m=\frac{y_2-y_1}{x_2-x_1}$

Taking the two points at (-2,3) and (1,-3), we find:

$m=\frac{-3-3}{1-(-2)}=\frac{-6}{3}=-2$

So the equation of the line is

$y=-2x-1$

Now we have to re-arrange it in the standard form, so in the form

$ax+bx=c$

where a, b and c are integer numbers.

To do that, we simply add +2x on both sides of the equation of the line in the slope-intercept form, and we get:

$y+2x=-1$

So, option 3).

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3 years ago

What is the slope of the line that passes through the points (1, -2) and (3, -5)?

Ksivusya [100]

Slope of the line is -3/2

Step-by-step explanation:

Step 1: Given points are (1, -2) and (3, -5)
Step 2: Equation for slope, m = (y2 - y1)/(x2 - x1)

⇒ m = (-5 - -2)/(3 - 1) = -3/2

4 0

3 years ago

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true

IgorLugansk [536]

Answer:

(a) 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

Step-by-step explanation:

We are given that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.75.

(a) Also, the average porosity for 20 specimens from the seam was 4.85.

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 average porosity = 4.85

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 20

$\mu$ = true average porosity

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 true 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 $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.85-1.96 \times {\frac{0.75}{\sqrt{20} } }$ , $4.85+1.96 \times {\frac{0.75}{\sqrt{20} } }$ ]

= [4.52 , 5.18]

Therefore, 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) Now, there is another seam based on 16 specimens with a sample average porosity of 4.56.

The pivotal quantity for 98% 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 average porosity = 4.56

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 16

$\mu$ = true average porosity

Here for constructing 98% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 98% confidence interval for the true mean, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98 {As the critical value of z at 1% level

of significance are -2.3263 & 2.3263}

P(-2.3263 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.3263$ ) = 0.98

P( $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.56-2.3263 \times {\frac{0.75}{\sqrt{16} } }$ , $4.56+2.3263 \times {\frac{0.75}{\sqrt{16} } }$ ]

= [4.12 , 4.99]

Therefore, 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

7 0

3 years ago