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2 years ago

Solve for X y = 0.25x + -11 Please explain and provide evidence

Mathematics

2 answers:

DedPeter [7]2 years ago

8 0

Answer:

Unfortunately it can't be solved.

Step-by-step explanation:

x is the all real number. which means it will have many solutions.

Orlov [11]2 years ago

8 0

Answer:

x = 4(y + 11).

Step-by-step explanation:

y = 0.25x + -11 Add 11 to both sides:

y + 11 = 0.25x + - 11 + 11

y + 11 = 0.25x Now multiply both sides by 40 to give x on its own:

4(y + 11) = 0.25x * 4

x = 4(y + 11).

Let's check if the answer is right by substituting 2 for x in:

y = 0.25x - 11

= 0.5 - 11 = -10.5.

Now substitute -10.5 for y and 2 for x in x = 4(y + 11):

2 = 4(-10.5 + 11) = 4 * 0.5 = 2.

So our answer is correct.

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I know the answer but I don’t know how to set it up

vladimir2022 [97]

Answer: 4 x 16 = 64

16 + 44 = 60

I think..

Step-by-step explanation:

7 0

3 years ago

Find the two smallest possible solutions to part 1a

bixtya [17]

<h3>Answer: A. 5/12, 25/12</h3>

============================

Work Shown:

12*sin(2pi/5*x)+10 = 16

12*sin(2pi/5*x) = 16-10

12*sin(2pi/5*x) = 6

sin(2pi/5*x) = 6/12

sin(2pi/5*x) = 0.5

2pi/5*x = arcsin(0.5)

2pi/5*x = pi/6+2pi*n or 2pi/5*x = 5pi/6+2pi*n

2/5*x = 1/6+2*n or 2/5*x = 5/6+2*n

x = (5/2)*(1/6+2*n) or x = (5/2)*(5/6+2*n)

x = 5/12+5n or x = 25/12+5n

these equations form the set of all solutions. The n is any integer.

--------

The two smallest positive solutions occur when n = 0, so,

x = 5/12+5n or x = 25/12+5n

x = 5/12+5*0 or x = 25/12+5*0

x = 5/12 or x = 25/12

--------

Plugging either x value into the expression 12*sin(2pi/5*x)+10 should yield 16, which would confirm the two answers.

7 0

2 years ago

Read 2 more answers

Evaluate the expression. (-3)^-2

Wewaii [24]

$~~(-3)^{-2}\\\\\\=\dfrac 1{(-3)^2}~~~~~~~~~~~~~~~~~~~~~~~~~;\left[k^{-n} = \dfrac 1{k^n},~~~ k \neq 0\right]\\\\\\=\dfrac 1{(-1)^2 \cdot 3^2}\\\\\\=\dfrac 1{1 \times 9}\\\\\\=\dfrac 19$

3 0

1 year ago

Read 2 more answers

Wal-Mart conducted a study to check the accuracy of checkout scanners at its stores. At each of the 60 randomly selected Wal-Mar

Mamont248 [21]

Answer:

The 95% confidence interval is

$0.7811 < p < 0.9529$

Generally the interval above can interpreted as

There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval

Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable

$n = 125 \ stores$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The number of stores that had more than 2 items price incorrectly is k = 52

Generally the sample proportion is mathematically represented as

$\^ p = \frac{ k }{ n }$

=> $\^ p = \frac{ 52 }{ 60 }$

=> $\^ p = 0.867$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.96 * \sqrt{\frac{ 0.867 (1- 0.867)}{60} }$

=> $E = 0.0859$

Generally 95% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.867 - 0.0859 < p < 0.867 + 0.0859$

=> $0.7811 < p < 0.9529$

Generally the interval above can interpreted as

There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval

Considering question b

Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable

Considering question c

From the question we are told that

The margin of error is E = 0.05

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$