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2 years ago

11

X 2 +6x+___ complete the square

1 answer:

vaieri [72.5K]2 years ago

6 0

9 is the blank.

(a+b)^2 = a^2 + 2ab + b^2

a = x
b = 3

(x+3)^2 = x^2 + 6x + 9

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A 22 gram sample of a substance that's used to preserve fruit and vegetables has a k-value of 0.1011. Find the substance's half-

Eddi Din [679]

<span>Formula for exponential decay is:

N = No * e^-kt

N = 1/2(45) = 22.5

No = 45

k = 0.1109
The solution to your problem is as follows:

Let's try to solve the half-life if it is 45 grams, after which, you will now have an idea on how to solve the problem.

22.5 = 45 * e^(-0.1109t)

1/2 = e^(-0.1109t)

ln(1/2) = -0.1109t

t = (ln(1/2))/(-0.1109) = 6.25 days

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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4 0

3 years ago

X=17-4y<br> y=x-2<br> answer by substitution

GaryK [48]

Answer:

(5,3)

Step-by-step explanation:

x=17-4y

x=17-4x+8

x=25-4x

5x=25

x=5

y=x-2

y=5-2

y=3

(5,3)

5 0

3 years ago

i need help on math

dolphi86 [110]

X=2
2x-3=-x+3

add three to both sides
add 1x to each side
should be left with 3x=6
divide both sides by 3 to get x alone
your answer then is x=2

7 0

3 years ago

PLEASE HELP ME ASAP! THERE'S A PICTURE OF THE QUESTION!!!!! 15 POINTS FOR THE CORRECT ANSWER!!! I really need this answer so ans

sweet-ann [11.9K]

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89°C

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6 0

2 years ago

Read 2 more answers

How many times must we toss a coin to ensure that a 0.95-confidence interval for the probability of heads on a single toss has l

musickatia [10]

Answer:

(1) 97

(2) 385

(3) 9604

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The margin of error in this interval is:

$MOE= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The formula to compute the sample size is:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}$

(1)

Given:

$\hat p = 0.50\\MOE=0.1\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.1^{2}}\\=96.04\\\approx97$

Thus, the minimum sample size required is 97.

(2)

Given:

$\hat p = 0.50\\MOE=0.05\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.05^{2}}\\=384.16\\\approx385$

Thus, the minimum sample size required is 385.

(3)

Given:

$\hat p = 0.50\\MOE=0.01\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.01^{2}}\\=9604$

Thus, the minimum sample size required is 9604.

8 0

2 years ago