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KATRIN_1 [288]
3 years ago
7

Select the equations whose graphs are parallel to the graph of the equation y=23x+5.

Mathematics
1 answer:
iren [92.7K]3 years ago
6 0

Answer:

2x-3y=9    and    4x-6y=30

Step-by-step explanation:

hello :  

note y=(2/3)x+5  no y=23x+5

2/3 is the slope....... all lines parallel to this line has same slope.

1)   2x-3y=9   means : 3y = 2x-9   divide by 3 : y=(2/3)x-3..same slope.

2)  4x-6y=30  means : 6y= 4x-30 divide by 6 : y=(4/6)x -30/6

but 4/6 = 2/3 so : y=(2/3)x-5..same slope.

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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
2 years ago
An acre measures 4840 square yards. In a city, a lot on which a single-
belka [17]

The lot is <u>9.2% of the an acre.</u>

<u>Given:</u>

Area of an acre = 4840 sq. yards

Area of lot = 40 by 100 ft = 40 ]\times 100 = 4,000 $ ft^{2}

Convert the area of the lot to feet (9 ft^{2} = yd^{2}):

<em>Therefore:</em>

Area of lot = \frac{4,000}{9} = 444.4 $ yd^{2}

Percentage occupied by the lot = area of lot / area of acre x 100

= \frac{444.4}{4840} \times 100\\\= 9.2

The lot therefore is <u>9.2% of the an acre</u>.

Learn more here:

brainly.com/question/20730855

7 0
2 years ago
What is the area of a circle if the diameter is 6 centimeters?
Klio2033 [76]

Answer:

18pi or 56.52 if you substitute pi with 3.14

Step-by-step explanation:

6 0
2 years ago
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Mr. Falco baked blueberry muffins several times in May using the recipe shown. He used 3 3/4 cups of milk in all. How many times
const2013 [10]

Answer:

use fraction caculator

Step-by-step explanation:

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3 0
2 years ago
Read 2 more answers
Specifications for a piece of material used in the manufacture of a bed mattress require that the piece be between a Lower and U
wolverine [178]

Answer:

Hello a value is missing in your question below is the value

samples of n = 10

answer : 24.63 <  μ  < 25.370

Step-by-step explanation:

Given :

Sample size = n = 10

Z∝/2 = Z0.0256 = 1.95

next calculate for margin of error

E = Z∝/2 * ( 0.6 /  / \sqrt{10} )

   = 0.370

therefore at  94.87% confidence interval the estimated population mean is

x - 0.370 < μ <  x + 0.370

25 - 0.370 < μ < 25 + 0.370

24.63 <  μ  < 25.370

4 0
2 years ago
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