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NISA [10]
3 years ago
6

SOMEONE HELP ME PLEASE

Mathematics
1 answer:
Novay_Z [31]3 years ago
3 0
Y - 5 = -1/2(x-1)

hope this helps!
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The mass of 5 m' of copper is 44 800 kg. Work out<br> the density of copper.
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Here bro this should help

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Step-by-step explanation: jcnk envkp crew mnvc XLMZ

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3 years ago
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Liula [17]

Answer:

2.56

Step-by-step explanation:

I'm pretty sure, I'm sorry if its wrong

5 0
2 years ago
Jane multiplies her number by three and subtracts 2.
katrin [286]

Answer:

Let the number they chose be 'x'.

Result of Jane = 3x -2

Result of Elliot = 2x+3

Given that,

The product of their result = 110

or, (3x-2)(2x+3) = 110\\or, 6x^{2} +9x-4x-6 = 110\\or, 6x^{2} +5x - 6 = 110\\or, 6x^{2} +5x - 6-110 = 0\\\\or, 6x^{2} +5x - 116 = 0\\or, 6x^{2} +29x - 24x - 116 = 0\\or, x (6x +29) -4(6x+29) = 0\\or, (x-4)(6x+29) = 0\\i.e. x=4 or x=\frac{-29}{6} \\So,~ the ~number ~they~ chose~was ~either ~4 ~or ~\frac{-29}{6}

3 0
3 years ago
Assume the sample below is a perfectly random sample of students at a school. How much greater is the mean of the
igor_vitrenko [27]

Answer:

0.75

Step-by-step explanation:

The table is not well presented (See Attachment)

There are at least two approaches to this question

Method 1:  

Steps

1. Calculate the mean of reported heights

Mean of reported heights = (61+68+57.5+48.5+75+65+80+68+69+63)/10

Mean of reported heights = 655/10

Mean of reported heights = 65.5

2. Calculate the mean of measured heights

Mean of measured heights = (62 + 68 + 56.5 + 47 + 72 + 65 + 78 + 67 + 69.5 + 62.5)/10

Mean of measured heights = 647.5/10

Mean of measured heights = 64.75

3. Get their difference

Difference = Mean of reported heights - Mean of measured heights

Difference = 65.5 - 64.75

Difference = 0,75

Method 2: Calculate the mean of their difference

Mean of difference = Sum of difference / Number of observations

Mean of difference = (-1 + 0 + 1 + 1.5 + 3 + 0 + 2 + 1 – 0.5 + 0.5)/10

Mean of difference = 7.5/10

Mean of difference = 0.75

Note that in both cases, the result is 0,75.

Hence, the reported heights at the school is 0.75 greater than the actual measured height

3 0
3 years ago
Read 2 more answers
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