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kicyunya [14]
3 years ago
15

Can you solve this?? It's, 5x-9y= -2

Mathematics
1 answer:
Andrej [43]3 years ago
8 0
The slope is m= 5/9 sorry if you weren’t looking for the slope
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Solve h(x) = 5 - 9x Find h(8)
Alex_Xolod [135]

h(x) = 5 - 9x

h(8) = 5 - 9×8 ( putting value x = 8)

h(8) = 5 - 72

h(8) = -67

Answer is -67.

4 0
3 years ago
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Which digit is the hundred place in the number of 742.891
zavuch27 [327]

Answer:

7

Step-by-step explanation:

if you meant hundredths it would be 9, but 7 is in the hundreds place

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3 years ago
PLEASE HELP FAST WORTH 30 points
Orlov [11]
I tried to explain but it’s hard HAHHA

4 0
3 years ago
The lifetime of a certain brand of battery is known to have a standard deviation of 9 hours. Suppose that a random sample of 150
Kobotan [32]

Answer:

The 90% confidence interval for the true mean lifetime of all batteries of this brand is between 39.3 hours and 41.7 hours. The lower limit is 39.3 hours while the upper limit is 41.7 hours.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.05 = 0.95, so Z = 1.645.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.645\frac{9}{\sqrt{150}} = 1.2

The lower end of the interval is the sample mean subtracted by M. So it is 40.5 - 1.2 = 39.3 hours

The upper end of the interval is the sample mean added to M. So it is 40.5 + 1.2 = 41.7 hours

The 90% confidence interval for the true mean lifetime of all batteries of this brand is between 39.3 hours and 41.7 hours. The lower limit is 39.3 hours while the upper limit is 41.7 hours.

7 0
3 years ago
Write the equation of the parabola in vertex form. <br> vertex (2,1) point (3,-3)
Mariana [72]

first off let's notice that hmmm the vertex is above the point (3,-3), so if we "assume" is a vertical parabola, then it'd be opening downwards.... anyhow that said, let's plug in those values.

\bf ~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf vertex~~(\stackrel{h}{2}~~,~~\stackrel{k}{1})\qquad \implies \qquad y = a(x-2)^2+1 \\\\\\ \stackrel{\textit{we also know that }}{(3~~,~~-3)} \begin{cases} x = 3\\ y = -3 \end{cases}\qquad \implies -3=a(3-2)^2+1 \\\\\\ -3=a(1)^2+1\implies -3=a+1\implies -4=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y = -4(x-2)^2+1~\hfill

6 0
3 years ago
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