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3 years ago

24=x^2−4x+3 completing the square

Mathematics

1 answer:

Mars2501 [29]3 years ago

8 0

Answer:

-3,7

Step-by-step explanation:

See the Image below:)

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If a fair coin is tossed 6 times, what is the probability, to the nearest thousandth, of getting exactly 6 tails?

xxTIMURxx [149]

<h2>Answer:</h2>

$\frac{1}{64}$

<h2>Step-by-step explanation:</h2>

$The\ probability\ of\ tails\ per\ flip\ is\ \frac{1}{2}$

$so\ the\ probability\ of\ getting\ six$

$tails\ is$

$\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{64}$

$The\ probability\ of\ getting\ tails\ every$

$time\ you\ multiply\ them\ is\ the$

$probobility\ of\ getting\ tails\ all\ six$

$times?$

<em>I hope this helps you</em>

3 0

2 years ago

Please help I’ll mark you as brainliest if correct

Dominik [7]

Answer:

25%

Step-by-step explanation:

5 0

3 years ago

CAN SOMEONE PLEASE HELP ME OUT

liberstina [14]

Answer:

Similarly: yes

Similarly Statement: LMNO ≈ ZWXY

Scale factor: 2/7

Answered by GAUTHMATH

6 0

3 years ago

Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in

Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the population variance is given as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

It is provided that:

<em>n</em> = 20

<em>s</em> = 3.9

Confidence level = 95%

⇒ <em>α</em> = 0.05

Compute the critical values of Chi-square:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907$

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

$\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45$

Thus, the 95% confidence interval for the population variance is (8.80, 32.45).

4 0

3 years ago

What percentage is .0019951

ICE Princess25 [194]

The most exact answer would be 0.19951% but the approximate answer would be 0.2%.

8 0

3 years ago

Read 2 more answers