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kow [346]
3 years ago
11

Ax+by=c solve for y plz help hurryyyyy

Mathematics
1 answer:
Leviafan [203]3 years ago
5 0
I believe that the answer for y is...

y= -ax+c/b
You might be interested in
In a sample of 70 stores of a certain​ company, 62 violated a scanner accuracy standard. It has been demonstrated that the condi
uysha [10]

Answer:

We need at least 243 stores.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error of the interval is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

For this problem, we have that:

n = 70, p = \frac{62}{70} = 0.886

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

Determine the number of stores that must be sampled in order to estimate the true proportion to within 0.04 with 95​% confidence using the​ large-sample method.

We need at least n stores.

n is found when M = 0.04. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.886*0.114}}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.886*0.114}

\sqrt{n} = \frac{1.96\sqrt{0.886*0.114}}{0.04}

(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.886*0.114}}{0.04})^{2}

n = 242.5

Rounding up

We need at least 243 stores.

5 0
3 years ago
What is the volume of a pyramid that is 15mm , 18mm , 21mm in cubic millimeters
Ugo [173]

Answer:

V=1890mm

Step-by-step explanation:

You must use its formula: which is

V=(Length x height x width) / 3

V=(15x18x21)/3

V=5670/3

V=1890mm

3 0
3 years ago
Read 2 more answers
The function h(t) = –16t2 + 96t + 6 represents an object projected into the air from a cannon. The maximum height reached by the
jolli1 [7]
If you've started pre-calculus, then you know that the derivative of  h(t)
is zero where h(t)  is maximum.

The derivative is            h'(t) = -32 t  +  96 .

At the maximum ...        h'(t) = 0

                                       32 t = 96 sec

                                           t  =  3 sec . 
___________________________________________

If you haven't had any calculus yet, then you don't know how to
take a derivative, and you don't know what it's good for anyway.

In that case, the question GIVES you the maximum height.
Just write it in place of  h(t), then solve the quadratic equation
and find out what  't'  must be at that height.

                                       150 ft = -16 t²  +  96  t  +  6 

Subtract 150ft from each side:    -16t²  +  96t  -  144  =  0 .

Before you attack that, you can divide each side by  -16,
making it a lot easier to handle:

                                                         t²  -  6t  +  9  =  0

I'm sure you can run with that equation now and solve it.    
The solution is the time after launch when the object reaches 150 ft.
It's 3 seconds.  
(Funny how the two widely different methods lead to the same answer.)

The answer is from AL2006

6 0
3 years ago
Graph the function. Then State the domain and range. Also increasing and decreasing​
masha68 [24]

Answer:

See below, I will let graphing part to yourself.

Step-by-step explanation:

First function: domain: [-9,\infty), range: (-\infty, -2], decreasing

Second function: domain: (-\infty, \infty), range: (-\infty, \infty), increasing

Third function: domain: [3, \infty), range: [-5,\infty), increasing

4 0
3 years ago
Read 2 more answers
~**Will mark brainliest **~<br> For the correct answers to all three questions
aleksandr82 [10.1K]

a)

\dfrac{42}{65}\cdot\dfrac{25}{36}\cdot\dfrac{26}{49}=\dfrac{2\cdot3\cdot7}{5\cdot13}\cdot\dfrac{5\cdot5}{2\cdot2\cdot3\cdot3}\cdot\dfrac{2\cdot13}{7\cdot7}=\dfrac{7}{13}\cdot\dfrac{5}{2\cdot3}\cdot\dfrac{2\cdot13}{7\cdot7}=\\\\\\=\dfrac{1}{1}\cdot\dfrac{5}{2\cdot3}\cdot\dfrac{2}{7}=\dfrac{5}{3}\cdot\dfrac{1}{7}=\dfrac{5}{21}

b)

\dfrac{21}{32}\cdot\dfrac{39}{120}\cdot\dfrac{40}{65}=\dfrac{21}{32}\cdot\dfrac{3\cdot13}{2\cdot2\cdot2\cdot3\cdot5}\cdot\dfrac{2\cdot2\cdot2\cdot5}{5\cdot13}=\\\\\\=\dfrac{3\cdot7}{32}\cdot\dfrac{13}{2\cdot2\cdot2\cdot5}\cdot\dfrac{2\cdot2\cdot2}{13}=\dfrac{21}{32}\cdot\dfrac{1}{5}\cdot\dfrac{1}{1}=\dfrac{21}{160}

c)

\dfrac{15}{90}\cdot\dfrac{36}{75}\cdot\dfrac{27}{42}=\dfrac{3\cdot5}{2\cdot3\cdot3\cdot5}\cdot\dfrac{2\cdot2\cdot3\cdot3}{3\cdot5\cdot5}\cdot\dfrac{3\cdot3\cdot3}{2\cdot3\cdot7}=\\\\\\=\dfrac{1}{2\cdot3}\cdot\dfrac{2\cdot2\cdot3}{5\cdot5}\cdot\dfrac{3\cdot3}{2\cdot7}=\dfrac{1}{1}\cdot\dfrac{2}{5\cdot5}\cdot\dfrac{3\cdot3}{2\cdot7}=\dfrac{1}{25}\cdot\dfrac{9}{7}=\dfrac{9}{175}

6 0
3 years ago
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