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Temka [501]
3 years ago
12

Which of the following is true?

Mathematics
1 answer:
mart [117]3 years ago
5 0

Answer:

C

Step-by-step explanation:

Consider all options:

A. \sqrt{2} is irrational number because it cannot be represented as a fraction with integer numerator and natural denominator. This option is false.

B. Number 0 is integer number, therefore, its rational number. This option is false.

C. Repeating decimal 1.\overline{3} can be represented as 1 \dfrac{1}{3} that is \dfrac{4}{3} and, therefore, this number is rational number. This number is not integer number, because this is a fraction when simplified to lowest terms. So, this option is true.

D. Number -\sqrt{16}=-4 and is integer, therefore, is rational number. So, this option is false.

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A school is organising a competition. There are 128 students from the Maths specialist class and 80 students from the English sp
Elden [556K]

Answer:

5 english students in each team

Step-by-step explanation:

So there are multiple ways to do this, you could do trial and error or you could factor out common numbers between the two.

Through trial and error, the lowest possible number of groups that could be divided between the number of students whilst still being able to maintain a whole # was 16 groups

  • 128 math students ÷16 groups = 8 math students/group
  • 80 english students ÷ 16 groups = 5 english students/group

The other way to solve this is to factor out common numbers between the two:

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2 years ago
A school knows that 10% of its students are left-handed. What is the
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Step-by-step explanation:

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2 years ago
A pair of dice was rolled many times
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Evet canım benim için çok teşekkür ederimmm
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2 years ago
The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand
Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

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 mean strength of 50 items = 74.28

            \sigma = population standard deviation = 2.15

            n = sample of items = 50

            \mu = population mean tensile strength after machine was adjusted

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

So, 95% confidence interval for the population 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

<u><em>95% confidence interval for</em></u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                 = [ 74.28-1.96 \times {\frac{2.15}{\sqrt{50} } } , 74.28+1.96 \times {\frac{2.15}{\sqrt{50} } } ]

                 = [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

<em>Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.</em>

8 0
3 years ago
What is the value of y in the equation 2(2y − 12) = 0? <br><br> 4<br><br> 6<br><br> 7<br><br> 8
Elena-2011 [213]
First you distribute making it 4y-24=0
second you isolate the variable making it 4y=24
finally you divide by four giving you the answer of y=6
4 0
3 years ago
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