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Sever21 [200]
3 years ago
14

A photography studio charges $50 that includes a sitting fee and 6 prints. Luigi increased his order to 11 prints and paid $65.

How much was the sitting fee? PLEASE HELP ASAP. Thank you!!
Mathematics
1 answer:
Effectus [21]3 years ago
7 0

Answer:

Sitting fee - 32$

Step-by-step explanation:

This is a system of equations(let x represent the sitting fee)

x+6y=50

x+11y=65

You want to isolate the x variable - x+6y-6y=50-6y ; x = 50-6y

Input this into the 2nd equation: 50-6y+11y=65 ; 50+5y=65

Subtract 50 from both sides. 5y=15 (Divided 5y on both sides) ; y=3

Now that y = 3 input this into any equation I choose the 1st one.

x+6(3) = 50 ; x + 18 = 50 (Subtract 18 on both sides to get x)

x = 32

Prove: 32 + 6(3) = 50 ; 32+18 = 50 ; 50 = 50 True



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Find the midpoint of the line segment with end coordinates of:(−2,−4) and (2,−10)
zhuklara [117]

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Step-by-step explanation:

By using mid point formula ,

(X,Y) =( X1+X2/2) , Y1 +Y2/2

(-2+2/2) , (-4-10/2)

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6 0
4 years ago
Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):
Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, \mu_X represent the population mean for Brand B and let \mu_Y represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

        P.Q. = \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } ~ t_n__1+n_2-2

where, Xbar = Sample mean for Brand B data = 36.9

            Ybar = Sample mean for Brand A data = 32.9

              n_1  = Sample size for Brand B data = 13

              n_2 = Sample size for Brand A data = 9

              s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2}  }{n_1+n_2-2} } = \sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} } = 3.013

Here, s^{2}_X and s^{2} _Y are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < t_2_0 < 2.528) = 0.98

P(-2.528 < \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } < 2.528) = 0.98

P(-2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (Xbar -Ybar) -(\mu_X-\mu_Y) < 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

P( (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (\mu_X-\mu_Y) < (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} , (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ]

[ (36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} , (36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

                     

4 0
3 years ago
Ben earns $9 per hour and $6 for each delivery he makes. He wants to earn more than $155 in an 8-hour workday. What is the least
hichkok12 [17]

Answer:

He will need 14 deliveries to reach his goal.

Step-by-step explanation:

$9/hr x 8hrs = $72.00

$155 - $72 = $83

$83/$6 = 13.83, but he cant make less than a whole delevery, so he would have to make 14 delivers to reach his goal.

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